Question 15 Marks
Prove that:
$\sin3\text{x}+\sin2\text{x}-\sin\text{x}=4\sin\text{x}\cos\Big(\frac{\text{x}}{2}\Big)\cos\Big(\frac{3\text{x}}{2}\Big)$
Answer $\text{L.H.S.}=\sin3\text{x}+\sin2\text{x}-\sin\text{x}$ $=\sin3\text{x}+(\sin2\text{x}-\sin\text{x})$
$=\sin3\text{x}+\Big[2\cos\Big(\frac{2\text{x}+\text{x}}{2}\Big)\sin\Big(\frac{2\text{x}-\text{x}}{2}\Big)\Big]$$\Big[\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\sin3\text{x}+\Big[2\cos\Big(\frac{3\text{x}}{2}\Big)+\sin\Big(\frac{\text{x}}{2}\Big)\Big]$
$=\sin3\text{x}+2\cos\frac{3\text{x}}{2}\sin\frac{\text{x}}{2}$
$=2\sin\frac{3\text{x}}{2}.\cos\frac{3\text{x}}{2}+2\cos\frac{3\text{x}}{2}\sin\frac{\text{x}}{2}$ $[\sin2\text{A}=2\sin\text{A}.\cos\text{B}]$
$=2\sin\cos\Big(\frac{3\text{x}}{2}\Big)\Big[\sin\Big(\frac{3\text{x}}{2}\Big)+\sin\Big(\frac{\text{x}}{2}\Big)\Big]$
$=2\cos\Big(\frac{3\text{x}}{2}\Big)\begin{bmatrix}2\sin\begin{Bmatrix}\frac{\Big(\frac{3\text{x}}{2}\Big)+\Big(\frac{\text{x}}{2}\Big)}{2}\end{Bmatrix}\end{bmatrix}\cos\begin{Bmatrix}\frac{\Big(\frac{3\text{x}}{2}\Big)-\Big(\frac{\text{x}}{2}\Big)}{2}\end{Bmatrix}$ $\Big[\sin\text{A}+\sin\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=2\cos\Big(\frac{3\text{x}}{2}\Big).2\sin\text{x}\cos\Big(\frac{\text{x}}{2}\Big)$
$=4\sin\text{x}\cos\Big(\frac{\text{x}}{2}\Big)\cos\Big(\frac{3\text{x}}{2}\Big)=\text{R.H.S.}$
View full question & answer→Question 25 Marks
Find the general solution for each of the following equations:
$\sec^22\text{x}=1-\tan2\text{x}$
Answer $\sec^22\text{x}=1-\tan2\text{x}$ $\Rightarrow1+\tan^22\text{x}=1-\tan2\text{x}$
$\Rightarrow\tan^22\text{x}+\tan2\text{x}=0$
$\Rightarrow\tan2\text{x}(\tan2\text{x}+1)=0$
$\Rightarrow\tan2\text{x}=0$ or $\tan2\text{x}+1=0$
Now, $\tan2\text{x}=0$
$\Rightarrow\tan2\text{x}=\tan0$
$\Rightarrow2\text{x}=\text{n}\pi+0,$ where $\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{2},$ where $\text{n}\in\text{Z}$
$\tan2\text{x}+1=0$
$\Rightarrow\tan2\text{x}=-1=-\tan\frac{\pi}{4}=\tan\Big(\pi-\frac{\pi}{4}\Big)=\tan\frac{3\pi}{4}$
$\Rightarrow2\text{x}=\text{n}\pi+\frac{3\pi}{8},$ where $\text{n}\in\text{Z}$
$\Rightarrow2\text{x}=\frac{\text{n}\pi}{2}+\frac{3\pi}{8}$ where $\text{n}\in\text{Z}$
Therefore, the general solution is $\frac{\text{n}\pi}{2}$ or $\frac{\text{n}\pi}{2}+\frac{3\pi}{8},\text{n}\in\text{Z}$
View full question & answer→Question 35 Marks
Find the general solution for each of the following equations:
$\sin2\text{x}+\cos\text{x}=0$
Answer $\sin2\text{x}+\cos\text{x}=0$ $\Rightarrow\sin2\text{x}+\cos\text{x}+\cos\text{x}=0$
$\Rightarrow\cos\text{x}(2\sin\text{x}+1)=0$
$\Rightarrow\cos\text{x}=0$ or $2\sin\text{x}+1=0$
Now, $\cos\text{x}=0\Rightarrow\cos\text{x}=(2\text{n}+1)\frac{\pi}{2},$ where $\text{n}\in\text{Z}$
$2\sin\text{x}+1=0$
$\Rightarrow\sin\text{x}=\frac{-1}{2}=-\sin\frac{\pi}{6}=\sin\Big(\pi+\frac{\pi}{6}\Big)$
$=\sin\Big(\pi+\frac{\pi}{6}\Big)=\sin\frac{7\pi}{6}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{7\pi}{6},$ where $\text{n}\in\text{Z}$
Therefore, the general solution is $(2\text{n}+1)\frac{\pi}{2}$ or $\text{n}\pi+(-1)^2\frac{7\pi}{6},\text{n}\in\text{Z}$
View full question & answer→Question 45 Marks
Prove that:
$(\cos\text{x}+\cos\text{y})^2+(\sin\text{x}-\sin\text{y})^2=4\cos^2\frac{\text{x}+\text{y}}{2}$
Answer $\text{L.H.S.}=(\cos\text{x}+\cos\text{y})^2+(\sin\text{x}-\sin\text{y})^2$ $=\cos^2\text{x}+\cos^2\text{y}+2\cos\text{x}\cos\text{y}+\sin^2\text{x}+\sin^2\text{y}-2\sin\text{x}\sin\text{y}$
$=(\cos^2\text{x}+\sin^2\text{x})+(\cos^2\text{y}+\sin^2\text{y})+2(\cos\text{x}\cos\text{y}-\sin\text{x}\sin\text{y})$
$=1+1+2\cos(\text{x}+\text{y})\ [\cos(\text{A}+\text{B})=(\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B})]$
$=2+2\cos(\text{x}+\text{y})$
$=2[1+\cos(\text{x}+\text{y})]$
$=2\Big[1+2\cos^2\Big(\frac{\text{x}+\text{y}}{2}\Big)-1\Big]\ [\cos2\text{A}=2\cos^2\text{A}-1]$
$=4\cos^2\frac{\text{x}+\text{y}}{2}=\text{R.H.S.}$
View full question & answer→Question 55 Marks
Prove that:
$\frac{(\sin7\text{x}+\sin5\text{x})+(\sin9\text{x}+\sin3\text{x})}{(\cos7\text{x}+\cos5\text{x})(\cos9\text{x}+\cos3\text{x})}=\tan6\text{x}$
Answer $\text{L.H.S.}=\frac{(\sin7\text{x}+\sin5\text{x})+(\sin9\text{x}+\sin3\text{x})}{(\cos7\text{x}+\cos5\text{x})(\cos9\text{x}+\cos3\text{x})}$ $=\frac{\Big[2\sin\Big(\frac{7\text{x}+5\text{x}}{2}\Big)\cos\Big(\frac{7\text{x}-5\text{x}}{2}\Big)\Big]+\Big[2\sin\Big(\frac{9\text{x}+3\text{x}}{2}\Big)\cos\Big(\frac{9\text{x}-3\text{x}}{2}\Big)\Big]}{\Big[2\cos\Big(\frac{7\text{x}+5\text{x}}{2}\Big)\cos\Big(\frac{7\text{x}-5\text{x}}{2}\Big)\Big]+\Big[2\cos\Big(\frac{9\text{x}+3\text{x}}{2}\Big)\cos\Big(\frac{9\text{x}-3\text{x}}{2}\Big)\Big]}$
$=\frac{2\sin6\text{x}\cos\text{x}+2\sin6\text{x}\cos3\text{x}}{2\cos6\text{x}\cos\text{x}+2\cos6\text{x}\cos3\text{x}}$
$=\frac{2\sin6\text{x}(\cos\text{x}+\cos3\text{x})}{2\cos6\text{x}(\cos\text{x}+\cos3\text{x})}$
$=\tan6\text{x}=\text{R.H.S.}$
View full question & answer→Question 65 Marks
Find $\sin\frac{\text{x}}{2},\cos\frac{\text{x}}{2}$ and $\tan\frac{\text{x}}{2}$ in each of the following :
$\tan\text{x}=-\frac{4}{3},$ x in quadrant II
AnswerHere, x is in quadrant II.
i.e. $\frac{\pi}{2}<\text{x}<\pi$
$\Rightarrow\frac{\pi}{4}<\frac{\text{x}}{2}<\frac{\pi}{2}$
Therefore, $\sin\frac{\text{x}}{2},\cos\frac{\text{x}}{2}$ and $\tan\frac{\text{x}}{2}$ are all positive.
It is given that $\tan\text{x}=-\frac{4}{3}.$
$\sec^2\text{x}=1+\tan^2\text{x}=1+\Big(\frac{-4}{3}\Big)^2=1+\frac{16}{9}=\frac{25}{9}$
$\therefore\cos^2\text{x}=\frac{9}{25}$
$\Rightarrow\cos\text{x}=\pm\frac{-3}{5}$
As x is in quadrant II, cos x is negative.
$\therefore\cos\text{x}=\frac{-3}{5}$
Now, $\cos\text{x}=2\cos^2\frac{\text{x}}{2}-1$
$\Rightarrow\frac{-3}{5}=2\cos^2\frac{\text{x}}{2}-1$
$\Rightarrow2\cos^2\frac{\text{x}}{2}=1-\frac{3}{5}$
$\Rightarrow2\cos^2\frac{\text{x}}{2}=\frac{2}{5}$
$\Rightarrow\cos^2\frac{\text{x}}{2}=\frac{1}{5}$
$\Rightarrow\cos^2\frac{\text{x}}{2}=\frac{1}{\sqrt{5}}$ $\Big[\therefore\cos\frac{\text{x}}{2}\text{ is positive}\Big]$
$\therefore\cos\frac{\text{x}}{2}=\frac{\sqrt{5}}{5}$
$\sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}=1$
$\Rightarrow\sin^2\frac{\text{x}}{2}+\Big(\frac{1}{\sqrt{5}}\Big)^2=1$
$\Rightarrow\sin^2\frac{\text{x}}{2}=1-\frac{1}{5}=\frac{4}{5}$
$\Rightarrow\sin\frac{\text{x}}{2}=\frac{2}{\sqrt{5}}$ $\Big[\therefore\sin\frac{\text{x}}{2}\text{ is positive}\Big]$
$\therefore\sin\frac{\text{x}}{2}=\frac{2\sqrt{5}}{5}$
$\tan\frac{\text{x}}{2}=\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}=\frac{\Big(\frac{2\sqrt{5}}{5}\Big)}{\Big(\frac{1}{\sqrt{5}}\Big)}=2$
Thus, the respective values of $\sin\frac{\text{x}}{2},\cos\frac{\text{x}}{2}$ and $\tan\frac{\text{x}}{2}$ are $\frac{2\sqrt{5}}{5},\frac{\sqrt{5}}{5}$ and $2.$
View full question & answer→Question 75 Marks
Prove that:
$2\sin^2\frac{3\pi}{4}+2\cos^2\frac{\pi}{4}+2\sec^2\frac{\pi}{3}=10$
AnswerL.H.S = $2\sin^2\frac{3\pi}{4}+2\cos^2\frac{\pi}{4}+2\sec^2\frac{\pi}{3}$
$=2\Big\{\sin\Big(\pi-\frac{\pi}{4}\Big)\Big\}^2+2\Big(\frac{1}{\sqrt{2}}\Big)^2+2(2)^2$ $=2\Big\{\sin\frac{\pi}{4}\Big\}^2+2\times\frac{1}{2}+8$ $=2\Big(\frac{1}{\sqrt{2}}\Big)^2+1+8$ $=1+1+8$ $=10$ = R.H.S. View full question & answer→Question 85 Marks
In a circle of diameter 40cm, the length of a chord is 20cm. Find the length of minor arc of the chord.
AnswerDiameter of the circle = 40cm Radius (r) of the circle $=\frac{40}{2}\text{cm}=20\text{cm}$ Let AB be a chord (length = 20cm) of the circle. 
In $\triangle\text{OAB}$ OA = OB = Radius of circle = 20cm Also, AB = 20cm Thus, $\triangle\text{OAB}$ is an equilateral triangle. $\therefore\theta=60^\circ=\frac{\pi}{3}\text{radian}$ We know that in a circle of radius r unit, if an are of length l unit subtends an angle $\theta$ radian at the centre, then $\theta=\frac{\text{l}}{\text{r}}.$ $\frac{\pi}{3}=\frac{\widehat{\text{AB}}}{20}\Rightarrow\widehat{\text{AB}}=\frac{20\pi}{3}\text{cm}$ Thus, the length of the minor are of the chord is $\frac{20\pi}{3}\text{cm}.$ View full question & answer→Question 95 Marks
Prove the following:
$\sin(\text{n}+1)\text{x}\sin(\text{n}+2)\text{x}+2\cos(\text{n}+1)\text{x}\cos(\text{n}+2)\text{x}=\cos\text{x}$
Answer$ \text{L.H.S.}=$ $=\frac{1}{2}[2\sin(\text{n}+1)\text{x}\sin(\text{n}+2)\text{x}+2\cos(\text{n}+1)\text{x}\cos(\text{n}+2)\text{x}]$ $=\frac{1}{2}\begin{bmatrix}\cos\{(\text{n}+1)\text{x}-(\text{n}+2)\text{x}\}-\cos\{(\text{n}+1)\text{x}+(\text{n}+2)\text{x}\}\\+\cos\{(\text{n}+1)\text{x}+(\text{n}+2)\text{x}\}+\cos\{(\text{n}+1)\text{x}-(\text{n}+2)\text{x}\}\end{bmatrix}$
$\begin{bmatrix}\therefore-2\sin\text{A}\sin\text{B}=\cos(\text{A}+\text{B})-\cos(\text{A}-\text{B})\\2\cos\text{A}\cos\text{B}=\cos(\text{A}+\text{B})+\cos(\text{A}-\text{B})\end{bmatrix}$
$=\frac{1}{2}\times2\cos\{(\text{n}+1)\text{x}-(\text{n}+2)\text{x}\}$
$=\cos(-\text{x})=\cos\text{x}$
$= \text{R.H.S.}$ View full question & answer→Question 105 Marks
Prove that:
$(\cos\text{x}-\cos\text{y})^2+(\sin\text{x}-\sin\text{y})^2=4\sin^2\frac{\text{x}-\text{y}}{2}$
Answer $\text{L.H.S.}=(\cos\text{x}-\cos\text{y})^2+(\sin\text{x}-\sin\text{y})^2$ $=\Big(-2\sin\frac{\text{x}+\text{y}}{2}\sin\frac{\text{x}-\text{y}}{2}\Big)^2+\Big(2\cos\frac{\text{x}+\text{y}}{2}\sin\frac{\text{x}-\text{y}}{2}\Big)^2$
$\bigg[\because\cos\text{A}-\cos\text{B}=-2\sin\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2}\text{and} \sin\text{B}=2\cos\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2}\bigg]$
$=\Big(4\sin^2\frac{\text{x}+\text{y}}{2}\sin^2\frac{\text{x}-\text{y}}{2}\Big)+\Big(4\cos^2\frac{\text{x}+\text{y}}{2}\sin^2\frac{\text{x}-\text{y}}{2}\Big)$
$=4\sin^2\frac{\text{x}-\text{y}}{2}\Big(\sin^2\frac{\text{x}+\text{y}}{2}+\cos^2\frac{\text{x}+\text{y}}{2}\Big)$
$=4\sin^2\frac{\text{x}-\text{y}}{2}=\text{R.H.S.}$
View full question & answer→Question 115 Marks
Prove the following:
$\sin2\text{x}+2\sin4\text{x}+\sin6\text{x}=4\cos^2\text{x}\sin4\text{x}$
Answer$\text{L.H.S.}=\sin2\text{x}+2\sin4\text{x}+\sin6\text{x}$ $=[\sin2\text{x}+\sin6\text{x}]+2\sin4\text{x}$
$=\Big[2\sin\Big(\frac{2\text{x}+6\text{x}}{2}\Big)\Big(\frac{2\text{x}-6\text{x}}{2}\Big)\Big]+2\sin4\text{x}$
$=\Big[\therefore\sin\text{A}+\sin\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=2\sin4\text{x}\cos(-2\text{x})+2\sin4\text{x}$
$=2\sin4\text{x}\cos2\text{x}+2\sin4\text{x}$ $=2\sin4\text{x}(\cos2\text{x}+1)$ $=2\sin4\text{x}(2\cos^2\text{x}-1+1)$ $=2\sin4\text{x}(2\cos^2\text{x})$ $=4\cos^2\text{x}\sin4\text{x}$ $= \text{R.H.S.}$ View full question & answer→Question 125 Marks
Find the general solution for each of the following equations:
$\sin\text{x}+\sin3\text{x}+\sin5\text{x}=0$
Answer Given: $\sin\text{x}+\sin3\text{x}+\sin5\text{x}=0$
$\Rightarrow2\sin\Big(\frac{5\text{x}+\text{x}}{2}\Big)\cos\Big(\frac{5\text{x}-\text{x}}{2}\Big)+\sin3\text{x}=0$
$\Rightarrow2\sin3\text{x}\cos2\text{x}+\sin3\text{x}=0$
$\Rightarrow\sin3\text{x}(2\cos2\text{x}+1)=0$
$\Rightarrow\sin3\text{x}=0$ or $2\cos2\text{x}+1=0$
$\Rightarrow3\text{x}=\text{n}\pi$ or $\cos2\text{x}=\frac{-1}{2}=\cos\frac{2\pi}{3},\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{3}$ where $2\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{3}$ where $\text{x}=\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$
View full question & answer→Question 135 Marks
Prove the following:
$\cot4\text{x}(\sin5\text{x}+\sin3\text{x})=\cot4\text{x}(\sin5\text{x}-\sin3\text{x})$
Answer$\text{L.H.S.}=\cot4\text{x}(\sin5\text{x}+\sin3\text{x})$ $=\frac{\cos4\text{x}}{\sin4\text{x}}[2\sin4\text{x}\cos\text{x}]$
$=2\cos4\text{x}\cos\text{x}$
$\text{R.H.S.}=\cot4\text{x}(\sin5\text{x}-\sin3\text{x})$
$\text{L.H.S.} = \text{R.H.S.}$ View full question & answer→Question 145 Marks
Find the principal and general solutions of the following equations:
$\text{cosec x}=-2$
Answer$\text{cosec x}=-2$
It is known that
$\Rightarrow\text{cosec x}=\frac{\pi}{6}=2$
$\therefore\text{cosec}\Big(\pi+\frac{\pi}{6}\Big)=-\text{cosec}\frac{\pi}{6}=-2$
$\text{cosec}\Big(2\pi-\frac{\pi}{6}\Big)=-\text{cosec}\frac{\pi}{6}=-2$ i.e. $\text{cosec}\frac{7\pi}{6}=-2$ or $\text{cosec x}=\frac{11\pi}{6}=-2$
Therefore, the principal solutions are $\text{x}=\frac{7\pi}{6}$ and $\frac{11\pi}{6}.$
Now, $\text{cosec x}=\text{cosec}\frac{7\pi}{6}$
$\Rightarrow\sin\text{x}=\sin\frac{7\pi}{6}\ \Big[\text{cosec x}=\frac{1}{\sin\text{x}}\Big]$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{7\pi}{6}$ where $\text{n}\in\text{Z}$
Therefore, the general solution is $\text{x}=\text{n}\pi+(-1)^\text{n}\frac{7\pi}{6}$ where $\text{n}\in\text{Z}$
View full question & answer→Question 155 Marks
Find the general solution for each of the following equations:
$\cos4\text{x}=\cos2\text{x}$
Answer$\cos4\text{x}=\cos2\text{x}$
$\Rightarrow\cos4\text{x}-\cos2\text{x}=0$
$\Rightarrow-2\sin\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)=0$
$\Big[\therefore\cos\text{A}-\cos\text{B=}-2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$\Rightarrow\sin3\text{x}\sin\text{x}=0$
$\Rightarrow\sin3\text{x}=0$ or $\sin\text{x}=0$
$\therefore3\text{x}=\text{n}\pi$ or $\text{x}=\text{n}\pi,$ where $\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{3}$ or $\text{x}=\text{n}\pi,$ where $\text{n}\in\text{Z}$
View full question & answer→Question 165 Marks
Prove the following:
$\cos6\text{x}=32\cos^6\text{x}-48\cos^4\text{x}+18\cos^2\text{x}-1$
Answer$\text{L.H.S.}=\cos6\text{x}$
$=\cos3(2\text{x})$
$=4[(2\cos^2\text{x}-1)^3-3(2\cos^2\text{x}-1)[\cos2\text{x}=2\cos^2\text{x}-1]$
$=4[(2\cos^2\text{x})^3-(1)^3-3(2\cos^2\text{x})^2+3(2\cos^2\text{x}=2\cos^2\text{x})]-\cos^2\text{x}+3$
$=4[8\cos^6\text{x}-1-12\cos^4\text{x}+6\cos^2\text{x}]-6\cos^2\text{x}+3$
$=32\cos^6\text{x}-4-48\cos^4\text{x}+24\cos^2\text{x}-6\cos^2\text{x}+3$
$=32\cos^6\text{x}-48\cos^4\text{x}+18\cos^2\text{x}-1=\text{R.H.S.}$
View full question & answer→Question 175 Marks
Find $\sin\frac{\text{x}}{2},\cos\frac{\text{x}}{2}$ and $\tan\frac{\text{x}}{2}$ in each of the following :
$\cos\text{x}=-\frac{1}{3},$ x in quadrant III
AnswerHere, $\cos\text{x}=-\frac{1}{3},$ x is in quadrant III. Now, $\pi<\text{x}<\frac{3\pi}{2}$ $\Rightarrow\frac{\pi}{2}<\frac{\pi}{2}<\frac{3\pi}{4}$ $\therefore\frac{\text{x}}{2}$ lies in second quadrant. $\therefore\sin\frac{\text{x}}{2}$ are positive and $\cos\frac{\text{x}}{2},\tan\frac{\text{x}}{2}$ are negative. Now, $\cos\frac{\text{x}}{2}=\sqrt{\frac{1+\cos\text{x}}{2}}=-\sqrt{\frac{1-\frac{1}{3}}{2}}=-\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{-\sqrt{3}}{3}$
$\sin\frac{\text{x}}{2}=\sqrt{\frac{1-\cos\text{x}}{2}}=\sqrt{\frac{1+\frac{1}{3}}{2}}=\sqrt{\frac{2}{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$ $\tan\frac{\text{x}}{2}=\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}=\frac{\frac{\sqrt{2}}{\sqrt{3}}}{-\frac{1}{\sqrt{3}}}=-\sqrt{2}$ View full question & answer→Question 185 Marks
Prove the following:
$\cos\Big(\frac{3\pi}{4}+\text{x}\Big)-\cos\Big(\frac{3\pi}{4}-\text{x}\Big)=-\sqrt{2}\sin\text{x}$
AnswerIt is known that $\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big).\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $\therefore\text{L.H.S.}=\cos\Big(\frac{3\pi}{4}+\text{x}\Big)-\cos\Big(\frac{3\pi}{4}-\text{x}\Big)$ $=-2\sin\Bigg\{\frac{\Big(\frac{3\pi}{4}+\text{x}\Big)+\Big(\frac{3\pi}{4}-\text{x}\Big)}{2}\Bigg\}.\sin\Bigg\{\frac{\Big(\frac{3\pi}{4}+\text{x}\Big)+\Big(\frac{3\pi}{4}-\text{x}\Big)}{2}\Bigg\}$
$=-2\sin\Big(\frac{3\pi}{4}\Big)\sin\text{x}$
$=-2\sin\Big(\pi-\frac{\pi}{4}\Big)\sin\text{x}$
$=-2\sin\frac{\pi}{4}\sin\text{x}$
$=-2\times\frac{1}{\sqrt{2}}\sin\text{x}$ $=-\sqrt{2}\sin\text{x}$ $= \text{R.H.S.}$ View full question & answer→Question 195 Marks
Find the general solution for each of the following equations:
$\cos3\text{x}+\cos\text{x}-\cos2\text{x}=0$
AnswerGiven: $\cos3\text{x}+\cos\text{x}-\cos2\text{x}=0$
$\Rightarrow2\cos\Big(\frac{3\text{x}+\text{x}}{2}\Big)\cos\Big(\frac{3\text{x}-\text{x}}{2}\Big)-\cos2\text{x}=0$
$\Rightarrow2\cos2\text{x}\cos\text{x}-\cos2\text{x}=0$
$\Rightarrow\cos2\text{x}(2\cos\text{x}-1)=0$
$\Rightarrow\cos2\text{x}=0$ or $2\cos\text{x}-1=0$
$\Rightarrow2\text{x}=(2\text{n}+1)\frac{\pi}{2}$ or $\cos\text{x}=\frac{1}{2}=\cos\frac{\pi}{3},\text{n}\in\text{Z}$
$\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{4}$ or $\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$
View full question & answer→Question 205 Marks
Prove that:
$\sin\text{x}+\sin3\text{x}+\sin5\text{x}+\sin7\text{x}=4\cos2\text{x}\sin4\text{x}\cos\text{x}$
AnswerIt is known that $\sin\text{A}+\sin\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big).\cos\Big(\frac{\text{A}-\text{B}}{2}\Big).$ $\text{L.H.S.}=\sin\text{x}+\sin3\text{x}+\sin5\text{x}+\sin7\text{x}$
$=(\sin\text{x}+\sin5\text{x})+(\sin3\text{x}+\sin7\text{x})$ $=2\sin\Big(\frac{\text{x}+5\text{x}}{2}\Big).\cos\Big(\frac{\text{x}-5\text{x}}{2}\Big)+2\sin\Big(\frac{3\text{x}+7\text{x}}{2}\Big)\cos\Big(\frac{3\text{x}-7\text{x}}{2}\Big)$ $=2\sin3\text{x}\cos(-2\text{x})+2\sin5\text{x}\cos(-2\text{x})$ $=2\sin3\text{x}\cos2\text{x}+2\sin5\text{x}\cos2\text{x}$ $=2\cos2\text{x}[\sin3\text{x}+\sin5\text{x}]$ $=2\cos2\text{x}\Big[2\sin\Big(\frac{3\text{x}+5\text{x}}{2}\Big).\cos\Big(\frac{3\text{x}-5\text{x}}{2}\Big)\Big]$ $=2\cos2\text{x}[2\sin4\text{x}.\cos(-\text{x})]$ $=4\cos2\text{x}\sin4\text{x}\cos\text{x}=\text{R.H.S.}$ View full question & answer→Question 215 Marks
Prove the following:
$\cos^22\text{x}-\cos^26\text{x}=\sin8\text{x}\sin4\text{x}$
AnswerIt is known that $\cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big).\cos\Big(\frac{\text{A}-\text{B}}{2}\Big).$ $\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big).\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $\therefore\text{L.H.S.}=\cos^22\text{x}-\cos^26\text{x}$ $=(\cos2\text{x}+\cos6\text{x})(\cos2\text{x}-6\text{x})$
$=\Big[2\cos\Big(\frac{2\text{x}+6\text{x}}{2}\Big)\cos\Big(\frac{2\text{x}-6\text{x}}{2}\Big)\Big]\Big[-2\sin\Big(\frac{2\text{x}+6\text{x}}{2}\Big)\sin\Big(\frac{2\text{x}-6\text{x}}{2}\Big)\Big]$
$=[2\cos4\text{x}\cos(-2\text{x})][-2\sin4\text{x}\sin(-2\text{x})]$
$=[2\cos4\text{x}\cos2\text{x}][-2\sin4\text{x})(\sin-2\text{x})]$
$=(2\cos4\text{x}\cos4\text{x})(2\sin2\text{x}\cos2\text{x})$ $=\sin8\text{x}\sin4\text{x}$ $=\text{R.H.S.}$ View full question & answer→Question 225 Marks
Prove that:
$\sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt3}{8}$
Answer$\text{LHS}=\sin20^\circ\sin40^\circ\sin80^\circ$
$=\ \frac{1}{2}(2\sin20^\circ\sin40^\circ)\sin80^\circ$
$=\ \frac{1}{2}[\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ)]\sin80^\circ$$[\because2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})$
$=\ \frac{1}{2}[\cos20^\circ-\cos60^\circ]\sin80^\circ$
$=\ \frac{1}{2}\Big[\cos20^\circ-\frac{1}{2}\Big]\sin80^\circ$
$=\ \frac{1}{2}[\cos20^\circ\sin80^\circ]-\frac{1}{4}\sin80^\circ$
$=\ \frac{1}{4}[2\cos20^\circ\sin80^\circ-\sin80^\circ]$
$=\ \frac{1}{4}[\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ]$$[\because2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$
$=\ \frac{1}{4}[\sin100^\circ+\sin60^\circ-\sin80^\circ]$
$=\ \frac{1}{4}\Big[\sin(180^\circ-80^\circ)+\frac{\sqrt3}{2}-\sin80^\circ\Big]$
$=\ \frac{1}{4}\Big[\sin80^\circ+\frac{\sqrt3}{2}-\sin80^\circ\Big]$
$= \frac{\sqrt3}{8}=\ \text{RHS}$
View full question & answer→Question 235 Marks
Prove that:
$\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac{3}{16}$
Answer$\text{LHS}=\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ$
$\sin20^\circ\sin40^\circ\sin80^\circ\times\frac{\sqrt3}{2}$$\Big[\because\ \sin60^\circ=\frac{\sqrt3}{2}\Big]$
$=\ \frac{\sqrt3}{2}\times\frac{1}{2}(2\sin20^\circ\sin40^\circ)\sin80^\circ$
$=\ \frac{\sqrt3}{4}[\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ)]\sin80^\circ$
$=\ \frac{\sqrt3}{4}[\cos20^\circ-\cos60^\circ]\sin80^\circ$
$=\ \frac{\sqrt3}{4}\Big[\cos20^\circ\sin80^\circ-\frac{1}{2}\sin80^\circ\Big]$
$=\ \frac{\sqrt3}{8}[2\cos20^\circ\sin80^\circ-\sin80^\circ]$
$=\ \frac{\sqrt3}{8}[\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ]$
$=\ \frac{\sqrt3}{8}[\sin100^\circ+\sin60^\circ-\sin80^\circ]$
$=\ \frac{\sqrt3}{8}[\sin80^\circ+\sin60^\circ-\sin80^\circ]$
$=\ \frac{\sqrt3}{8}\times\sin60^\circ=\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}$
$=\ \frac{3}{16}=\text{RHS}$
View full question & answer→Question 245 Marks
Prove that:
$\frac{1}{\sin\text{(x}-\text{b})\sin\text{(x}-\text{b)}}=\frac{\cot\text{(x}-\text{b)}-\cot\text{(x}-\text{b)}}{\sin\text{(a}-\text{b)}}$
Answer$\text{L.H.S}\ \frac{1}{\sin\text{(x}-\text{a)}\sin\text{(x}-\text{b)}}$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(a}-\text{b)}}{\sin\text{(x}-\text{b})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(x}-\text{b)-}(\text{x}-\text{b)}}{\sin\text{(x}-\text{a})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(x}-\text{b)}\cos(\text{x}-\text{a})-\cos(\text{x}-\text{b})\sin(\text{x}-\text{a)}}{\sin\text{(x}-\text{a})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin(\text{x}-\text{b)}\cos(\text{x}-\text{a})}{\sin\text{(x}-\text{a)}\sin(\text{x}-\text{b})}-\frac{\sin(\text{x}-\text{b)}\cos(\text{x}-\text{a})}{\sin\text{(x}-\text{a)}\sin(\text{x}-\text{b})}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\cot(\text {x}-\text{a}-\cot(\text{x}-\text{b})\Big]$
$=\frac{\cot(\text{x}-\text{a})-\cot(\text{x}-\text{b})}{\sin(\text{a}-\text{b})}$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.
View full question & answer→Question 255 Marks
Prove the following identities:
$\frac{2\sin\text{x}\cos\text{x}-\cos\text{x}}{1-\sin\text{x}+\sin^2\text{x}-\cos^2\text{x}}=\cot\text{x}$
Answer$\text{L.H.S}=\frac{2\sin\text{x}\cos\text{x}-\cos\text{x}}{1-\sin+\sin^2\cos^2\text{x}}$
$=\frac{\cos\text{x}(2\sin\text{x}-1)}{1-\cos^2\text{x}+\sin^2\text{x}-\sin\text{x}}$
$=\frac{\cos\text{x}(2\sin\text{x}-1)}{\sin^2\text{x}+\sin^2\text{x}-\sin\text{x}}$ $(\because1-\cos^2\text{x}=\sin^2\text{x})$
$=\frac{\cos\text{x}(2\sin\text{x}-1)}{2\sin^2\text{x}-\sin\text{x}}$
$=\frac{\cos\text{x}(2\sin\text{x}-1)}{\sin\text{x}(2\sin\text{x}-1)}$
$=\frac{\cos\text{x}}{\sin\text{x}}$
$=\cot\text{x}$
$=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 265 Marks
Solve the following equations:
$\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$
Answer$\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$
$(\sin\text{x}+\sin3\text{x})-3\sin2\text{x}-(\cos\text{x}+\cos3\text{x})+3\cos2\text{x}=0$
$2\sin2\text{x}\cos\text{x}-3\sin2\text{x}-2\cos2\text{x}\cos\text{x}+3\cos2\text{x}=0$
$\sin2\text{x}(2\cos\text{x}-3)-\cos2\text{x}(2\cos\text{x}-3)=0$
$(2\cos\text{x}-3)(\sin2\text{x}-\cos2\text{x})=0$
$\cos\text{x}=\frac{3}{2}$ or $\sin2\text{x}-\cos2\text{x}-\cos2\text{x}=0$
but $\cos\text{x}\in[-11]\Rightarrow\cos\text{x}\not=\frac{3}{2}$
$\sin2\text{x}=\cos2\text{x}$
$2\text{x}=\text{n}\pi+\frac{\pi}{4}$
$\text{x}=\frac{\text{n}\pi}{2}+\frac{\pi}{8}$
View full question & answer→Question 275 Marks
Solve the following equations:
$4\sin\text{x}\cos\text{x}+2\sin\text{x}+2\cos\text{x}+1=0$
Answer$4\sin\text{x}\cos\text{x}+2\sin\text{x}+2\cos\text{x}+1=0$
$\Rightarrow2\sin\text{x}(2\cos\text{x}+1)+1(2\cos\text{x}+1)=0$
$\Rightarrow(2\sin\text{x}+1)(2\cos\text{x+1})=0$
$\Rightarrow2\sin\text{x}+1=0$ or $2\cos\text{x}=-\frac{1}{2}$
$\Rightarrow\sin\text{x}=\sin\frac{7\pi}{6}$ or $\cos\text{x}=\frac{2\pi}{3}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{7\pi}{6}$ or $\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{Z}$
View full question & answer→Question 285 Marks
Prove that:
$\cos20^\circ\cos40^\circ\cos80^\circ=\frac{1}{8}$
Answer$\text{LHS}=\cos20^\circ\cos40^\circ\cos80^\circ$
$=\ \frac{1}{2}(2\cos20^\circ\cos40^\circ)\cos80^\circ$
$=\ \frac{1}{2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20^\circ)]\cos80^\circ$$[\because\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$
$=\ \frac{1}{2}[\cos60^\circ+\cos20^\circ]\cos80^\circ$
$=\ \frac{1}{2}\Big[\frac{1}{2}+\cos20^\circ\Big]\cos80^\circ$
$=\ \frac{1}{2}[\cos80^\circ+2\cos20^\circ\cos80^\circ]$
$=\ \frac{1}{4}[\cos80^\circ+\cos(80^\circ+20^\circ)+\cos(20^\circ-80^\circ)]$
$=\ \frac{1}{4}[\cos80^\circ+\cos100^\circ+\cos60^\circ]$
$=\ \frac{1}{4}[\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ]$
$=\ \frac{1}{4}[\cos80^\circ-\cos80^\circ+\cos60^\circ]$
$= \frac{1}{4}\Big[\frac{1}{2}\Big]=\frac{1}{8}=\text{RHS}$
View full question & answer→Question 295 Marks
Prove the following identities:
$(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2=\sec^2\alpha\sec^2\beta$
Answer$\text{L.H.S}=(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2$
$=1(\tan\alpha+\tan\beta)^2+2.1\tan\alpha\tan\beta\\\ \ +(\tan\alpha)^2+(\tan\beta)^2-1\tan\alpha.\tan\beta$ $(\text{Using(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab and (a-b)}^2=\text{a}^2+\text{b}^2-2\text{ab})$
$=1+\tan^2\alpha+\tan^2\beta+2\tan\alpha\tan\beta+\tan^2\alpha+\tan^2\beta-2\tan\alpha\tan\beta$
$=1+\tan^2\alpha+\tan^2\alpha+\tan^2\beta+\tan^2\beta$
$=\sec^2\alpha+\tan^2\beta(1+\tan^2\alpha)$$(\because1+\tan^2\alpha=\sec^2\alpha)$
$=\sec^2\alpha+\tan^2\beta.\sec^2\alpha$
$=\sec^2\alpha(1+\tan^2\beta)$
$=\sec^2\alpha.\sec^2\beta$ $(\because1+\tan^2\beta=\sec^2\beta)$
$=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 305 Marks
Find the general solutions of the following equations:
$\cos4\text{x}=\cos2\text{x}$
AnswerWe have,
$\cos4\text{x}=\cos2\text{x}$
$\Rightarrow\cos4\text{x}=\cos2\text{x}=0$
$\Rightarrow2\sin\text{x}.\sin3\text{x}=0$
$\Rightarrow\text{Either}$
$\sin\text{x}=0$ or $\sin3\text{x}=0$
$\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $3\text{x}=\text{m}\pi,\text{m}\in$
Thus,
$\text{x}=\text{n}\pi$ or $\text{m}\frac{\pi}{3},\text{n,m}\in\text{z}$
View full question & answer→Question 315 Marks
Solve the following equations:
$\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$
Answer$\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$
$(\sin2\text{x}+\sin6\text{x})-\sin4\text{x}=0$
$2.\sin\Big(\frac{8\text{x}}{2}\Big).\cos\Big(\frac{4\text{x}}{2}\Big)-\sin4\text{x}=0$
$2\sin4\text{x}.\cos2\text{x}-\sin4\text{x}=0$
$\sin4\text{x}(2\cos2\text{x}-1)=0$
$\sin4\text{x}=0$ or $2\cos2\text{x}-1=0$
$4\text{x}=\text{n}(\pi)$ or $\cos2\text{x}=\frac{1}{2}$
$\text{x}=[\frac{\text{n}\pi}{4}]$ or $\cos2\text{x}=\cos[\frac{\pi}{3}]$
$\text{x}=[\frac{\text{n}\pi}{4}]$ or $\text{x}=\text{n}(\pi)\pm[\frac{\pi}{6}]$
View full question & answer→Question 325 Marks
$\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
AnswerThen,
Consider the RHS of the equation $\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
$\text{RHS}=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
$=\frac{\text{k}(\sin\text{B}-\sin\text{C})}{\text{k}\sin\text{A}}\cos\Big(\frac{\pi-(\text{B + C})}{2}\Big)$ $(\because\text{A + B + C}=\pi)$
$=\frac{2\sin\big(\frac{\text{B}-\text{C}}{2}\big)\cos\big(\frac{\text{B + C}}{2}\big)}{\sin\text{A}}$
$=\frac{\sin\big(\frac{\text{B}-\text{C}}{2}\big)2\cos\big(\frac{\text{B + C}}{2}\big)}{\sin\text{A}}\sin\Big(\frac{\text{B + C}}{2}\Big)$
$=\frac{\sin\big(\frac{\text{B}-\text{C}}{2}\big)\sin(\text{B + C})}{\sin\text{A}}$
$=\frac{\sin\big(\frac{\text{B} - \text{C}}{2}\big)\sin(\pi-\text{A})}{\sin\text{A}}$
$=\frac{\sin\text{A}\sin\big(\frac{\text{B}-\text{C}}{2}\big)}{\sin\text{A}}$
$=\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\text{LHS}$
Hence proved.
View full question & answer→Question 335 Marks
If than $\alpha=\text{x}+1,\tan\beta=\text{x}-1,$ prove that $2\cot(\alpha-\beta)=\text{x}^2$
AnswerWe have,
$\tan\alpha=\text{x}+1\text{ and }\tan\beta=\text{x}-1$
Now, $2\cot(\alpha-\beta)$
$=\frac{2}{\tan(\alpha-\beta)}$
$=\frac{\frac{2}{\tan\alpha-\tan\beta}}{1+\tan\alpha\tan\beta}$
$=\frac{2(1+\tan\alpha\tan\beta)}{\tan\alpha-\tan\beta}$
$=\frac{2[1+(\text{x}+1)(\text{x}-1)]}{\text{x}+1-(\text{x}-1)}$
$=\frac{2[1+\text{x}^2-1]}{\text{x}+1-\text{x}+1}$
$=\frac{2\text{x}\text{x}^2}{2}=\text{x}^2$
$\therefore2+\cot(\alpha+\beta )=\text{x}^2$
Hence proved.
View full question & answer→Question 345 Marks
If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that
$6\text{T}_{10}-15\text{T}_8+10\text{T}_6-1=0$
Answer$=-6\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})+6\sin^4\cos^4\text{x}\\\ \ \ +9\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})-3\sin^2\text{x}\cos^2\text{x}$
$=3\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})=6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$
$=3\sin^2\text{x}\cos^2\text{x}((\sin^2\text{x})^2+(\cos^2\text{x})^2\\\ \ +2\sin^2\text{x}\cos^2\text{x }2\sin^2\text{x}\cos^2)\\\ \ +6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $($Adding and subtracting $2\sin^2\text{x}\cos^2\text{x})$$$
$=3\sin^2\text{x}\cos^2\text{x}((\sin^2\text{x}+\cos^2\text{x})^2-2\sin^2\text{x}\cos^2\text{x})\\\ \ +6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$
$=3\sin^2\text{x}\cos^2\text{x}(1-2\sin^2\text{x}\cos^2\text{x})+6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$
$=3\sin^2\text{x}\cos^2\text{x}-6\sin^4\text{x}\cos^4+6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$
$=0$
$=\text{R.H.S}$
$\text{Proved}.$
View full question & answer→Question 355 Marks
Prove that $\Bigg|\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}\Bigg|$ $=-\frac{2}{\cos\text{x}},$ where $\frac{\pi}{2}<\text{x}<\pi$
Answer$\text{L.H.S}=\Bigg|\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}\Bigg|$
$=\Bigg|\frac{(\sqrt{1-\sin\text{x}})^2+(1+\sin\text{x})^2}{\sqrt{(1+\sin\text{x})(1-\sin\text{x})}}\Bigg|$
$=\Bigg|\frac{1-\sin\text{x}+1+\sin\text{x}}{\sqrt{1-\sin^2\text{x}}}\Bigg|$
$=\Big|\frac{2}{\cos\text{x}}\Big|$ $\Big(\because1-\sin^2\text{x}=\cos^2\text{x}\Rightarrow\sqrt{1-\sin^2\text{x}=\cos\text{x}}\Big)$
$=\frac{-2}{\cos\text{x}}$ $\Big(\frac{\pi}{2}<\text{x}<\pi\Rightarrow\cos\text{x}<0\Big)$
$=\text{R.H.S}$
View full question & answer→Question 365 Marks
$\text{If}\ \cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D}),$
prove that $\tan\text{A}\tan\text{B}\tan\text{C}+\tan\text{D}=0$
AnswerWe have,
$\cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D})$
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}...(\text{i})$
Now,
$\frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}$
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}+1=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}+1$
$\Rightarrow\ \frac{\cos(\text{A+B})+\cos(\text{A}-\text{B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})+\sin(\text{C}-\text{D})}{\sin(\text{C}-\text{D})}...(\text{ii})$
Again,
$\frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}$ [By equation (i)]
$\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}-1=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}-1$
$\Rightarrow\ \frac{\cos(\text{A+B})-\cos(\text{A}-\text{B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})-\sin(\text{C}-\text{D})}{\sin(\text{C}-\text{D})}...(\text{iii})$
Dividing equation (ii) by equation (iii), we get
$\frac{\cos(\text{A+B})+\cos(\text{A}-\text{B})}{\cos(\text{A+B})-\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})+\sin(\text{C}-\text{D})}{\sin(\text{C+D})-\sin(\text{C}-\text{D})}$
$\Rightarrow\ \frac{2\cos\big\{\frac{\text{A+B+A}-\text{B}}{2}\big\}\cos\big\{\frac{\text{A+B}-\text{A+B}}{2}\big\}}{-2\sin\big\{\frac{\text{A+B+A}-\text{B}}{2}\big\}\sin\big\{\frac{\text{A+B}-\text{A+B}}{2}\big\}}\\ \ \ \ =\frac{2\sin\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}\cos\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}}{2\sin\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}\cos\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}}$
$\Rightarrow\ \frac{\cos\text{A}\cos\text{B}}{-\sin\text{A}\sin\text{B}}=\frac{\sin\text{C}\cos\text{D}}{\sin\text{D}\cos\text{C}}$
$\Rightarrow\ \frac{1}{-\tan\text{A}\tan\text{B}}=\frac{\sin\text{C}\cos\text{D}}{\cos\text{C}\sin\text{D}}$
$\Rightarrow\ \frac{-1}{\tan\text{A}\tan\text{B}}=\frac{\tan\text{C}}{\tan\text{D}}$
$\Rightarrow\ -\tan\text{D}=\tan\text{A}\tan\text{B}\tan\text{C}$
$\Rightarrow\ \tan\text{A}\tan\text{B}\tan\text{C}=-\tan\text{D}$
$\Rightarrow\ \tan\text{A}\tan\text{B}\tan\text{C}+\tan\text{D}=0$ Hence proved.
View full question & answer→Question 375 Marks
Prove that:
$\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ=3$
Answer$\text{LHS}=\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ$
$=\ (\tan20^\circ\tan40^\circ\tan80^\circ)\sqrt3$$[\because\ \tan60^\circ=\sqrt3]$
$=\ \Big(\frac{\sin20^\circ\sin40^\circ\sin80^\circ}{\cos20^\circ\cos40^\circ\cos80^\circ}\Big)\sqrt3$
$=\ \frac{(2\sin20^\circ\sin40^\circ)\sin80^\circ\times\sqrt3}{(2\cos20^\circ\cos40^\circ)\cos80^\circ}$
Applying
$\Rightarrow\ 2\sin\text{A}\sin\text{B}-\cos(\text{A}-\text{B})-\cos(\text{A+B})$
$2\cos\text{A}\cos\text{B}-\cos(\text{A+B})+\cos(\text{A}-\text{B})$
$=\ \frac{(\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ))\sin80^\circ\times\sqrt3}{(\cos(20^\circ+40^\circ)+\cos(40^\circ-20^\circ))\cos80^\circ}$
$=\ \frac{(\cos20^\circ-\cos60^\circ)\sin80^\circ\times\sqrt3}{(\cos60^\circ+\cos20^\circ)\cos80^\circ}$
$=\ \frac{\Big(\cos20^\circ-\frac{1}{2}\Big)\sin80^\circ\times\sqrt3}{\Big(\frac{1}{2}+\cos20^\circ\Big)\cos80^\circ}$
$=\ \frac{(2\sin80^\circ\cos20^\circ-\sin80^\circ)\sqrt3}{\cos80^\circ+2\cos20^\circ\cos80^\circ}$
$\Rightarrow\ 2\sin\text{A}\cos\text{B}-\sin(\text{A+B})+\sin(\text{A}-\text{B})$
$=\ \frac{(\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ)\sqrt3}{\cos80^\circ+(\cos(20^\circ+80^\circ)+\cos(80^\circ-20^\circ))}$
$=\ \frac{(\sin100^\circ+\sin60^\circ-\sin80^\circ)\sqrt3}{\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ}$
$=\ \frac{\Big(\sin(180^\circ+\frac{\sqrt3}{2}-\sin80^\circ)\Big)\sqrt3}{\cos80^\circ-\cos80^\circ+\cos60^\circ}$
$=\ \frac{\frac{3}{2}}{\frac{1}{2}}=3=\text{RHS}$
View full question & answer→Question 385 Marks
Find the value of the expression $\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$
[Hint: Simplify the expression to $2\Big(\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}\Big)=2\Big[\Big(\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}8{}\Big)^2-2\cos^2\frac{\pi}{8}\cos^2\frac{3\pi}{8}\Big]$
Answer$\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$
$=\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\Big(\pi-\frac{3\pi}{8}\Big)+\cos^4\Big(\pi-\frac{\pi}{8}\Big)$
$=\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\frac{\pi}{8}$
$=2\Big[\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}\Big]$
$=2\Big[\cos^4\frac{\pi}{8}+\cos^4\Big(\frac{\pi}{2}-\frac{\pi}{8}\Big)\Big]$
$=2\Big[\cos^4\frac{\pi}{8}+\sin^4\frac{\pi}{8}\Big]$
$=2\Big[\Big(\cos^2\frac{\pi}{8}+\sin^2\frac{\pi}{8}\Big)^2-2\cos^2\frac{\pi}{8}.\sin^2\frac{\pi}{8}\Big]$
$=2\Big[1-2\cos^2\frac{\pi}{8}.\sin^2\frac{\pi}{8}\Big]$
$=2-\Big(2\sin\frac{\pi}{8}.\cos\frac{\pi}{8}\Big)^2$
$=2-\Big(\sin\frac{2\pi}{8}\Big)^2=2-\Big(\frac{1}{\sqrt{2}}\Big)^2=2-\frac{1}{2}=\frac{3}{2}$
View full question & answer→Question 395 Marks
Find the maximum and minimum values of each of the following trigonometrical expressions:
$\sin\text{x}-\cos\text{x}+1$
AnswerLet $\text{f}(\theta)=\sin\theta-\cos\theta+1$
$\text{f}(\theta)=\sin\theta+(-1)\cos\theta+1$
$(-1)\cos\theta+\sin\theta+1$
We know that,
$-\sqrt{(-1)^2+(1)^2}\leq\cos\theta+\sin\theta\leq\sqrt{(-1)^2+(1)^2}$
$\Rightarrow-\sqrt{1+1}\leq-\cos\theta+\sin\theta\leq\sqrt{1+1}$
$\Rightarrow-\sqrt{2}-\cos\theta+\sin\theta\leq\sqrt{2}$
$\Rightarrow-\sqrt{2}+1\leq-\cos\theta+\sin\theta+1\leq\sqrt{2}+1$
$\Rightarrow1-\sqrt{2}\leq\text{f}(\theta)\leq1+\sqrt{2}$
Hence, minimum and maximum values of $\sin\theta-\cos\theta+1-\sqrt{2}$ are $1 +\sqrt{2}$ respectively.
View full question & answer→Question 405 Marks
Find the distance from the eye at which a coin of 2cm diameter should be held so as to conceal the full moon whose angular diameter is 31'.
AnswerLet, r be the distance at which coin in placed. So that it completely conceals the full moon.
Let, E be the eye of the observer.
Now,
$\theta=31'= \Big(\frac{31}{60}\Big)^{\circ}$
$=\frac{31}{60}\times\Big(\frac{\pi}{180}\Big)^{\circ}$
Also,
$\theta= \frac{\text{arc}}{\text{radius}}$
$=\frac{31\pi}{60\times180}=\frac{0.02}{\text{r}}$
$=2.217\text{m}$
The coin should be placed at a distance of 2.217m from the eye.
View full question & answer→Question 415 Marks
Prove that:$\sin^2\text{B}=\sin^2\text{A}+\sin^2\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\sin\text{(A}-\text{B)} $
Answer$\text{R.H.S}=\sin^2\text{A}+\sin^2\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\sin\text{(A}-\text{B)} $
$=\sin^2\text{A}+\sin\text{(A}-\text{B)}\Big[\sin\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\Big] $
$=\sin^2\text{A}+\sin\text{(A}-\text{B)}\Big[-\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}\ \Big]$
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}\Big[\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\ \Big]$
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}(\sin(\text{A}+\text{B))} $
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}\sin(\text{A}+\text{B)} $
$=\sin^2\text{A}-\sin(\sin^2\text{A}-\sin^2\text{B})$
$=\sin^2\text{A}-\sin\sin^2\text{A}+\sin^2\text{B}$$\Big[\because\sin(\text{A}-\text{B)}\sin\text{(A+}\text{B)}=\sin^2\text{A}-\sin^2\text{B}\Big]$
$=\sin^2\text{B}$
$=\text{L.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$
Hence proved.
View full question & answer→Question 425 Marks
Prove the following identities:
$\Big(\frac{1}{\sec^2\text{x}-\cos^\text{x}}+\frac{1}{\text{Cosec}^2\text{x}-\sin^2\text{x}}\Big)\sin^2\text{x}\cos^2\text{x}=\frac{1-\sin^2\text{x}\cos^2\text{x}}{2+\sin^2\text{x}\cos^2\text{x}}$
Answer$\text{L.H.S}=\Big(\frac{1}{\sec^2\text{x}-\cos^2\text{x}}+\frac{1}{\text{cosec}^2\text{x}-\sin^2\text{x}}\Big)\sin^2\text{x}\cos^2\text{x}$
$=\Bigg(\frac{1}{\frac{1}{\cos^2\text{x}}-\cos^2\text{x}}+\frac{1}{\sin^2\text{x}-\sin^2\text{x}}\Bigg)\sin^2\text{x}\cos^2\text{x}$
$\Bigg(\frac{1}{\frac{1-\cos^4\text{x}}{\cos^2\text{x}}}+\frac{1}{\frac{1-\sin^4\text{x}}{\sin^2\text{x}}}\Bigg)\sin^2\text{x}\cos^2\text{x}$
$=\Big(\frac{\cos^2\text{x}}{(1-\cos^2\text{x})(1+\cos^2\text{x})}+\frac{\sin^2\text{x}}{(1-\sin^2\text{x})(1+\sin^2\text{x})}\Big)\sin^2\text{x}\cos^2\text{x}$$\left(\begin{array}{c}\text{Using}1-\text{a}^4=1-(\text{a}^2)^2\\ =(1-\text{a}^2)(1+\text{a}^2)\end{array}\right)$
$=\Big(\frac{\cos^2\text{x}}{\sin^2\text{x}(1+\cos^2\text{x})}+\frac{\sin^2\text{x}}{\cos^2\text{x}(1+\sin^2)}\Big)\sin^2\text{x}\cos^2\text{x}$$\left(\begin{array}{c}\text{Using }1-\cos^2\text{x}=\sin^2\text{x}\\\&1-\sin^2\text{x}=\cos^2\text{x}\end{array}\right)$
$=\Big(\frac{\cos^4\text{x}(1+\sin^\text{x})+\sin^4\text{x}(1+\cos^2\text{x})}{\sin^2\text{x}\cos^2\text{x}(1+\cos^2\text{x})(1+\sin^2\text{x})}\Big)\sin^2\text{x}\cos^2\text{x}$
$=\frac{\cos^4\text{x}+\sin^2\text{x}\cos^4\text{x}+\sin^4\text{x}+\cos^2\text{x}\sin^4\text{x}}{(1+\cos^2\text{x})(1+\sin^2\text{x})}$
$=\frac{(\cos^2\text{x})^2+(\sin^2\text{x})^2+2\cos^2\text{x}\sin^2\text{x}-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^4\text{x}+\cos^2\text{x}\sin^4\text{x}}{(1+\cos^2\text{x})(1+\sin^2\text{x})}$ $(\text{adding an subtracting }2\cos^2\text{x}\sin^4\text{x})$
$=\frac{(\cos^2+\sin^2\text{x})^2-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^2\text{x}(\cos^2\text{x}+\sin^2\text{x})}{1+\sin^2\text{x}+\cos^2\text{x}+\sin^2\text{x}\cos^2\text{x}}$
$=\frac{1^2-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^2\text{x}.1}{1+1+\sin^2\text{x}+\cos^2\text{x}}$
$=\frac{1-\sin^2\text{x}\cos^2\text{x}}{2+\sin^2\text{x}\cos^2\text{x}}$
$=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 435 Marks
Prove that:
$\sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}=4\cos\frac{\text{A}}{2}\cos\frac{2\text{A}}{2}\cos4\text{A}$
AnswerWe have,
$\text{LHS}=\sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}$
$=\ (\sin2\text{A}+\sin\text{A})+(\sin5\text{A}+\sin4\text{A})$
$=\ \Big[2\sin\Big(\frac{2\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{2\text{A}-\text{A}}{2}\Big)\Big]+\Big[2\sin\Big(\frac{5\text{A}+4\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-4\text{A}}{2}\Big)\Big]$
$=\ 2\sin\frac{3\text{A}}{2}\cos\frac{\text{A}}{2}+2\sin\frac{9\text{A}}{2}\cos\frac{\text{A}}{2}$
$=\ 2\cos\frac{\text{A}}{2}\Big[\sin\frac{3\text{A}}{2}+\sin\frac{9\text{A}}{2}\Big]$
$=\ 2\cos\frac{\text{A}}{2}\Big[\sin\frac{9\text{A}}{2}+\sin\frac{3\text{A}}{2}\Big]$
$=\ 2\cos\frac{\text{A}}{2}\Big[2\sin\Big\{\frac{1}{2}\Big(\frac{9\text{A}}{2}+\frac{3\text{A}}{2}\Big)\Big\}\cos\Big\{\frac{1}{2}\Big(\frac{9\text{A}}{2}-\frac{3\text{A}}{2}\Big)\Big\}\Big]$
$=\ 4\cos\frac{\text{A}}{2}\Big[\sin\frac{12\text{A}}{4}\cos\frac{6\text{A}}{4}\Big]$
$=\ 4\cos\frac{\text{A}}{2}\sin3\text{A}\cos\frac{3\text{A}}{2}$
$=\ 4\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}\sin3\text{A}$
$=\ \text{RHS}$
$\therefore\ \sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}=4\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}\sin3\text{A}.$
Hence proved.
View full question & answer→Question 445 Marks
Prove that:
$\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ=\frac{\sqrt3}{16}$
Answer$\text{LHS}=\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ$
$\Big(\sin10^\circ\sin50^\circ\sin70^\circ\frac{\sqrt3}{2}\Big)$$\Big[\because\ \sin60^\circ=\frac{\sqrt3}{2}\Big]$
$=\ \sin(90^\circ-80^\circ)\sin(90^\circ-40^\circ)\sin(90^\circ-20^\circ)\frac{\sqrt3}{2}$
$=\ \cos80^\circ\cos40^\circ\cos20^\circ\frac{\sqrt3}{2}$
$=\ \frac{\sqrt3}{2\times2}(2\cos40^\circ\cos20^\circ)\cos80^\circ$$[\because\ 2\cos\text{A}\cos\text{B}=(\cos\text{A+B})+\cos(\text{A}-\text{B})]$
$=\ \frac{\sqrt3}{2\times2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20^\circ)]\cos80^\circ$
$=\ \frac{\sqrt3}{2\times2}[\cos60^\circ+\cos20^\circ]\cos80^\circ$
$=\ \frac{\sqrt3}{2\times2}\Big[\frac{1}{2}+\cos20^\circ\Big]\cos80^\circ$
$=\ \frac{\sqrt3}{4}\Big[\frac{1}{2}\cos80^\circ+\cos20^\circ\cos80^\circ\Big]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+2\cos20^\circ\cos80^\circ]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos(80^\circ+20^\circ)+\cos(80^\circ-20^\circ)]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos100^\circ+\cos60^\circ]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ]$
$=\ \frac{\sqrt3}{8}[\cos60^\circ]=\frac{\sqrt3}{16}=\text { RHS}$
View full question & answer→Question 455 Marks
If $\text{x}=\sec\phi-\tan\phi$ and $\text{y}=\text{cosec}\phi+\cot\phi$ then show that $\text{xy}+\text{x}-\text{y}+1=0$
[Hint: Find xy + 1 and then show that x - y = -(xy + 1)]
Answer$\text{x}=\sec\phi-\tan\phi\Rightarrow\text{x}=\frac{1-\sin\phi}{\cos\phi}$
$\text{y}=\text{cosec}\phi+\cot\phi\Rightarrow\text{y}=\frac{1+\cos\phi}{\sin\phi}$
$\Rightarrow\text{xy + x}-\text{y}=\frac{1-\sin\phi1+\cos\phi}{\cos\phi\sin\phi}+\frac{1-\sin\phi}{\cos\phi}-\frac{1+\cos\phi}{\sin\phi}$
$=\frac{(1-\sin\phi)(1+\cos\phi)+(1-\sin\phi)\sin\phi-\cos\phi(1+\cos\phi)}{\sin\phi\cos\phi}$
$=\frac{1-\sin\phi+\cos\phi-\sin\phi\cos\phi+\sin\phi-\sin^2\phi-\cos\phi-\cos^2\phi}{\sin\phi\cos\phi}$
$=\frac{1-\sin\phi\cos\phi-(\sin^2\phi+\cos^2\phi)}{\sin\phi\cos\phi}=-1$
$\therefore\text{xy + x}-\text{y}-1=0$
View full question & answer→Question 465 Marks
Prove that:
$\cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}=4\cos\text{A}\cos2\text{A}\cos4\text{A}$
AnswerWe have,
$\text{LHS}=\cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}$
$=\ [\cos3\text{A}+\cos\text{A}]+[\cos7\text{A}+\cos5\text{A}]$
$=\ \Big[2\cos\Big(\frac{3\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{3\text{A}-\text{A}}{2}\Big)\Big]+\Big[2\cos\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-5\text{A}}{2}\Big)\Big]$
$=\ 2\cos2\text{A}\cos\text{A}+2\cos6\text{A}\cos4\text{A}$
$=\ 2\cos\text{A}[\cos2\text{A}+\cos6\text{A}]$
$=\ 2\cos\text{A}[\cos2\text{A}+\cos6\text{A}]$
$=\ 2\cos2\text{A}\Big[2\cos\Big(\frac{6\text{A}+2\text{A}}{2}\Big)\cos\Big(\frac{6\text{A}-\text{A}}{2}\Big)\Big]$
$=\ 4\cos\text{A}[\cos4\text{A}\cos2\text{A}]$
$=\ \text{RHS}$
$\therefore\ \cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}=4\cos\text{A}\cos2\text{A}\cos4\text{A}$
Hence proved.
View full question & answer→Question 475 Marks
Sketch the graphs of the following trigonometric functions:
$\text{u(x)}=\cos^2\frac{\text{x}}{2}$
Answer$\text{u(x)}=\cos^2\frac{\text{x}}{2}$
$\text{y}=\cos^2\frac{\text{x}}{2}$
The following graph is:
View full question & answer→Question 485 Marks
Prove that
$\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}\cos\frac{16\pi}{15}=\frac{1}{16}$
Answer$\sin(\text{a+b})=\sin(\text{a})\cos(\text{b})+\sin(\text{b})\cos(\text{a})\ .....(1)$
$\text{for}\ \text{a}=\text{b},\sin(2\text{a})=2\sin(\text{a})\cos(\text{a})\ .....(2)$
$\text{let}\ \text{a}=16\frac{\text{pi}}{15}\ .....(3)$
$(\text{so}2\text{a}=32\frac{\text{pi}}{15})$
then using (3) in (2),we have
$\sin(2\text{a})=2\sin(\text{a})\cos(\text{a})$
$=2(2\sin(\frac{\text{a}}{2})\cos(\frac{\text{a}}{2}))\cos(\text{a})$
$=2(2(2\sin(\frac{\text{a}}{4})\cos\frac{\text{a}}{4}))\cos(\frac{\text{a}}{2}))\cos(\text{a})$
$=2(2(2(2\sin(\frac{\text{a}}{8})\cos(\frac{\text{a}}{8}))\cos(\frac{\text{a}}{4}))\cos(\frac{\text{a}}{2}))\cos(\text{a})$
$=16\sin(\frac{\text{a}}{8})(\cos(\frac{\text{a}}{8})\cos(\frac{\text{a}}{4})\cos(\frac{\text{a}}{2})\cos(\text{a})$
now note,
$\sin(2\text{a})=\sin(2\frac{\text{pi}}{15})$ and $\sin(\frac{\text{a}}{8})=\sin(2\frac{\text{pi}}{15})$
so,
$\cos(\frac{\text{a}}{8})\cos(\frac{\text{a}}{4})\cos(\frac{\text{a}}{2})\cos(\text{a})=\frac{1}{16}$
or, relacing a with $16\frac{\text{pi}}{15},$
$\cos\Big(2\frac{\text{pi}}{15}\Big)\times\cos\Big(4\frac{\text{pi}}{15}\Big)\times\cos\Big(8\frac{\text{pi}}{15}\Big)\times\cos\Big(16\frac{\text{pi}}{15}\Big)=\frac{1}{16}$
View full question & answer→Question 495 Marks
Prove the following identities:
$\frac{\cos\text{x}}{1-\sin\text{x}}=\frac{1+\cos\text{x}+\sin\text{x}}{1+\cos\text{x}-\sin\text{x}}$
Answer$\text{R.H.S}=\frac{1+\cos\text{x}+\sin\text{x}}{1+\cos\text{x}-\sin\text{x}}$
$=\frac{\big(\big(1+\cos\text{x}\big)+\sin\text{x}\big)}{\big(1+\cos\text{x}\big)-\sin\text{x}}\times\frac{\big(\big(1+\cos\text{x}\big)+\sin\text{x}\big)}{\big(1+\cos\text{x}+\sin\text{x}\big)}$
$=\frac{\big(\big(1+\cos\text{x}\big)+\sin\text{x}\big)^{2}}{\big(1+\cos\text{x}\big)^{2}-\sin^{2}\text{x}}$$\Big(\text{using}\big(\text{a+b}\big)\big(\text{a+b}\big)=\big(\text{a+b}\big)^{2}\&\big(\text{a+b}\big)\big(\text{a-b}\big)=\text{a}^{2}\text{b}^{2}\Big)$
$=\frac{\big(1+\cos\text{x}\big)^{2}+\sin^{2}\text{x}+2\sin\text{x}\big(1+\cos\text{x}\big)}{1+\cos^{2}\text{x}+2\cos\text{x}-\sin^{2}\text{x}}$ $\big(\text{using}\big(\text{a+b}\big)^{2}=\text{a}^{2}+\text{b}^{2}+2\text{ab}\big)$
$=\frac{1+\cos^{2}\text{x}+2.1\cos\text{x}+\sin^{2}\text{x}+2\sin\text{x}\big(1+\cos\text{x}\big)}{1+\cos{2}\text{x}+2\cos\text{x}-\big(1-\cos^{2}\text{x}\big)}$$\big(\text{using}\sin^{2}\text{x}=1-\cos^{2}\text{x}\big)$
$=\frac{1+1+2\cos\text{x}+2\sin\text{x}\big(1+\cos\text{x}\big)}{1-1+\cos^{2}\text{x}+\cos^{2}\text{x}+2\cos\text{x}}$ $\big(\text{using}\sin^{2}\text{x}+\cos^{2}\text{x}\big)=1$
$=\frac{2+2\cos\text{x}+2\sin\text{x}\big(1+\cos\text{x}\big)}{2\cos^{2}\text{x}+2\cos\text{x}}$
$=\frac{2\big(1+\cos\text{x}\big)+2\sin\text{x}\big(1+\cos\text{x}\big)}{2\cos\text{x}\big(\cos\text{x}+1\big)}$
$=\frac{\big(1+\cos\text{x}\big)\big(2+2\sin\text{x}\big)}{2\cos\text{x}\big(1+\cos\text{x}\big)}$
$=\frac{1+\sin\text{x}}{\cos\text{x}}$
$=\frac{1+\sin\text{x}}{\cos\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}$
$\text{L.H.S}=\frac{\cos\text{x}}{1-\sin\text{x}}$
View full question & answer→Question 505 Marks
Prove that
$\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}=\frac{1}{64}$
Answer$\text{LHS}=\cos\frac{\pi}{65}.\cos\frac{2\pi}{65}.\cos\frac{4\pi}{65}.\cos\frac{8\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
Divide and multiply by $2\sin\frac{\pi}{65},$ we get
$=\frac{2.\sin\frac{\pi}{65}}{2\sin\frac{\pi}{65}}.\cos\frac{\pi}{65}.\cos\frac{2\pi}{65}.\cos\frac{4\pi}{65}.\cos\frac{8\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{2\pi}{65}}{2.2\sin\frac{\pi}{65}}.\cos\frac{2\pi}{65}.\cos\frac{4\pi}{65}.\cos\frac{6\pi}{65}.\cos\frac{8\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{4\pi}{65}}{2.4\sin\frac{\pi}{65}}.\cos\frac{4\pi}{65}.\cos\frac{8\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{8\pi}{65}}{2.8\sin\frac{\pi}{65}}.\cos\frac{8\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{16\pi}{65}}{2.16\sin\frac{\pi}{65}}.\cos\frac{16\pi}{65}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{32\pi}{65}}{2.32\sin\frac{\pi}{65}}.\cos\frac{32\pi}{65}$
$=\frac{2.\sin\frac{64\pi}{65}}{2.64\sin\frac{\pi}{65}}$
$=\frac{1}{64}.\frac{\sin\Big(\pi-\frac{\pi}{65}\Big)}{\sin\frac{\pi}{65}}$
$=\frac{1}{64}\frac{\sin\frac{\pi}{65}}{\sin\frac{\pi}{65}}$
$=\frac{1}{64}=\text{RHS}$
View full question & answer→