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Trigonometric Ratios of Multiple and Submultiple Angles · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 16 questions.

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Question 11 Mark
If $\tan(\text{A+B})=\text{p}$ and $\tan\text{(A - B)}=\text{q},$ then write the value of $\tan2\text{B}.$
Answer
We have, $\tan(\text{A+B})=\text{p}$ and $\tan(\text{A - B})=\text{q}$ Now, $\tan2\text{B}=\tan\big[(\text{A+B})-(\text{A}-\text{B})\big]$ $=\frac{\tan(\text{A+B})-\tan\text{(A - B)}}{1+\tan(\text{A+B})\times\tan(\text{A - B})}$ $=\frac{\text{p - q}}{1+\text{pq}}$ $\therefore\ \tan2\text{B}=\frac{\text{p - q}}{1+\text{pq}}$
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Question 21 Mark
Write the maximum and minimum values of $3\cos\text{x}+4\sin\text{x}+5.$
Answer
Let $\text{f(x)}=3\cos\text{x}+4\sin\text{x}+5$ We know that, $-\sqrt{3^2+4^2}\le3\cos\text{x}+4\sin\text{x}\le\sqrt{3^2+4^2}$ $\Rightarrow-\sqrt{9+16}\le3\cos\text{x}+4\sin\text{x}\le\sqrt{9+16}$ $\Rightarrow-\sqrt{25}\le3\cos\text{x}+4\sin\text{x}\le\sqrt{25}$ $\Rightarrow-5\le3\cos\text{x}+4\sin\text{x}\le5$ $\Rightarrow-5+5\le3\cos\text{x}+4\sin\text{x}+5\le5+5$ $\Rightarrow0\le3\cos\text{x}+4\sin\text{x}+5\le10$ Hence, minimum and maximum value of $3\cos\text{x}+4\sin\text{x}+5$ are 0 and 10 respectively.
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Question 31 Mark
Evaluate the following: $\cos47^\circ\cos13^\circ-\sin47^\circ\sin13^\circ$
Answer
$=\cos(47^\circ+13^\circ)$ $[\cos(\text{A+B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}]$ $=\cos60^\circ$ $=\frac{1}{2}$
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Question 41 Mark
Evaluate the following: $\cos80^\circ\cos20^\circ+\sin80^\circ\sin20^\circ$
Answer
$=\cos(80^\circ-20^\circ)$$[\cos(\text{A-B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}]$ $=\cos60^\circ$ $=\frac{1}{2}$
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Question 51 Mark
If $\text{x}\cos\theta=\text{y}\cos\Big(\theta+\frac{2\pi}{3}\Big)=\text{z}\Big(\theta+\frac{4\pi}{3}\Big),$ then write the value of $\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}.$
Answer
Let, $\text{x}\cos\theta=\text{y}\cos\Big(\theta+\frac{2\pi}{3}\Big)=\text{z}\cos\Big(\theta+\frac{4\pi}{3}\Big)=\text{k}$ $\Rightarrow\text{x}\cos\theta=\text{k},\text{y}\cos\Big(\theta+\frac{2\pi}{3}\Big)=\text{k}$ and, $\text{z}\cos\Big(\theta+\frac{4\pi}{3}\Big)=\text{k}$ $\Rightarrow\frac{1}{\text{x}}=\frac{\cos\theta}{\text{k}},\frac{1}{\text{y}}=\frac{\cos\Big(\frac{2\pi}{3}+\theta\Big)}{\text{k}}$ and, $\frac{1}{\text{z}}=\frac{\cos\Big(\frac{4\pi}{3}+\theta\Big)}{\text{k}}$ Now, $\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=\frac{\cos\theta}{\text{k}}+\frac{\cos\big(\frac{2\pi}{3}+\theta\Big)}{\text{k}}+\frac{\cos\Big(\frac{4\pi}{3}+\theta\Big)}{\text{k}}$ $=\frac{1}{\text{k}}\Big[\cos\theta+\cos\Big\{\frac{\pi}{2}+\Big(\frac{\pi}{6}+\theta\Big)\Big\}\cos\Big\{\pi+\Big(\frac{\pi}{3}+\theta\Big)\Big\}\Big]$ $=\frac{1}{\text{k}}\Big[\cos\theta-\sin\Big(\frac{\pi}{6}+\theta\Big)-\cos\Big(\frac{\pi}{3}+\theta\Big)\Big]$ $=\frac{1}{\text{k}}\Big[\cos\theta-\Big(\sin\frac{\pi}{6}\cos\theta+\cos\frac{\pi}{6}\sin\theta\Big)-(\cos\frac\pi3\times\cos\theta-\sin\frac{\pi}{3}\times\sin\theta\Big)\Big]$ $=\frac{1}{\text{k}}\Big[\cos\theta-\Big(\frac12\cos\theta+\frac{\sqrt{3}}{2}\sin\theta\Big)-\Big(\frac12\cos\theta-\frac{\sqrt{3}}{2}\sin\theta\Big)\Big]$ $=\frac{1}{\text{k}}\Big[\cos\theta-\frac12\cos\theta-\frac{\sqrt{3}}{2}\sin\theta-\frac12\cos\theta+\frac{\sqrt{3}}{2}\sin\theta\Big]$ $=0$
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Question 61 Mark
Evaluate the following: $\sin36^\circ\cos9^\circ+\cos36^\circ\sin9^\circ$
Answer
$=\sin(36^\circ+9^\circ)$ $[\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}]$ $=\sin45^\circ$ $=\frac{1}{\sqrt{2}}$
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Question 71 Mark
If $12\sin\text{x}-9\sin^2\text{x}$ attains its maximum value at $\text{x}=\alpha,$ then write the value of $\sin\alpha.$
Answer
$12\sin\text{x}-9\sin^2\text{x}$ has maximum value as 4, so $12\sin\alpha-9\sin^2\alpha=4$ $9\sin^2\alpha-612\sin\alpha+4=0$ $9\sin^2\alpha-6\sin\alpha-6\sin\alpha+4=0$ $3\sin\alpha(3\sin\alpha-2)-2(3\sin\alpha-2)=0$ $(3\sin\alpha-2)(3\sin\alpha-2)=0$$3\sin\alpha-2=0$
$\sin\alpha=\frac23$
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Question 81 Mark
If $\sin\alpha-\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b},$ then write the value of $\cos(\alpha+\beta).$
Answer
We have, $\sin\alpha-\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b}$ Now, $\text{a}^2+\text{b}^2=(\sin\alpha-\sin\beta)^2+(\cos\alpha+\cos\beta)^2$ $=\sin^2\alpha+\sin^2\beta-2\sin\alpha\sin\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta$ $=(\sin^2\alpha+\cos^2\alpha)+(\sin^2\beta+\cos^2\beta)+2[\cos\alpha\cos\beta-\sin\alpha\sin\beta]$ $=1+1+2\cos(\alpha+\beta)$ $=2+2\cos(\alpha+\beta)$ $\Rightarrow\text{a}^2+\text{b}^2=2+2\cos(\alpha+\beta)$ $\Rightarrow\text{a}^2+\text{b}^2-2=2\cos(\alpha+\beta)$ $\Rightarrow2\cos(\alpha+\beta)=\text{a}^2+\text{b}^2-2$ $\Rightarrow\cos(\alpha+\beta)=\frac{\text{a}^2+\text{b}^2-2}{2}$
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Question 91 Mark
If $\alpha+\beta-\gamma=\pi,$ and $\sin^2\alpha+\sin^2\beta-\sin^2\gamma=\lambda\sin\alpha\sin\beta\cos\gamma,$ then write the value of $\lambda.$
Answer
We have: $\alpha+\beta-\gamma=\pi$ Now, $\sin^2\alpha+\sin^2\beta-\sin^2\gamma$ $=\frac{1-\cos2\alpha}{2}+\frac{1\cos2\beta}{2}-\frac{(1-\cos2\gamma)}{2}$ $\Big(\frac12+\frac12+\frac12\Big)-\Big(\frac{\cos2\alpha}{2}+\frac{\cos2\beta}{2}-\frac{\cos2\gamma}{2}\Big)$ $=\frac12-\frac{1}{2}\Big[2\cos\frac{2\alpha+2\beta}{2}\cos\frac{2\alpha+2\beta}{2}-\cos2\gamma\Big]$ $=\frac12-\frac12[2\cos(\alpha+\beta)\cos(\alpha-\beta)-\cos2\gamma]$ $=\frac12-\frac12[2\cos(\pi+\gamma)\cos(\alpha-\beta)-\cos2\gamma]$ $=\frac12-\frac12[2\cos\gamma\cos(\alpha-\beta)-\cos2\gamma]$ $=\frac12-\frac12\big[2\cos\gamma\cos(\alpha-\beta)-\big(2\cos^2\gamma-1\big]$ $=\frac12+\frac12\big[2\cos\gamma\cos(\alpha-\beta)+2\cos^2\gamma-1\big]$ $=\frac12-\frac12\big[2\{\cos(\alpha+\beta)\cos(\alpha-\beta)-\cos2\gamma\}\big]$ $=\cos\gamma\cos(\alpha-\beta)+2\cos^2\gamma$ $=\cos\gamma[\cos(\alpha-\beta)+\cos\gamma]$ $=\cos\Big[\cos(\alpha-\beta)+\cos\{-(\pi-(\alpha+\beta))\}\Big]$ $=\cos\gamma[\cos(\alpha-\beta)-\cos(\alpha+\beta)]$ $=\cos\gamma[2\sin\alpha\sin\beta]$ $=2\cos\gamma\sin\alpha\sin\beta$ $=2\sin\alpha\sin\beta\sin\gamma$ $\therefore\sin^2\alpha+\sin^2\beta-\sin^2\gamma=2\sin\alpha\sin\beta\cos\gamma$ $\therefore\lambda=2$
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Question 101 Mark
If $\text{a}=\text{b}\cos\frac{2\pi}{3}=\text{c}\cos\frac{4\pi}{3},$ then write the value of ab + bc + ca.
Answer
Let, $\text{a}=\text{b}\cos\frac{2\pi}{3}=\text{c}\cos\frac{4\pi}{3}=\text{k}$ $\Rightarrow\text{a}=\text{k},\text{b}=\frac{\text{k}}{\cos\frac{2\pi}{3}}$ and $\text{c}=\frac{\text{k}}{\cos\frac{4\pi}{3}}$ Now, ab + bc + ca $=\text{k}\times\frac{\text{k}}{\cos\frac{2\pi}{3}}+\frac{\text{k}}{\cos\frac{2\pi}{3}}\times\frac{\text{k}}{\cos\frac{4\pi}{3}}+\frac{\text{k}}{\cos\frac{4\pi}{3}}\times\text{k}$ $=\text{k}^2\sec\frac{2\pi}{3}+\text{k}^2\sec\frac{2\pi}{3}\sec\frac{4\pi}{3}+\text{k}^2\sec\frac{4\pi}{3}$ $=\text{k}^2\sec\Big(\frac{\pi}{2}+\frac\pi6\Big)+\text{k}^2\sec\Big(\frac{\pi}{2}+\frac\pi6\Big)\sec\Big(\pi+\frac{\pi}{3}\Big)+\text{k}^2\sec\Big(\pi+\frac{\pi}{3}\Big)$ $=-\text{k}^2\text{cosec}\frac{\pi}{6}+\text{k}^2\text{cosec}\frac{\pi}{6}\sec\frac{\pi}{3}-\text{k}^2\sec\frac{\pi}{3}$ $=-\text{k}^2\times2+\text{k}^2\times2\times2-\text{k}^2\times2$ $=-4\text{k}^2+4\text{k}^2=0$ $\therefore\text{ ab + bc + ca}=0$
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Question 111 Mark
Evaluate the following: $\sin78^\circ\cos18^\circ-\cos78^\circ\sin18^\circ$
Answer
$=\sin(78^\circ-18^\circ)$ $[\sin(\text{A-B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}]$ $=\sin60^\circ$ $=\frac{\sqrt{3}}{2}$
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Question 121 Mark
If $\tan\alpha=\frac{1}{1+2^{-\text{x}}}$ and $\tan\beta=\frac{1}{1+2^{\text{x+1}}},$ then write the value of $\alpha+\beta$ lying in the interval $\Big(0,\frac\pi2\Big).$
Answer
We have, $\tan\alpha=\frac{1}{1+2^{-\text{x}}}$ and $\tan\beta=\frac{1}{1+2^{\text{x+1}}}$ Now, $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\times\tan\beta}$ $=\frac{\frac{1}{1+2^{-\text{x}}}+\frac{1}{1+2^{\text{x+1}}}}{1-\frac{1}{1+2^{-\text{x}}}\times\frac{1}{1+2^{\text{x+1}}}}$ $=\frac{\frac{1+2^{\text{x+1}}+1+2^{-\text{x}}}{(1+2^{-\text{x}})(1+2^{\text{x+1}})}}{1-\frac{1}{(1+2^{-\text{x}})(1+2^{\text{x+1}})}}$ $=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{1+2^{\text{x}+1}+2^{-\text{x}}+2}}{1-\frac{1}{1+2^{\text{x}+1}+2^{-\text{x}}+2}}$ $=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{3+2^{\text{x}+1}+2^{-\text{x}}}}{1-\frac{1}{3+2^{\text{x}+1}+2^{-\text{x}}}}$ $=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{3+2^{\text{x}+1}+2^{-\text{x}}}}{\frac{3+2^{\text{x}+1}+2^{-\text{x}}-1}{3+2^{\text{x}+1}+2^{-\text{x}}}}$ $=\frac{2+2^{\text{x+1}}+2^{-\text{x}}}{2+2^{\text{x+1}}+2^{-\text{x}}}$ $=1$ $\Rightarrow\tan(\alpha+\beta)=1=\tan\Big(\frac{\pi}{4}\Big)$ $\Rightarrow\tan(\alpha+\beta)=\tan\Big(\frac\pi4\Big)$ $\Rightarrow\alpha+\beta=\frac\pi4$
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Question 131 Mark
Write the maximum values of $12\sin\text{x}-9\sin^2\text{x}.$
Answer
$12\sin\text{x}-9\sin^2\text{x}=3\sin\text{x}(4-3\sin\text{x})$ We know that $0\le\sin\text{x}\le1$ $\therefore$ Maximum value of $12\sin\text{x}-9\sin^2\text{x}$ $=12\Big(\frac23\Big)-9\Big(23\Big)^2$ $=\frac{24}{3}-\frac{36}{9}$ $=\frac{36}{9}$ $=4$ $\because$ for other than $\sin\text{x}=\frac{4}{3}$ it is less than 4 Maximum value of $12\sin\text{x}-9\sin^2\text{x}$ $=4$
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Question 141 Mark
If A + B = C, then write the value of $\tan\text{A}\tan\text{B}\tan\text{C}.$
Answer
We have, A + B = C $\Rightarrow\tan(\text{A+B})=\tan\text{C}$ $\Rightarrow\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}=\tan\text{C}$ $\Rightarrow\tan\text{A}+\tan\text{B}=\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}$ $\Rightarrow\tan\text{A}\tan\text{B}\tan\text{C}=\tan\text{C}-\tan\text{A}-\tan\text{B}$
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Question 151 Mark
Write the interval in which the value of $5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3$ lie.
Answer
If $\text{f(x)}=5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3,$ then $\text{f(x)}=5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3$ $=5\cos+3\Big(\cos\theta\cos\frac\pi3-\sin\text{x}\sin\frac\pi3\Big)+3$ $=5\cos\text{x}+\frac32\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3$ $=\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3\cdots(\text{i})$ Now, $-\sqrt{\Big(\frac{13}{2}\Big)^2+\Big(\frac{3\sqrt{3}}{2}\Big)^3}\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}\le\sqrt{\Big(\frac{13}{2}\Big)^2+\Big(\frac{3\sqrt{3}}{2}\Big)^2}$ $\Rightarrow-7\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}\le7,$ for all x $\Rightarrow-7+3\le\frac{13}{2}\sin\text{x}+3\le7+3$ $\Rightarrow-4\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3\le10$ $\Rightarrow-4\le5\cos\text{x}-3\cos\Big(\text{x}+\frac\pi3\Big)+3\le10$ [Using 1] Hence, the required in terval is [-4, 10]
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Question 161 Mark
If $\frac{\cos(\text{x - y})}{\cos(\text{x+y})}=\frac{\text{m}}{\text{n}},$ then write the value of $\tan\text{x}\tan\text{y}.$
Answer
We have, $\frac{\cos(\text{x - y})}{\cos(\text{x+y})}=\frac{\text{m}}{\text{n}}$ $\Rightarrow\frac{\cos\text{x}\cos\text{y}+\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}-\sin\text{x}\sin\text{y}}=\frac{\text{m}}{\text{n}}$ $\Rightarrow\frac{\frac{\cos\text{x}\cos\text{y}+}{\cos\text{x}\cos\text{y}}+\frac{\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}}}{\frac{\cos\text{x}\cos\text{y}}{\cos\text{x}\cos\text{y}}-\frac{\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}}}=\frac{\text{m}}{\text{n}}$ $[$Dividing and multiplying numerator and denominator by $\cos\text{x}\cos\text{y}]$ $\Rightarrow\frac{1+\tan\text{x}\tan\text{y}}{1-\tan\text{x}\tan\text{y}}=\frac{\text{m}}{\text{n}}$ $\Rightarrow\text{n+n}\tan\text{x}\tan\text{y}=\text{m - m}\tan\text{x}\tan\text{y}$ $\Rightarrow\tan\text{x}\tan\text{y}(\text{n+m})=\text{m - n}$ $\Rightarrow\tan\text{x}\tan\text{y}=\frac{\text{m - n}}{\text{m+n}}$
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