Class 11: Values of Trigonometric Functions at Multiples and Submultiples of an Angle

$\displaystyle \textbf{Question 1: }\text{Prove that }\sin5x=5\sin x-20\sin^3x+16\sin^5x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sin5x$
$\displaystyle =\sin(4x+x)$
$\displaystyle =\sin4x\cos x+\cos4x\sin x$
$\displaystyle =2\sin2x\cos2x\cos x+(1-2\sin^22x)\sin x$
$\displaystyle =4\sin x\cos^2x\cos2x+\sin x-2\sin^22x\sin x$
$\displaystyle \text{Using }\cos2x=1-2\sin^2x\text{ and }\sin2x=2\sin x\cos x,$
$\displaystyle \text{LHS}=4\sin x\cos^2x(1-2\sin^2x)+\sin x$
$\displaystyle \qquad-8\sin^3x\cos^2x$
$\displaystyle =4\sin x(1-\sin^2x)(1-2\sin^2x)+\sin x$
$\displaystyle \qquad-8\sin^3x(1-\sin^2x)$
$\displaystyle =4\sin x(1-3\sin^2x+2\sin^4x)+\sin x$
$\displaystyle \qquad-8\sin^3x+8\sin^5x$
$\displaystyle =4\sin x-12\sin^3x+8\sin^5x+\sin x-8\sin^3x+8\sin^5x$
$\displaystyle =5\sin x-20\sin^3x+16\sin^5x$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Prove that }4(\cos^310^\circ+\sin^320^\circ)=3(\cos10^\circ+\sin20^\circ).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Using }\cos3\theta=4\cos^3\theta-3\cos\theta,$
$\displaystyle 4\cos^3\theta=\cos3\theta+3\cos\theta$
$\displaystyle \text{Putting }\theta=10^\circ,$
$\displaystyle 4\cos^310^\circ=\cos30^\circ+3\cos10^\circ \qquad\ldots(1)$
$\displaystyle \text{Also, }\sin3\theta=3\sin\theta-4\sin^3\theta,$
$\displaystyle 4\sin^3\theta=3\sin\theta-\sin3\theta$
$\displaystyle \text{Putting }\theta=20^\circ,$
$\displaystyle 4\sin^320^\circ=3\sin20^\circ-\sin60^\circ \qquad\ldots(2)$
$\displaystyle \text{Adding (1) and (2),}$
$\displaystyle 4(\cos^310^\circ+\sin^320^\circ)$
$\displaystyle =3(\cos10^\circ+\sin20^\circ)+\cos30^\circ-\sin60^\circ$
$\displaystyle \text{Since }\cos30^\circ=\sin60^\circ,$
$\displaystyle 4(\cos^310^\circ+\sin^320^\circ)=3(\cos10^\circ+\sin20^\circ)$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Prove that }\cos^3x\sin3x+\sin^3x\cos3x=\frac{3}{4}\sin4x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cos^3x\sin3x+\sin^3x\cos3x$
$\displaystyle \text{Using }\cos3x=4\cos^3x-3\cos x,$
$\displaystyle \cos^3x=\frac{\cos3x+3\cos x}{4}$
$\displaystyle \text{Also, using }\sin3x=3\sin x-4\sin^3x,$
$\displaystyle \sin^3x=\frac{3\sin x-\sin3x}{4}$
$\displaystyle \therefore \text{LHS}=\left(\frac{\cos3x+3\cos x}{4}\right)\sin3x$
$\displaystyle \qquad+\left(\frac{3\sin x-\sin3x}{4}\right)\cos3x$
$\displaystyle =\frac{1}{4}\left[\cos3x\sin3x+3\cos x\sin3x\right.$
$\displaystyle \left.\qquad+3\sin x\cos3x-\sin3x\cos3x\right]$
$\displaystyle =\frac{3}{4}\left(\cos x\sin3x+\sin x\cos3x\right)$
$\displaystyle =\frac{3}{4}\sin(3x+x)$
$\displaystyle =\frac{3}{4}\sin4x$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Prove that}$
$\displaystyle \tan x\tan\left(x+\frac{\pi}{3}\right)+\tan x\tan\left(x-\frac{\pi}{3}\right) +\tan\left(x+\frac{\pi}{3}\right)\tan\left(x-\frac{\pi}{3}\right)=-3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\tan x\tan(x+60^\circ)+\tan x\tan(x-60^\circ)$
$\displaystyle \qquad+\tan(x+60^\circ)\tan(x-60^\circ)$
$\displaystyle =\tan x\left(\frac{\tan x+\sqrt3}{1-\sqrt3\tan x}\right)$
$\displaystyle \qquad+\tan x\left(\frac{\tan x-\sqrt3}{1+\sqrt3\tan x}\right)$
$\displaystyle \qquad+\left(\frac{\tan x+\sqrt3}{1-\sqrt3\tan x}\right)\left(\frac{\tan x-\sqrt3}{1+\sqrt3\tan x}\right)$
$\displaystyle =\tan x\left[\frac{\tan x+\sqrt3}{1-\sqrt3\tan x}+\frac{\tan x-\sqrt3}{1+\sqrt3\tan x}\right]$
$\displaystyle \qquad+\frac{\tan^2x-3}{1-3\tan^2x}$
$\displaystyle =\tan x\left[\frac{(\tan x+\sqrt3)(1+\sqrt3\tan x)}{1-3\tan^2x}\right.$
$\displaystyle \left.\qquad+\frac{(\tan x-\sqrt3)(1-\sqrt3\tan x)}{1-3\tan^2x}\right]$
$\displaystyle \qquad+\frac{\tan^2x-3}{1-3\tan^2x}$
$\displaystyle =\tan x\left[\frac{8\tan x}{1-3\tan^2x}\right]+\frac{\tan^2x-3}{1-3\tan^2x}$
$\displaystyle =\frac{8\tan^2x+\tan^2x-3}{1-3\tan^2x}$
$\displaystyle =\frac{9\tan^2x-3}{1-3\tan^2x}$
$\displaystyle =\frac{-3(1-3\tan^2x)}{1-3\tan^2x}$
$\displaystyle =-3$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Prove that }\tan x+\tan\left(\frac{\pi}{3}+x\right)-\tan\left(\frac{\pi}{3}-x\right) =3\tan3x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\tan x+\tan(60^\circ+x)-\tan(60^\circ-x)$
$\displaystyle =\tan x+\frac{\tan60^\circ+\tan x}{1-\tan60^\circ\tan x}$
$\displaystyle \qquad-\frac{\tan60^\circ-\tan x}{1+\tan60^\circ\tan x}$
$\displaystyle =\tan x+\frac{\sqrt3+\tan x}{1-\sqrt3\tan x}$
$\displaystyle \qquad-\frac{\sqrt3-\tan x}{1+\sqrt3\tan x}$
$\displaystyle =\tan x+\frac{(\sqrt3+\tan x)(1+\sqrt3\tan x)}{1-3\tan^2x}$
$\displaystyle \qquad-\frac{(\sqrt3-\tan x)(1-\sqrt3\tan x)}{1-3\tan^2x}$
$\displaystyle =\tan x+\frac{(\sqrt3+\tan x)(1+\sqrt3\tan x)}{1-3\tan^2x}$
$\displaystyle \qquad-\frac{\sqrt3-\tan x-3\tan x+\sqrt3\tan^2x}{1-3\tan^2x}$
$\displaystyle =\tan x+\frac{8\tan x}{1-3\tan^2x}$
$\displaystyle =\frac{\tan x(1-3\tan^2x)+8\tan x}{1-3\tan^2x}$
$\displaystyle =\frac{9\tan x-3\tan^3x}{1-3\tan^2x}$
$\displaystyle =3\left(\frac{3\tan x-\tan^3x}{1-3\tan^2x}\right)$
$\displaystyle \text{Using }\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x},$
$\displaystyle \text{LHS}=3\tan3x$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Prove that }\cot x+\cot\left(\frac{\pi}{3}+x\right)-\cot\left(\frac{\pi}{3}-x\right) =3\cot3x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cot x+\cot(60^\circ+x)-\cot(60^\circ-x)$
$\displaystyle =\frac{1}{\tan x}+\frac{1}{\tan(60^\circ+x)}-\frac{1}{\tan(60^\circ-x)}$
$\displaystyle =\frac{1}{\tan x}+\frac{1-\tan60^\circ\tan x}{\tan60^\circ+\tan x}$
$\displaystyle \qquad-\frac{1+\tan60^\circ\tan x}{\tan60^\circ-\tan x}$
$\displaystyle =\frac{1}{\tan x}+\frac{1-\sqrt3\tan x}{\sqrt3+\tan x}$
$\displaystyle \qquad-\frac{1+\sqrt3\tan x}{\sqrt3-\tan x}$
$\displaystyle =\frac{1}{\tan x}+\frac{(1-\sqrt3\tan x)(\sqrt3-\tan x)}{3-\tan^2x}$
$\displaystyle \qquad-\frac{(1+\sqrt3\tan x)(\sqrt3+\tan x)}{3-\tan^2x}$
$\displaystyle =\frac{1}{\tan x}+\frac{-8\tan x}{3-\tan^2x}$
$\displaystyle =\frac{1}{\tan x}-\frac{8\tan x}{3-\tan^2x}$
$\displaystyle =\frac{3-\tan^2x-8\tan^2x}{\tan x(3-\tan^2x)}$
$\displaystyle =\frac{3-9\tan^2x}{3\tan x-\tan^3x}$
$\displaystyle =3\left(\frac{1-3\tan^2x}{3\tan x-\tan^3x}\right)$
$\displaystyle \text{Using }\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x},$
$\displaystyle \text{LHS}=\frac{3}{\tan3x}$
$\displaystyle =3\cot3x$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Prove that }\cot x+\cot\left(\frac{\pi}{3}+x\right)+\cot\left(\frac{2\pi}{3}+x\right) =3\cot3x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cot x+\cot(60^\circ+x)+\cot(120^\circ+x)$
$\displaystyle =\cot x+\cot(60^\circ+x)-\cot\left[180^\circ-(120^\circ+x)\right]$
$\displaystyle =\cot x+\cot(60^\circ+x)-\cot(60^\circ-x)$
$\displaystyle =\frac{1}{\tan x}+\frac{1}{\tan(60^\circ+x)}-\frac{1}{\tan(60^\circ-x)}$
$\displaystyle =\frac{1}{\tan x}+\frac{1-\tan60^\circ\tan x}{\tan60^\circ+\tan x}$
$\displaystyle \qquad-\frac{1+\tan60^\circ\tan x}{\tan60^\circ-\tan x}$
$\displaystyle =\frac{1}{\tan x}+\frac{1-\sqrt3\tan x}{\sqrt3+\tan x}$
$\displaystyle \qquad-\frac{1+\sqrt3\tan x}{\sqrt3-\tan x}$
$\displaystyle =\frac{1}{\tan x}+\frac{(1-\sqrt3\tan x)(\sqrt3-\tan x)}{3-\tan^2x}$
$\displaystyle \qquad-\frac{(1+\sqrt3\tan x)(\sqrt3+\tan x)}{3-\tan^2x}$
$\displaystyle =\frac{1}{\tan x}-\frac{8\tan x}{3-\tan^2x}$
$\displaystyle =\frac{3-\tan^2x-8\tan^2x}{\tan x(3-\tan^2x)}$
$\displaystyle =\frac{3-9\tan^2x}{3\tan x-\tan^3x}$
$\displaystyle =3\left(\frac{1-3\tan^2x}{3\tan x-\tan^3x}\right)$
$\displaystyle \text{Using }\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x},$
$\displaystyle \text{LHS}=\frac{3}{\tan3x}$
$\displaystyle =3\cot3x$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Prove that }\sin5x=5\cos^4x\sin x-10\cos^2x\sin^3x+\sin^5x.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin5x=\sin(3x+2x)$
$\displaystyle =\sin3x\cos2x+\cos3x\sin2x$
$\displaystyle \text{Now, }\sin3x=\sin(2x+x)$
$\displaystyle =\sin2x\cos x+\cos2x\sin x$
$\displaystyle =2\sin x\cos^2x+(\cos^2x-\sin^2x)\sin x$
$\displaystyle =3\sin x\cos^2x-\sin^3x$
$\displaystyle \therefore \sin3x\cos2x=(3\sin x\cos^2x-\sin^3x)(\cos^2x-\sin^2x)$
$\displaystyle =3\sin x\cos^4x-4\sin^3x\cos^2x+\sin^5x$
$\displaystyle \text{Also, }\cos3x=\cos(2x+x)$
$\displaystyle =\cos2x\cos x-\sin2x\sin x$
$\displaystyle =(\cos^2x-\sin^2x)\cos x-2\sin^2x\cos x$
$\displaystyle =\cos^3x-3\sin^2x\cos x$
$\displaystyle \therefore \cos3x\sin2x=(\cos^3x-3\sin^2x\cos x)(2\sin x\cos x)$
$\displaystyle =2\cos^4x\sin x-6\sin^3x\cos^2x$
$\displaystyle \therefore \sin5x=3\sin x\cos^4x-4\sin^3x\cos^2x+\sin^5x$
$\displaystyle \qquad+2\cos^4x\sin x-6\sin^3x\cos^2x$
$\displaystyle =5\cos^4x\sin x-10\cos^2x\sin^3x+\sin^5x$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Prove that }\sin^3x+\sin^3\left(\frac{2\pi}{3}+x\right) +\sin^3\left(\frac{4\pi}{3}+x\right)=-\frac{3}{4}\sin3x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sin^3x+\sin^3(120^\circ+x)+\sin^3(240^\circ+x)$
$\displaystyle =\sin^3x+\left(\frac{\sqrt3}{2}\cos x-\frac{1}{2}\sin x\right)^3$
$\displaystyle \qquad+\left(-\frac{\sqrt3}{2}\cos x-\frac{1}{2}\sin x\right)^3$
$\displaystyle =\sin^3x+\frac{1}{8}(\sqrt3\cos x-\sin x)^3$
$\displaystyle \qquad-\frac{1}{8}(\sqrt3\cos x+\sin x)^3$
$\displaystyle \text{Let }a=\sqrt3\cos x\text{ and }b=\sin x.$
$\displaystyle \therefore \text{LHS}=\sin^3x+\frac{1}{8}\left[(a-b)^3-(a+b)^3\right]$
$\displaystyle \text{Using }(a-b)^3-(a+b)^3=-6a^2b-2b^3,$
$\displaystyle \text{LHS}=\sin^3x+\frac{1}{8}\left[-6(3\cos^2x)\sin x-2\sin^3x\right]$
$\displaystyle =\sin^3x-\frac{18}{8}\sin x\cos^2x-\frac{2}{8}\sin^3x$
$\displaystyle =\frac{3}{4}\sin^3x-\frac{9}{4}\sin x\cos^2x$
$\displaystyle =-\frac{3}{4}\left(3\sin x\cos^2x-\sin^3x\right)$
$\displaystyle \text{Since }\sin3x=3\sin x\cos^2x-\sin^3x,$
$\displaystyle \text{LHS}=-\frac{3}{4}\sin3x$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Prove that }\left|\sin x\sin\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{3}+x\right)\right| \leq\frac{1}{4}\text{ for all values of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\left|\sin x\sin(60^\circ-x)\sin(60^\circ+x)\right|$
$\displaystyle =\left|\sin x(\sin60^\circ\cos x-\cos60^\circ\sin x)\right.$
$\displaystyle \left.\qquad\times(\sin60^\circ\cos x+\cos60^\circ\sin x)\right|$
$\displaystyle =\left|\sin x\left(\frac{\sqrt3}{2}\cos x-\frac{1}{2}\sin x\right)\right.$
$\displaystyle \left.\qquad\times\left(\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x\right)\right|$
$\displaystyle =\left|\sin x\left(\frac{3}{4}\cos^2x-\frac{1}{4}\sin^2x\right)\right|$
$\displaystyle =\left|\frac{1}{4}\sin x(3\cos^2x-\sin^2x)\right|$
$\displaystyle =\left|\frac{1}{4}\sin x(3-4\sin^2x)\right|$
$\displaystyle =\left|\frac{1}{4}(3\sin x-4\sin^3x)\right|$
$\displaystyle =\frac{1}{4}|\sin3x|$
$\displaystyle \text{Since }|\sin3x|\leq1,$
$\displaystyle \therefore \text{LHS}\leq\frac{1}{4}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Prove that }\left|\cos x\cos\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{3}+x\right)\right| \leq\frac{1}{4}\text{ for all values of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\left|\cos x\cos(60^\circ-x)\cos(60^\circ+x)\right|$
$\displaystyle =\left|\cos x(\cos60^\circ\cos x+\sin60^\circ\sin x)\right.$
$\displaystyle \left.\qquad\times(\cos60^\circ\cos x-\sin60^\circ\sin x)\right|$
$\displaystyle =\left|\cos x\left(\frac{1}{2}\cos x+\frac{\sqrt3}{2}\sin x\right)\right.$
$\displaystyle \left.\qquad\times\left(\frac{1}{2}\cos x-\frac{\sqrt3}{2}\sin x\right)\right|$
$\displaystyle =\left|\cos x\left(\frac{1}{4}\cos^2x-\frac{3}{4}\sin^2x\right)\right|$
$\displaystyle =\left|\frac{1}{4}\cos x(\cos^2x-3\sin^2x)\right|$
$\displaystyle =\left|\frac{1}{4}\cos x(4\cos^2x-3)\right|$
$\displaystyle =\left|\frac{1}{4}(4\cos^3x-3\cos x)\right|$
$\displaystyle =\frac{1}{4}|\cos3x|$
$\displaystyle \text{Since }|\cos3x|\leq1,$
$\displaystyle \therefore \text{LHS}\leq\frac{1}{4}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$