Question 13 Marks
Prove that: $\cot^2\text{x}-\tan^2\text{x}=4\cot2\text{x}\ \text{cosec}\ 2\text{x}$
Answer$\text{LHS}=\cot^2\text{x}=\tan^2\text{x}$ $=\frac{\cos^2\text{x}}{\sin^2\text{x}}-\frac{\sin^2\text{x}}{\cos^2\text{x}}$ $=\frac{(\cos^2\text{x})^2-(\sin^2\text{x})^2}{\sin^2\text{x}\cos^2\text{x}}$ $=\frac{(\cos^2\text{x}+\sin^2\text{x})(\cos^2\text{x}-\sin^2\text{x})}{(\sin\text{x}\cos\text{x}^2)}$ $[\because\text{a}^2-\text{b}^2-=(\text{a+b})(\text{a-b})]$ $=\frac{\cos2\text{x}}{\frac{1}{4}(2\sin\text{x}\cos\text{x})^2}$ $[\because\cos2\text{x}=\cos^2\text{x}-\sin^2\text{x}]$ $=\frac{4\cos2\text{x}}{\sin^22\text{x}}$ $=\frac{4\cos2\text{x}}{\sin^2\text{x}}.\frac{1}{\sin2\text{x}}$ $\Big[\because\text{cosec}\ \theta=\frac{1}{\sin\theta}\Big]$ $=4\cot1\text{x}.\text{cosec}2\text{x}=\text{RHS}$
View full question & answer→Question 23 Marks
If $\tan\text{x}=\frac{\text{b}}{\text{a}},$ then find the value of $\sqrt{\frac{\text{a+b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$
Answer$\sqrt{\frac{\text{a+b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}=\frac{(\text{a+b})+(\text{a}-\text{b})}{\sqrt{(\text{a}-\text{b})(\text{a+b})}}$ $=\frac{2\text{a}}{\sqrt{\text{a}^2-\text{b}^2}}$ $=\frac{2}{\sqrt{1-\Big(\frac{\text{b}}{\text{a}}\Big)^2}}$ $=\frac{2}{\sqrt{\tan^2\text{x}}}$ $\Big[\because\tan\text{x}=\frac{\text{b}}{\text{a}}\Big]$ $=\frac{2\cos\text{x}}{\sqrt{\cos^2\text{x}}\sin^2\text{x}}$ $=\frac{2\cos\text{x}}{\sqrt{\cos2\text{x}}}$
View full question & answer→Question 33 Marks
Prove that $\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{\pi}{5}\cos\frac{8\pi}{5}=\frac{-1}{16}$
Answer$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5}=\frac{\sin\frac{2^4\pi}{5}}{2^4\sin\frac{\pi}{5}}$ $\Big[\because\cos\text{A}\cos2\text{A}\cos2^2\text{A}\cos2^3\text{A}...\cos2^{\text{n}-1}\text{A}=\frac{\sin2^\text{n}\text{A}}{2^\text{n}\sin\text{A}}\Big]$ $=\frac{\sin\frac{16\pi}{5}}{16\sin\frac{\pi}{5}}$ $=\frac{\sin\Big(3\pi+\frac{\pi}{5}\Big)}{16\sin\frac{\pi}{5}}$ $=\frac{1\big\{3\pi+\frac{\pi}{5}\big\}}{16\sin\frac{\pi}{5}}$ $=\frac{-1}{16}$
View full question & answer→Question 43 Marks
Prove that: $\big(\cos\alpha+\cos\beta^2\big)+\big(\sin\alpha+\sin\beta\big)^2=2\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$
Answer$\text{LHS}=\big(\cos\lambda+\cos\beta\big)^2+\big(\sin\lambda+\sin\beta\big)^2$ $=\cos^2\lambda+\cos^2\beta+2\cos\lambda\cos\beta+\sin^2\lambda+\sin^2\beta+2\sin\lambda+\sin\beta$ $=\Big(\cos^2\lambda+\sin^2\lambda\Big)+\Big(\cos^2\beta+\sin^2\beta\Big)+2\Big(\cos\lambda\cos\beta+\sin\lambda\sin\beta\Big)$ $=1+1+2\cos(\lambda-\beta)$ $=2+2\cos(\lambda-\beta)$ $=2\big(1+\cos(\lambda-\beta)\big)$ $=2.2\cos^2\Big(\frac{\lambda-\beta}{2}\Big)$ $=4\cos^2\Big(\frac{\lambda-\beta}{2}\Big)=\text{RHS}$
View full question & answer→Question 53 Marks
Prove that: $\sin4\text{x}=4\sin\text{x}\cos^3\text{x}-4\cos\text{x}\sin^3\text{x}$
Answer$\text{LHS}=\sin4\text{x}$ $=\sin2.2\text{x}$ $=2\sin2\text{x}\cos2\text{x}$ $=2(\sin\text{x}\cos\text{x}).(\cos^2\text{x}-\sin^2\text{x})$ $=4\sin\text{x}\cos^3\text{x}-4\sin^3\text{x}\cos\text{x}=\text{RHS}$
View full question & answer→Question 63 Marks
Prove that: $\cos4\text{x}=1-8\cos^2\text{x}+8\cos^4\text{x}$
Answer$\text{LHS}=\cos4\text{x}$ $=\cos2.2\text{x}$ $=2\cos^22\text{x}=1$ $[\because\cos2\theta=1\cos^2\theta-1]$ $=2(2\cos2\text{x}-1)^2-1$ $=2(4\cos^4\text{x}-4\cos^2\text{x}+1)-1$ $8\cos^4\text{x}-8\cos^4\text{x}+1$ $=1=8\cos^2\text{x}+8\cos^4\text{x}=\text{RHS}$
View full question & answer→Question 73 Marks
Prove that: $\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}=\frac{5}{16}$
Answer$\text{LHS}=\sin36^\circ.\sin72^\circ.\sin108^\circ.\sin144^\circ$ $\begin{bmatrix}\because\sin144^\circ=\sin(180^\circ-366^\circ)=\sin36^\circ\\\sin108^\circ=\sin(180^\circ-78^\circ)=\sin72^\circ\end{bmatrix}$ $\sin36^\circ.\sin72^\circ.\sin72^\circ.\sin36^\circ$ $=\frac{1}{4}(2\sin36.\sin72^\circ)^2$ $=\frac{1}{4}(2\sin36.\cos18^\circ)^2$ $[\because\sin72^\circ=108^\circ]$ $=\frac{4}{4}\Bigg(\frac{\sqrt{10-2\sqrt{5}}}{4}.\frac{\sqrt{10+2\sqrt{5}}}{4}\Bigg)^2$ $=\frac{1}{64}\big(10-2\sqrt{5}\big)\big(10+2\sqrt{5}\big)$ $=\frac{100-20}{64\times4}$ $=\frac{80}{256}$ $=\frac{5}{16}$ $=\text{RHS}$
View full question & answer→Question 83 Marks
Prove that: $\cos7^\circ\cos14^\circ\cos28^\circ\cos56^\circ=\frac{\sin68^\circ}{16\cos83^\circ}$
Answer$\text{LHS}=\cos7^\circ\cos14^\circ\cos28^\circ\cos56^\circ$ Divide and miltiply by $2\sin7^\circ,$ weget $\frac{1}{2\sin7^\circ}.2\sin7^\circ.\cos14^\circ.\cos28^\circ.\cos56^\circ$ $=\frac{2\sin14^\circ}{2.2\sin7^\circ}.2\cos14^\circ.\cos28^\circ.\cos56^\circ$ $=\frac{2\sin28^\circ}{2.4\sin7^\circ}.\cos28^\circ.\cos56^\circ$ $=\frac{2\sin56^\circ}{2.8\sin7^\circ}.\cos56^\circ$ $=\frac{\sin112^\circ}{16\sin7^\circ}$ $=\frac{\sin(180^\circ-68^\circ)}{16\sin(90^\circ-83^\circ)}$ $=\frac{\sin68^\circ}{16\cos83^\circ}=\text{RHS}$
View full question & answer→Question 93 Marks
Prove that: $4(\cos^310^\circ+\sin^320^\circ)=3(\cos10^\circ+\sin20^\circ)$
AnswerConsider the LHS of the given equation $4(\cos^310^\circ+\sin^320^\circ)=3(\cos10^\circ+\sin20^\circ)$ $\text{LHS} = 4(\cos^310^\circ+\sin^320^\circ)$ since $\sin30=\cos60^\circ=\frac{1}{2}$ and $\sin60^\circ=\cos30^\circ=\frac{\sqrt{3}}{2}$ $\Rightarrow\sin3.20^\circ=\cos3.10^\circ$ $\Rightarrow3\sin20^\circ-4\sin^320^\circ=4\cos^310^\circ-3\cos10^\circ$ $\Rightarrow4(\cos^310^\circ+\sin^320^\circ)=3(\cos10^\circ+\sin20^\circ)$ $=\text{RHS}$
View full question & answer→Question 103 Marks
If sin $\sin\text{x}=\frac{\sqrt{5}}{3}$ and x lies in IInd quadrant, find the values of $\cos\frac{\text{x}}{2},\sin\frac{\text{x}}{2}$ and $\tan\frac{\text{x}}{2}$
Answer$\because$ x lies in $II^{nd}$ quad. $\Rightarrow\frac{\pi}{2}<\text{x}<\pi$ $\Rightarrow\frac{\pi}{4}<\text{x}\frac{\text{x}}{2}<\frac{\pi}{2}$ which menns $\frac{\text{x}}{2}$ lone in first quad.
Now, $\sin\text{x}=\frac{\sqrt{5}}{3}=\frac{\text{p}}{\text{h}}\Rightarrow\text{p}=\sqrt{5}\Rightarrow\text{b}=2$ so, $\cos\text{x}=\frac{\text{b}}{\text{h}}=\frac{-2}{3}$ (-ve due to $II^{nd}$ quad) Thus, $\cos\frac{\text{x}}{2}=\sqrt{\frac{1+\cos\text{x}}{}}2=\sqrt{\frac{1-\frac{2}{3}}{2}}=\frac{1}{\sqrt{6}}$
$\sin\frac{\text{x}}{2}=\sqrt{\frac{1-\cos\text{x}}{2}}=\sqrt{\frac{1+\frac{2}{3}}{2}}=\sqrt{\frac{5}{6}}$
$\tan\frac{\text{x}}{2}=\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}=\frac{\sqrt{\frac{5}{6}}}{\sqrt{\frac{1}{\sqrt{6}}}}=\sqrt{5}$
View full question & answer→Question 113 Marks
If $\sin\text{x}=\frac{4}{5}$ and $0<\text{x}<\frac{\pi}{2},$ find the value of sin 4x.
Answer$\sin\theta=\frac{4}{5}=\frac{\text{p}}{\text{h}}\Rightarrow\text{p}=4\Rightarrow\text{b}=3$ $\text{h}=5$ Now, $\sin\theta=2\sin\theta.\cos\theta=2.\frac{4}{5}.\frac{3}{5}=\frac{24}{25}$ $\cos2\theta=\cos^2\theta=\sin^2\theta=\Big(\frac{3}{5}\Big)^2-\Big(\frac{4}{5}\Big)^2=\frac{-7}{25}$ so, $\sin4\theta=\sin2.2\theta=2\sin2\theta.\cos2\theta$ $=2.\frac{24}{25}.\Big(\frac{-7}{25}\Big)$ $=\frac{-336}{625}$
View full question & answer→Question 123 Marks
$\cot\text{x}+\cot(\frac{\pi}{3}+\text{x})+\cot(\frac{2\pi}{3}+\text{x})=3\cot3\text{x}$
Answer$\text{LHS}=\cot\text{x}+\cot(60^\circ+\text{x})-\cot(60^\circ-\text{x})$ $=\frac{1}{\tan\text{x}}+\frac{1}{\tan(60^\circ+\text{x})}-\frac{1}{\tan(60^\circ-\text{x})}$ $=\frac{1}{\tan\text{x}}+\frac{1-\sqrt{3}\tan\text{x}}{\sqrt{3}+\tan\text{x}}-\frac{1+\sqrt{3}\tan\text{x}}{\sqrt{3}\tan\text{x}}$ $=\frac{1}{\tan\text{x}}-\frac{8\tan\text{x}}{3-\tan^2\text{x}}$ $=\frac{3-\tan^2\text{x}8\tan^2\text{x}}{3\tan\text{x}-\tan^3\text{x}}$ $\frac{3-9\tan^2\text{x}}{3\tan\text{x}-\tan^3\text{x}}$ $=3\Big(\frac{1-3\tan^2\text{x}}{3\tan\text{x}-\tan^3\text{x}}\Big)$ $=\frac{3}{\tan3\text{x}}$ $=3\cot3\text{x}=\text{RHS}$ $\text{LHS = RHS}$
View full question & answer→Question 133 Marks
Prove that: $\tan(\frac{\pi}{4}+\text{x})+\tan(\frac{\pi}{4}-\text{x})=2\sec2\text{x}$
Answer$\text{LHS}=\tan\Big(\frac{\pi}{4}+\text{x}\Big)+\tan\frac{\pi}{4}-\text{x}\Big)$ $=\frac{\tan\frac{\pi}{4}+\tan\text{x}}{1-\tan\frac{\pi}{4}\tan\text{x}}+\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\tan\text{x}}$ $=\frac{1+\tan\text{x}}{1-\tan\text{x}}+\frac{1-\tan\text{x}}{1+\tan\text{x}}$ $\Big[\because\tan\frac{\pi}{4}=1\Big]$ $=\frac{(1+\tan^2\text{x}+2\text{x})+(1+\tan^2\text{x}-1\tan\text{x})}{(1-\tan\text{x})(1+\tan\text{x})}$ $=\frac{1(1+\tan^2\text{x})}{1-\tan^2\text{x}}$ $=\frac{1\sec^2\text{x}}{1-\frac{\sin^2\text{x}}{\cos^2\text{x}}}$ $[\because\sec^2\text{x}=1+\tan^2\text{x}]$ $=\frac{2\sec^2\text{x}.\cos^2\text{x}}{\cos^2\text{x}-\sin62\text{x}}$ $[\because\sec-\frac{1}{\cos\text{x}}]$ $=\frac{2}{\cos2\text{x}}$ $=2\sec2\text{x}=\text{RHS}$
View full question & answer→Question 143 Marks
If $\text{a}\cos2\text{x}+\text{b}\sin2\text{x}=\text{c}$ has $\alpha$ and $\beta$ as its roots, then prove that, $\tan\alpha\tan\beta=\frac{\text{c}-\text{a}}{\text{c+a}}$
Answer$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$ $\sin^2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$ substitute these valuse in the given equation, it reduces to $\text{a}(1-\tan^2)+\text{b}(2\tan\theta)=\text{c}(1+\tan^2\theta)$ $(\text{c+a})\tan^2\theta+2\text{b}\tan\theta+\text{C-a}=0$ As $\alpha$ and $\beta$ are roots Product of roots, $\tan\alpha\tan\beta=\frac{\text{c-a}}{\text{c+a}}$
View full question & answer→Question 153 Marks
Prove that: $\sin^2(\frac{\pi}{8}+\frac{\text{x}}{2})-\sin^2(\frac{\pi}{8}-\frac{\text{x}}{2})=\frac{1}{\sqrt{2}}\sin\text{x}$
Answer$\text{LHS}=\sin^2\Big(\frac{\pi}{8}+\frac{\text{x}}{2}\Big)-\sin^2\Big(\frac{\pi}{8}-\frac{\text{x}}{2}\Big)$ $=\Big[\sin\Big(\frac{\pi}{8}+\frac{\text{x}}{2}\Big)+\sin\Big(\frac{\pi}{8}-\frac{\text{x}}{\pi}\Big)\Big]\Big[\frac{\pi}{8}+\frac{\text{x}}{2}\Big)-\sin\Big(\frac{\pi}{8}-\frac{\text{x}}{2}\Big)\Big]$ $=\Big[\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}+\cos\frac{\pi}{8},\sin\frac{\text{x}}{2}-\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}+\cos\frac{\pi}{8}\sin\frac{\text{x}}{2}\Big]$ $=\Big[\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}+\cos\frac{\pi}{\text{x}},\sin\frac{\text{x}}{2}-\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}+\cos\frac{\pi}{8},\sin\frac{\text{x}}{2}\Big]$ $=\Big(2\sin\frac{\pi}{8},\cos\frac{\text{x}}{2}\Big)\Big(2\cos\frac{\pi}{2},\sin\frac{\text{x}}{2}\Big)$ $=2\sin\frac{\pi}{8},\cos\frac{\pi}{2}2\sin\frac{\text{x}}{2},\cos\frac{\text{x}}{2}$ $=\sin2.\frac{\pi}{8},\sin2.\frac{\text{x}}{2}$ $=\sin\frac{\pi}{4},\sin\text{x}$ $=\frac{1}{\sqrt{2}}\sin\text{x}=\text{RHS}$
View full question & answer→Question 163 Marks
Prove that: $\cos36^\circ\cos42^\circ\cos60^\circ\cos78^\circ=\frac{1}{16}$
Answer$\text{LHS}=\cos36^\circ.\cos42^\circ.\cos60^\circ.\cos78^\circ$ $=\frac{1}{2}\cos36^\circ.\cos60^\circ\big(2\cos42^\circ.\cos78^\circ\big)$ $=\frac{1}{2}\bigg(\frac{\sqrt{5}+1}{4}\bigg).\frac{1}{2}(\cos120^\circ+\cos36^\circ)$ $=\frac{\big(\sqrt{5}+1\big)}{16}\Big(\frac{-1}{2}+\frac{\sqrt{5}+1}{4}\Big)$ $=\frac{\sqrt{5}+1}{16}\Big(\frac{-2+\sqrt{5}+1}{4}\Big)$ $=\frac{\big(\sqrt{5}+1\big)\big(\sqrt{5}-1\big)}{64}$ $=\frac{5-1}{64}$ $\frac{1}{16}$ $=\text{RHS}$
View full question & answer→Question 173 Marks
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b}$ prove that $\cos(\alpha-\beta)=\frac{\text{a}^2+\text{b}^2-2}{2}$
AnswerWe have, $\sin\alpha+\sin\beta=\text{a}\ \&\cos\alpha+\cos\beta=\text{b}$ squaring and adding, we get $\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta+\cos^2\beta+2\cos\alpha\cos\beta=\text{a}^2+\text{b}^2$ $\Rightarrow1+1+2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=\text{a}^2+\text{b}^2$ $\Rightarrow2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=\text{a}^2+\text{b}^2$ $\therefore2\cos(\alpha-\beta)=\text{a}^2+\text{b}^2-2$ Thus, $\cos(\alpha+\beta)=\frac{\text{a}^2+\text{b}^2-2}{2}$
View full question & answer→Question 183 Marks
$\tan\text{x}+\tan\big(\frac{\pi}{3}+\text{x}\big)-\tan\big(\frac{\pi}{3}-\text{x}\big)=3\tan3\text{x}$
Answer$\frac{\pi}{3}=60^\circ$ $\text{LHS}=\tan\text{x}+\tan(60^\circ+\text{x})=\tan(60^\circ-\text{x})$ $=\tan\text{x}+\big(\frac{\tan60^\circ+\tan\text{x}}{1-\tan60^\circ\tan\text{x}}\big)-\big(\frac{\tan60^\circ-\tan\text{x}}{1+\tan60^\circ\tan\text{x}}\big)$ $\Big[\tan(\text{x}+\text{y})=\frac{\tan60^\circ-\tan\text{x}}{1+\tan60^\circ\tan\text{x}}$ and $\tan(\text{(x}-\text{y})=\frac{\tan\text{x}\tan\text{y}}{1+\tan\text{x}\tan\text{y}}\Big]$ $=\tan\text{x}+\frac{\sqrt{3}+\tan\text{x}}{1-\sqrt{3}\tan\text{x}}-\frac{\sqrt{3}-\tan\text{x}}{1+\sqrt{3\tan\text{x}}}$ $=\tan\text{x}+\frac{\sqrt{3}+3\tan\text{x}+\tan\text{x}+\sqrt{3}\tan^3\text{x}+\sqrt{3}+3\tan\text{x}+\tan\text{x}-\sqrt{3}\tan^2\text{x}}{\big(1-\sqrt{3}\tan\text{x}\big)\big(1+\sqrt{3}\tan\text{x}\big)}$ $=\tan\text{x}+\frac{8\tan\text{x}}{1-3\tan^2\text{x}}$ $=\frac{\tan\text{x}-3\tan^3\text{x}+8\tan\text{x}}{1-3\tan^2\text{x}}$ $=\frac{9\tan\text{x}-3\tan^3\text{x}}{1-3\tan^2\text{x}}$ $=3\Big(\frac{3\tan\text{x}-\tan^3\text{x}}{1-3\tan^2\text{x}}\Big)\Big(\because\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\Big)$ $=3\tan3\text{x}=\text{RHS}$ Hence proved.
View full question & answer→Question 193 Marks
Prove that: $3\text{x}+\sin2\text{x}-\sin\text{x}=4\sin\text{x}\cos\frac{\text{x}}{2}\cos\frac{3\text{x}}{2}$
Answer$\sin3\text{x}+\sin2\text{x}-\sin\text{x}$ $=(\sin3\text{x}-\sin\text{x})+\sin2\text{x}$ $=2\cos\Big(\frac{3\text{x}+\text{x}}{2}\Big)\sin\Big(\frac{3\text{x}-\text{x}}{\text{2}}\Big)+2\sin\text{x}\cos\text{x}$ $\begin{bmatrix}\because\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\\\sin2\text{x}=2\sin\text{x}\cos\text{x}\end{bmatrix}$ $=2\cos(2\text{x})\sin\text{x}+2\sin\text{x}\cos\text{x}$ $=2\sin\text{x}[\cos(2\text{x})+\cos\text{x}]$ $=2\sin\text{x}\Big[2\cos\Big(\frac{2\text{x}+\text{x}}{2}\Big)+\cos\Big(\frac{2\text{x}-\text{x}}{2}\Big)\Big]$ $\Big[\because\cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$ $=4\sin\text{x}\cos\frac{3\text{x}}{2}\cos\frac{\text{x}}{2}$
View full question & answer→Question 203 Marks
If $\cos\alpha+\cos\beta=\frac{1}{3}$ and $\sin\alpha+\sin\beta=\frac{1}{4},$ prove that $\cos\frac{\alpha-\beta}{2}=\pm\frac{5}{24}$
AnswerWe have, $\cos\alpha+\cos\beta=\frac{1}{3}$ and $\sin\alpha+\sin\beta=\frac{1}{4}$ Squaring and adding, we get $(\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta)+(\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta)=\frac{1}{9}+\frac{1}{16}$ $\Rightarrow1+1+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=\frac{25}{144}$ $\Rightarrow2\cos(\alpha-\beta)=\frac{25}{144}-2=\frac{-263}{144}$ $\Rightarrow\cos(\alpha-\beta)=\frac{-263}{288}$ Now, $\cos\Big(\frac{\alpha-\beta}{2}\Big)=\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}$ $=\sqrt{\frac{1-\frac{263}{288}}{2}}=\sqrt{\frac{25}{576}}$ $=\pm\frac{5}{24}$ $\therefore\cos\Big(\frac{\alpha-\beta}{2}\Big)=\pm\frac{5}{25}$
View full question & answer→Question 213 Marks
If $\cos\text{x}=-\frac{3}{5}$ and x lies in IInd quadrant, find the values of sin $2x$ and $\sin\frac{\text{x}}{2}$
Answer$\because$ x lies in $II^{nd}$ quadrant. $\Rightarrow\frac{\pi}{2}<\times<\pi$
$\Rightarrow\pi<2\times<2\pi\Rightarrow2\text{x}\ \text{lies in I}^\text{st}\text{quad.}$
Also, $\cos\text{x}=\frac{-3}{5}=\frac{\text{b}}{\text{h}}\Rightarrow\text{b}=3\text{h}=5$
$\Rightarrow\text{p}=4$ So, $\sin\text{x}=\frac{\text{p}}{\text{h}}=\frac{4}{5}$
$\therefore\sin2\text{x}=2\sin\text{x}\cos\text{x}$
$=2.\frac{4}{5}.\Big(\frac{-3}{5}\Big)=\frac{-24}{25}$
$\sin\frac{\text{x}}{2}=\frac{\sin\text{x}}{2\cos\frac{\text{x}}{2}}$ or $\sqrt{\frac{1-\cos\text{x}}{2}}$
$=\sqrt{\frac{1-\big(1-\frac{3}{5}\big)}{2}}$
$=\frac{2}{\sqrt{5}}$
View full question & answer→Question 223 Marks
If $\tan\text{A}=\frac{1}{7}$ and $\tan\text{B}=\frac{1}{3},$ show that $\cos2\text{A}=\sin4\text{B}$
AnswerWe have, $\tan\text{A}=\frac{1}{7}\&\tan\text{B}=\frac{1}{3}$ $\therefore\cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}=\frac{1-\Big(\frac{1}{7}\Big)^2}{1+\Big(\frac{1}{7}\Big)^2}=\frac{48}{\frac{49}{\frac{50}{49}}}$ $=\frac{48}{50}=\frac{24}{25}\ ...{\text{(A)}}$ Also, $\sin4\text{B}=\sin2.2.\text{B}$ $=2\sin2\text{B}.\cos2\text{B}$ $=2.\Big(\frac{2\tan\text{B}}{1+\tan^2\text{B}}\Big).\Big(\frac{1-\tan^2\text{B}}{1+\tan^2\text{B}}\Big)$ $=4.\Bigg(\frac{\frac{1}{3}}{1+\frac{1}{9}}\Bigg).\Bigg(\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\Bigg)$ $=\frac{4.\frac{1}{3}.\frac{8}{9}}{\frac{10}{9}\times\frac{10}{9}}$ $=\frac{32\times3}{100}$ $=\frac{8\times3}{25}=\frac{24}{25}\ ...(\text{B})$ From(A) & (B) $\cos2\text{A}=\sin4\text{B}$
View full question & answer→Question 233 Marks
Prove that: $\cos4\text{x}-\cos4\alpha=8(\cos\text{x}-\cos\alpha)(\cos\text{x}+\cos\alpha)\$\cos\text{x}+\sin\alpha)(\cos\text{x}-\sin\alpha)$
Answer$\cos4\text{x}-\cos4\alpha=2\cos^22\theta-2\cos^22\alpha$ $=2(\cos2\text{x}+\cos2\alpha)(\cos2\text{x}-\cos2\alpha)$ $=2(2\cos^2\text{x}-1+1-2\sin^2\alpha)(2\cos^2\text{x}-1-2\cos^2\alpha+1)$ $=8(\cos^2\text{x}-\sin^2\alpha)(\cos^2\text{x}-\cos^2\alpha)$ $=8(\cos\text{x}-\sin\alpha)(\cos\text{x}+\sin\alpha)(\cos\text{x}-\cos\alpha)$
View full question & answer→Question 243 Marks
Prove that: $\sin^242^\circ-\cos^278^\circ=\frac{\sqrt{15}+1}{8}$
Answer$\text{LHS}=\sin^242^\circ-\cos^278^\circ$ $=\sin^2(90-48)-\cos^2(90-12)$ $=\cos^248^\circ-\sin^212^\circ$ $=\cos(48+12).\cos(48-12)$$\big[\because\cos(\text{A+B}).\cos(\text{A}-\text{B}=\cos^2\text{A}-\sin^2\text{B})\big]$ $=\cos60^\circ.\cos36^\circ$ $=\frac{1}{2}.\frac{\sqrt{5}+1}{4}$ $\Big[\because\cos36^\circ=\frac{\sqrt{5}+1}{4}\Big]$ $=\frac{\sqrt{5}+1}{8}$ $=\text{RHS}$
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prove that: $\cos^6\text{x}=\cos2\text{x}=\cos2\text{x}(1-\frac{1}{4}\sin^22\text{x})$
Answer$\text{LHS}=\cos^6\text{x}-\sin^6\text{x}$ $=(\cos^2\text{x})^3-(\sin^2\text{x})^3$ $=(\cos^2\text{x}-\sin^2\text{x})(\cos^4\text{x}+\sin^2\text{x}.\cos^2\text{x}+\sin^4\text{x})$ $\big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\big]$ $=\cos2\text{x}(\cos^4\text{x}+2\sin^2+\sin^4\text{x}-\sin^2\text{x}\cos^2\text{x})$ $\therefore\cos^2\text{x}-\sin^2\text{x}=\cos^2\text{x}=\&$ adding and subtracling $\sin^2\text{x}\cos^2\text{x}$ $=\cos2\text{x}\big[(\sin^2\text{x}+\cos^2\text{x})^2-\frac{4}{4}\sin^2\text{x}\cos^2\text{x}$ $=\cos2\text{x}\big[1-\frac{1}{4}(2\sin\text{A}\cos\text{x})^2\big]$ $=\cos2\text{x}\big[1-\frac{1}{4}\sin^22\text{x}\big]=\text{RHS}$
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Prove that: $\cos^3\text{x}\sin3\text{x}+\sin^3\text{x}\cos3\text{x}=\frac{3}{4}\sin4\text{x}$
Answer$\cos^3\text{x}\sin3\text{x}+\sin^3\text{x}\cos3\text{x}=\frac{3}{4}\sin4\text{x}$ $\text{LHS}=\cos^3\text{x}\sin3\text{x}+\sin^3\text{x}\cos3\text{x}$ $=\Big(\frac{\cos3\text{x}+3\cos\text{x}}{4}\Big)\sin3\text{x}+\Big(\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big)\cos3\text{x}$ $\begin{Bmatrix}\because\sin3\text{x}=3\sin\text{x}-4\sin^3\text{x}\\\cos3\text{x}=4\cos^3\text{x}-3\cos\text{x}\end{Bmatrix}$ $=\frac{1}{4}[3(\sin3\text{x}\cos\text{x}+\sin\text{x}\cos3\text{x})\\+\cos3\text{x}\sin3\text{x}-\sin3\text{x}\cos3\text{x}]$ $=\frac{1}{4}[3\sin(3\text{x}+\text{x})+0]$ $\frac{3}{4}\sin4\text{x}$ So, $\cos^3\text{x}\sin3\text{x}+\sin^3\text{x}\cos3\text{x}=\frac{3}{4}\sin4\text{x}$ $=\text{RHS}$
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Show thet: $(\sin^6\text{x}+\cos^6\text{x})-3(\sin^4\text{x}+\cos^4\text{x})+1=0$
Answer$\text{LHS}=2(\sin^6\text{x}+\cos^6\text{x})-3(\sin^4\text{x}+\cos^4\text{x})+1$ $=2\big((\sin^2)^3+(\cos^2\text{x})^3\big)-3(\sin^4\text{x}+\cos^4\text{x})+1$ $\big[\because\text{a}^3+\text{b}^3=(\text{a+b})\big(\text{a}^2\text{ab+b}^2\big)\big]$ $=2[(\sin^2\text{x}+\cos^4\text{x})(\sin^4\text{x}-\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x})]-3(\sin^4\text{x}+\cos^4\text{x})+1$ $=-[\sin^4\text{x}+\cos^4\text{x}+2\sin^2\text{x}\cos^2\text{x}]+1$ $=-[\sin^2\text{x}+\cos^2\text{x}]+1$ $=-1+1$ $=0=\text{RHS}$
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