Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\sin x=\frac{1}{4}, x$ in quadrant II
[question] Question. Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\sin x=\frac{1}{4}, x$ in quadrant II [/question] [solution] solution: Here, x is in quadrant II. $\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$ Therefore, $\sin \frac{x}{2}, \cos \frac{x}{2}$, and $\tan \frac{x}{2}$ are all positive. It is given that $\sin x=\frac{1}{4}$. $\cos ^{2} x=1-\sin ^{2} x=1-\left(\frac{1}{4}\right)^{2}=1-\frac{1}{16}=\frac{15}{16}$ $\Rightarrow \cos x=-\frac{\sqrt{15}}{4}...
Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\cos x=-\frac{1}{3}, x$ in quadrant III
[question] Question. Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\cos x=-\frac{1}{3}, x$ in quadrant III [/question] [solution] solution: Here, $x$ is in quadrant III. $\Rightarrow \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}$ Therefore, $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are negative, whereas $\sin \frac{x}{2}$ is positive. It is given that $\cos \mathrm{x}=-\frac{1}{3}$. $\cos x=1-2 \sin ^{2} \frac{x}{2}$ $\Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}$ $\Ri...
$\tan x=-\frac{4}{3}, x$ in quadrant II
[question] Question. $\tan x=-\frac{4}{3}, x$ in quadrant II [/question] [solution] solution: Here, x is in quadrant II. $\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$ Therefore, $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are all positive. It is given that $\tan x=-\frac{4}{3}$. $\sec ^{2} x=1+\tan ^{2} x=1+\left(\frac{-4}{3}\right)^{2}=1+\frac{16}{9}=\frac{25}{9}$ $\therefore \cos ^{2} x=\frac{9}{25}$ $\Rightarrow \cos x=\pm \frac{3}{5}$ As $x$ is in quadrant II, $\cos x...
Prove that: $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$
[question] Question. Prove that: $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$ [/question] [solution] solution: L.H.S. $=\sin 3 x+\sin 2 x-\sin x$ $=\sin 3 x+(\sin 2 x-\sin x)$ $=\sin 3 x+\left[2 \cos \left(\frac{2 x+x}{2}\right) \sin \left(\frac{2 x-x}{2}\right)\right] \quad\left[\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$ $=\sin 3 x+\left[2 \cos \left(\frac{3 x}{2}\right) \sin \left(\frac{x}{2}\right)\right]$ $=\sin 3 x+2 \c...
Prove that: $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$
[question] Question. Prove that: $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$ [/question] [solution] solution: It is known that $\sin \mathrm{A}+\sin \mathrm{B}=2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right), \cos \mathrm{A}+\cos \mathrm{B}=2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$ L.H.S. $=\frac{(\sin 7 \mathrm{x}+\...
Prove that: $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$
[question] Question. Prove that: $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$ [/question] [solution] solution: L.H.S. $=(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}$ $=\cos ^{2} x+\cos ^{2} y-2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y$ $=\left(\cos ^{2} x+\sin ^{2} x\right)+\left(\cos ^{2} y+\sin ^{2} y\right)-2[\cos x \cos y+\sin x \sin y]$ $=1+1-2[\cos (x-y)] \quad[\cos (A-B)=\cos A \cos B+\sin A \sin B]$ $=2[1-\cos (x-y)]$ $=2\left[1-\left\{1-2 \sin ^{2}\left(\f...
Prove that: $(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2} \frac{x+y}{2}$
[question] Question. Prove that: $(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2} \frac{x+y}{2}$ [/question] [solution] solution: L.H.S. $=(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}$ $=\cos ^{2} x+\cos ^{2} y+2 \cos x \cos y+\sin ^{2} x+\sin ^{2} y-2 \sin x \sin y$ $=\left(\cos ^{2} x+\sin ^{2} x\right)+\left(\cos ^{2} y+\sin ^{2} y\right)+2(\cos x \cos y-\sin x \sin y)$ $=1+1+2 \cos (x+y) \quad[\cos (A+B)=(\cos A \cos B-\sin A \sin B)]$ $=2+2 \cos (x+y)$ $=2[1+\cos (x+y)]$ $=2\left[1+2 \cos ^{...
Prove that: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
[question] Question. Prove that: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$ [/question] [solution] solution: L.H.S. $=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$ $=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+2 \cos \left(\frac{\frac{3 \pi}{13}+\frac{5 \pi}{13}}{2}\right) \cos \left(\frac{\frac{3 \pi}{13}-\frac{5 \pi}{13}}{2}\right)\left[\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-...
Prove that $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
[question] Question. Prove that $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$ [/question] [solution] solution: It is known that $\tan 2 \mathrm{~A}=\frac{2 \tan \mathrm{A}}{1-\tan ^{2} \mathrm{~A}}$. $\therefore$ L.H.S. $=\tan 4 x=\tan 2(2 x)$ $=\frac{2 \tan 2 x}{1-\tan ^{2}(2 x)}$ $=\frac{2\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)}{1-\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)^{2}}$ $=\frac{\left(\frac{4 \tan x}{1-\tan ^{2} x}\right)}{\left[1-\frac{4 \tan ^...
Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$
[question] Question. Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$ [/question] [solution] solution: L.H.S. $=\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}$ $=\frac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x}$ $=\frac{2 \cos \left(\frac{4 \mathrm{x}+2 \mathrm{x}}{2}\right) \cos \left(\frac{4 \mathrm{x}-2 \mathrm{x}}{2}\right)+\cos 3 \mathrm{x}}{2 \sin \left(\frac{4 \mathrm{x}+2 \mathrm{x}}{2}\right) \cos \left(\frac{4 \mathrm{x...
Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
[question] Question. Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$ [/question] [solution] solution: It is known that $\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos ^{2} A-\sin ^{2} A=\cos 2 A$ $\therefore$ L.H.S. $=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$ $=\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}$ $=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}$ $=-2 \times(-\sin x)$ $=2 \sin x=$ R.H....
Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
[question] Question. Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$ [/question] [solution] solution: It is known that $\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $\therefore$ L.H.S. $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$ $=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}...
Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$
[question] Question. Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$ [/question] [solution] solution: It is known that $\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $\therefore$ L.H.S. $=\frac{\sin x-\sin y}{\cos x+\cos y}$ $=\frac{2 \cos \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x...
Prove that $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
[question] Question. Prove that $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$ [/question] [solution] solution: It is known that $\sin \mathrm{A}+\sin \mathrm{B}=2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right), \cos \mathrm{A}+\cos \mathrm{B}=2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$ $\therefore$ L.H.S. $=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$ $=\frac{2 \sin \left(\fra...
Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$
[question] Question. Prove that $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$ [/question] [solution] solution: It is known that $\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$ $\therefore$ L.H.S $=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}$ $=\frac{-2 \sin \left(\frac{9 x+5 x}{2}\right) \cdot \sin \left(\frac{9 x-5 x}{2}\right)}{2 \cos \left(\frac{17 ...
Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$
[question] Question. Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$ [/question] [solution] solution: It is known that $\cos \mathrm{A}-\cos \mathrm{B}=-2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$. $\therefore$ L.H.S. $=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$ $=-2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}...
$\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
[question] Question. $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$ [/question] [solution] solution: L.H.S. $=\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$ $=\sin x \cos x[\tan x+\cot x]$ $=\sin x \cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$ $=(\sin x \cos x)\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right]$ $=1=\mathrm{R} . \math...
Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$
[question] Question. Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$ [/question] [solution] solution: L.H.S. $=\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}$ $=\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}$ $=\frac{-\cos ^{2} x}{-\sin ^{2} x}$ $=\cot ^{2} x$ $=$ R.HS. [/solution]...
Prove that: $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
[question] Question. Prove that: $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$ [/question] [solution] solution: $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$ $=\frac{1}{2}\left[2 \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\right]+\frac{1}{2}\left[-2 \sin \left(\frac{\pi}{4}-x\righ...
Prove that $2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}=10$
[question] Question. Prove that $2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}=10$ [/question] [solution] solution: L.H.S $=2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}$ $=2\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}^{2}+2\left(\frac{1}{\sqrt{2}}\right)^{2}+2(2)^{2}$ $=2\left\{\sin \frac{\pi}{4}\right\}^{2}+2 \times \frac{1}{2}+8$ $=2\left(\frac{1}{\sqrt{2}}\right)^{2}+1+8$ $=1+1+8$ $=10$ $=$ R.H.S [/solution]...
Prove that $\cot ^{2} \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}=6$
[question] Question. Prove that $\cot ^{2} \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}=6$ [/question] [solution] solution: L.H.S. $=\cot ^{2} \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}$ $=(\sqrt{3})^{2}+\operatorname{cosec}\left(\pi-\frac{\pi}{6}\right)+3\left(\frac{1}{\sqrt{3}}\right)^{2}$ $=3+\operatorname{cosec} \frac{\pi}{6}+3 \times \frac{1}{3}$ $=3+2+1=6$ $=$ R.H.S [/solution]...
Prove that $2 \sin ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$
[question] Question. Prove that $2 \sin ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$ [/question] [solution] solution: L.H.S. $=2 \sin ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}$ $=2\left(\frac{1}{2}\right)^{2}+\operatorname{cosec}^{2}\left(\pi+\frac{\pi}{6}\right)\left(\frac{1}{2}\right)^{2}$ $=2 \times \frac{1}{4}+\left(-\operatorname{cosec} \frac{\pi}{6}\right)^{2}\left(\frac{1}{4}\right)$ $=\frac{1}{2...
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
[question] Question. $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$ [/question] [solution] solution: L.H.S. $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$ $=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$ $=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$ $=$ R.H.S. [/solution]...