NCERT Solutions · Class 11 Mathematics Trigonometric Functions

Product-to-sum, then supplementary angles.Use \(\displaystyle 2\cos A\cos B=\cos(A+B)+\cos(A-B)\) with \(\displaystyle A=\dfrac{\pi}{13},\;B=\dfrac{9\pi}{13}\):\[2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} =\cos\frac{10\pi}{13}+\cos\left(-\frac{8\pi}{13}\right) =\cos\frac{10\pi}{13}+\cos\frac{8\pi}{13}, \]since cosine is an even function.Now use \(\displaystyle \cos(\pi-\theta)=-\cos\theta\). Because \(\displaystyle \dfrac{10\pi}{13}=\pi-\dfrac{3\pi}{13}\) and \(\displaystyle \dfrac{8\pi}{13}=\pi-\dfrac{5\pi}{13}\),\[\cos\frac{10\pi}{13}=-\cos\frac{3\pi}{13},\qquad \cos\frac{8\pi}{13}=-\cos\frac{5\pi}{13}. \]Substituting into the left-hand side:\[L.H.S.=-\cos\frac{3\pi}{13}-\cos\frac{5\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}=0 . \]Hence \(\displaystyle 2\cos\dfrac{\pi}{13}\cos\dfrac{9\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{5\pi}{13}=0\).

Sum-to-product inside each bracket.The two rules needed are\[\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},\qquad \cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}. \]With \(\displaystyle A=3x,\;B=x\) they give\[\sin 3x+\sin x=2\sin 2x\cos x,\qquad \cos 3x-\cos x=-2\sin 2x\sin x . \]Substituting into the left-hand side:\[L.H.S.=(2\sin 2x\cos x)\sin x+(-2\sin 2x\sin x)\cos x \] \[=2\sin 2x\sin x\cos x-2\sin 2x\sin x\cos x=0 . \]Hence \(\displaystyle (\sin 3x+\sin x)\sin x+(\cos 3x-\cos x)\cos x=0\).

Expand, use \(\displaystyle \sin^{2}+\cos^{2}=1\), then the cosine addition formula.Expanding both squares:\[(\cos x+\cos y)^{2}=\cos^{2}x+2\cos x\cos y+\cos^{2}y, \] \[(\sin x-\sin y)^{2}=\sin^{2}x-2\sin x\sin y+\sin^{2}y . \]Adding, and grouping \(\displaystyle (\cos^{2}x+\sin^{2}x)=1\) with \(\displaystyle (\cos^{2}y+\sin^{2}y)=1\):\[L.H.S.=2+2\left(\cos x\cos y-\sin x\sin y\right). \]The bracket is exactly \(\displaystyle \cos(x+y)\), so\[L.H.S.=2+2\cos(x+y)=2\bigl(1+\cos(x+y)\bigr). \]Finally \(\displaystyle 1+\cos 2\theta=2\cos^{2}\theta\) with \(\displaystyle \theta=\dfrac{x+y}{2}\) gives\[L.H.S.=2\cdot 2\cos^{2}\frac{x+y}{2}=4\cos^{2}\frac{x+y}{2}. \]Hence \(\displaystyle (\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4\cos^{2}\dfrac{x+y}{2}\).

Expand, use \(\displaystyle \sin^{2}+\cos^{2}=1\), then the cosine subtraction formula.Expanding both squares:\[(\cos x-\cos y)^{2}=\cos^{2}x-2\cos x\cos y+\cos^{2}y, \] \[(\sin x-\sin y)^{2}=\sin^{2}x-2\sin x\sin y+\sin^{2}y . \]Adding and using \(\displaystyle \cos^{2}x+\sin^{2}x=1\), \(\displaystyle \cos^{2}y+\sin^{2}y=1\):\[L.H.S.=2-2\left(\cos x\cos y+\sin x\sin y\right)=2-2\cos(x-y). \]Now use \(\displaystyle 1-\cos 2\theta=2\sin^{2}\theta\) with \(\displaystyle \theta=\dfrac{x-y}{2}\):\[L.H.S.=2\bigl(1-\cos(x-y)\bigr)=2\cdot 2\sin^{2}\frac{x-y}{2}=4\sin^{2}\frac{x-y}{2}. \]Hence \(\displaystyle (\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4\sin^{2}\dfrac{x-y}{2}\).

Pair the terms so that each pair has half-sum \(\displaystyle 4x\).Use \(\displaystyle \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\) on the pairs \(\displaystyle (\sin x,\sin 7x)\) and \(\displaystyle (\sin 3x,\sin 5x)\):\[\sin 7x+\sin x=2\sin 4x\cos 3x,\qquad \sin 5x+\sin 3x=2\sin 4x\cos x . \]Adding and taking out \(\displaystyle 2\sin 4x\):\[L.H.S.=2\sin 4x\,(\cos 3x+\cos x). \]Now \(\displaystyle \cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\) gives \(\displaystyle \cos 3x+\cos x=2\cos 2x\cos x\), so\[L.H.S.=2\sin 4x\cdot 2\cos 2x\cos x=4\cos x\cos 2x\sin 4x . \]Hence \(\displaystyle \sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x\).

Sum-to-product on each bracket; every pair has half-sum \(\displaystyle 6x\).For the numerator, \(\displaystyle \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\):\[\sin 7x+\sin 5x=2\sin 6x\cos x,\qquad \sin 9x+\sin 3x=2\sin 6x\cos 3x, \] \[\text{Numerator}=2\sin 6x\,(\cos x+\cos 3x). \]For the denominator, \(\displaystyle \cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\):\[\cos 7x+\cos 5x=2\cos 6x\cos x,\qquad \cos 9x+\cos 3x=2\cos 6x\cos 3x, \] \[\text{Denominator}=2\cos 6x\,(\cos x+\cos 3x). \]The factor \(\displaystyle 2(\cos x+\cos 3x)\) is common, so it cancels:\[L.H.S.=\frac{\sin 6x}{\cos 6x}=\tan 6x . \]Hence the given expression equals \(\displaystyle \tan 6x\).

Group \(\displaystyle \sin 3x-\sin x\) first, then take out \(\displaystyle \sin x\).Use \(\displaystyle \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\) with \(\displaystyle A=3x,\;B=x\):\[\sin 3x-\sin x=2\cos 2x\sin x . \]Also \(\displaystyle \sin 2x=2\sin x\cos x\). Hence\[L.H.S.=2\sin x\cos 2x+2\sin x\cos x=2\sin x\,(\cos 2x+\cos x). \]Now \(\displaystyle \cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\) with \(\displaystyle A=2x,\;B=x\) gives\[\cos 2x+\cos x=2\cos\frac{3x}{2}\cos\frac{x}{2}. \]Therefore\[L.H.S.=2\sin x\cdot 2\cos\frac{3x}{2}\cos\frac{x}{2}=4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}. \]Hence \(\displaystyle \sin 3x+\sin 2x-\sin x=4\sin x\cos\dfrac{x}{2}\cos\dfrac{3x}{2}\).

Half-angle formulae, with the quadrant of \(\displaystyle x/2\) fixing the signs.Step $\displaystyle 1$ — locate \(\displaystyle x/2\). Since \(\displaystyle x\) lies in quadrant II,\[\frac{\pi}{2}<x<\pi \quad\Longrightarrow\quad \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}, \]so \(\displaystyle x/2\) lies in quadrant I and all three ratios are positive.Step $\displaystyle 2$ — get \(\displaystyle \cos x\). With \(\displaystyle \tan x=-\dfrac43\), \(\displaystyle \sec^{2}x=1+\tan^{2}x=1+\dfrac{16}{9}=\dfrac{25}{9}\), so \(\displaystyle \sec x=\pm\dfrac53\). In quadrant II cosine is negative, hence \(\displaystyle \sec x=-\dfrac53\) and\[\cos x=-\frac{3}{5}. \]Step $\displaystyle 3$ — apply the half-angle formulae.\[\sin^{2}\frac{x}{2}=\frac{1-\cos x}{2}=\frac{1+\frac35}{2}=\frac{4}{5} \quad\Longrightarrow\quad \sin\frac{x}{2}=\frac{2}{\sqrt5}, \]\[\cos^{2}\frac{x}{2}=\frac{1+\cos x}{2}=\frac{1-\frac35}{2}=\frac{1}{5} \quad\Longrightarrow\quad \cos\frac{x}{2}=\frac{1}{\sqrt5}, \]both positive roots being taken by Step 1. Then\[\tan\frac{x}{2}=\frac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}=\frac{2/\sqrt5}{1/\sqrt5}=2 . \]\(\displaystyle \sin\dfrac{x}{2}=\dfrac{2}{\sqrt5},\quad \cos\dfrac{x}{2}=\dfrac{1}{\sqrt5},\quad \tan\dfrac{x}{2}=2\).

Half-angle formulae, with the quadrant of \(\displaystyle x/2\) fixing the signs.Step $\displaystyle 1$ — locate \(\displaystyle x/2\). Since \(\displaystyle x\) lies in quadrant III,\[\pi<x<\frac{3\pi}{2} \quad\Longrightarrow\quad \frac{\pi}{2}<\frac{x}{2}<\frac{3\pi}{4}, \]so \(\displaystyle x/2\) lies in quadrant II: \(\displaystyle \sin\dfrac{x}{2}>0\), while \(\displaystyle \cos\dfrac{x}{2}<0\) and \(\displaystyle \tan\dfrac{x}{2}<0\).Step $\displaystyle 2$ — apply the half-angle formulae with \(\displaystyle \cos x=-\dfrac13\).\[\sin^{2}\frac{x}{2}=\frac{1-\cos x}{2}=\frac{1+\frac13}{2}=\frac{2}{3} \quad\Longrightarrow\quad \sin\frac{x}{2}=+\sqrt{\frac{2}{3}}=\frac{\sqrt6}{3}, \]\[\cos^{2}\frac{x}{2}=\frac{1+\cos x}{2}=\frac{1-\frac13}{2}=\frac{1}{3} \quad\Longrightarrow\quad \cos\frac{x}{2}=-\frac{1}{\sqrt3}=-\frac{\sqrt3}{3}, \]the signs coming from Step 1. Then\[\tan\frac{x}{2}=\frac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}} =\frac{\sqrt{2}/\sqrt3}{-1/\sqrt3}=-\sqrt2 . \]\(\displaystyle \sin\dfrac{x}{2}=\dfrac{\sqrt6}{3},\quad \cos\dfrac{x}{2}=-\dfrac{\sqrt3}{3},\quad \tan\dfrac{x}{2}=-\sqrt2\).

Half-angle formulae, with the quadrant of \(\displaystyle x/2\) fixing the signs.Step $\displaystyle 1$ — locate \(\displaystyle x/2\). Since \(\displaystyle x\) lies in quadrant II,\[\frac{\pi}{2}<x<\pi \quad\Longrightarrow\quad \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}, \]so \(\displaystyle x/2\) lies in quadrant I and all three ratios are positive.Step $\displaystyle 2$ — get \(\displaystyle \cos x\). From \(\displaystyle \sin x=\dfrac14\), \(\displaystyle \cos^{2}x=1-\dfrac{1}{16}=\dfrac{15}{16}\); cosine is negative in quadrant II, so\[\cos x=-\frac{\sqrt{15}}{4}. \]Step $\displaystyle 3$ — apply the half-angle formulae.\[\sin^{2}\frac{x}{2}=\frac{1-\cos x}{2}=\frac{1+\dfrac{\sqrt{15}}{4}}{2} =\frac{4+\sqrt{15}}{8}=\frac{8+2\sqrt{15}}{16}=\frac{(\sqrt5+\sqrt3)^{2}}{16}, \]using \(\displaystyle (\sqrt5+\sqrt3)^{2}=5+3+2\sqrt{15}=8+2\sqrt{15}\). Taking the positive root,\[\sin\frac{x}{2}=\frac{\sqrt5+\sqrt3}{4}. \]Similarly,\[\cos^{2}\frac{x}{2}=\frac{1+\cos x}{2}=\frac{4-\sqrt{15}}{8}=\frac{8-2\sqrt{15}}{16}=\frac{(\sqrt5-\sqrt3)^{2}}{16} \quad\Longrightarrow\quad \cos\frac{x}{2}=\frac{\sqrt5-\sqrt3}{4}. \]Finally, rationalising by multiplying numerator and denominator by \(\displaystyle (\sqrt5+\sqrt3)\),\[\tan\frac{x}{2}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3} =\frac{(\sqrt5+\sqrt3)^{2}}{5-3}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}. \]\(\displaystyle \sin\dfrac{x}{2}=\dfrac{\sqrt5+\sqrt3}{4},\quad \cos\dfrac{x}{2}=\dfrac{\sqrt5-\sqrt3}{4},\quad \tan\dfrac{x}{2}=4+\sqrt{15}\).