NCERT Solutions · Class 11 Mathematics Trigonometric Functions

Pair the outer terms, then factor.Group \(\displaystyle \cos 4x\) with \(\displaystyle \cos 2x\) and \(\displaystyle \sin 4x\) with \(\displaystyle \sin 2x\), because their half-sum is \(\displaystyle 3x\) in both cases.Numerator, using \(\displaystyle \cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\):\[(\cos 4x+\cos 2x)+\cos 3x=2\cos 3x\cos x+\cos 3x=\cos 3x\,(2\cos x+1). \]Denominator, using \(\displaystyle \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\):\[(\sin 4x+\sin 2x)+\sin 3x=2\sin 3x\cos x+\sin 3x=\sin 3x\,(2\cos x+1). \]Dividing, the common factor \(\displaystyle (2\cos x+1)\) cancels:\[\frac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x} =\frac{\cos 3x}{\sin 3x}=\cot 3x . \]Hence the given expression equals \(\displaystyle \cot 3x\).

Write \(\displaystyle 3x=2x+x\) and use the cotangent addition formula.The rule is\[\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}. \]Take \(\displaystyle A=2x,\;B=x\), so that \(\displaystyle A+B=3x\):\[\cot 3x=\frac{\cot 2x\cot x-1}{\cot 2x+\cot x}. \]Cross-multiplying,\[\cot 3x\,(\cot 2x+\cot x)=\cot 2x\cot x-1, \] \[\cot 3x\cot 2x+\cot 3x\cot x=\cot x\cot 2x-1 . \]Moving the two products on the left across:\[\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1 . \]Hence \(\displaystyle \cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1\).

Apply the double-angle formula for tangent twice.The rule is \(\displaystyle \tan 2\theta=\dfrac{2\tan\theta}{1-\tan^{2}\theta}\). Write \(\displaystyle t=\tan x\), so\[\tan 2x=\frac{2t}{1-t^{2}} . \]Now use the same rule on \(\displaystyle 4x=2(2x)\):\[\tan 4x=\frac{2\tan 2x}{1-\tan^{2}2x} =\frac{\dfrac{4t}{1-t^{2}}}{1-\dfrac{4t^{2}}{(1-t^{2})^{2}}} . \]Multiply numerator and denominator by \(\displaystyle (1-t^{2})^{2}\):\[\tan 4x=\frac{4t(1-t^{2})}{(1-t^{2})^{2}-4t^{2}} . \]Expanding the denominator,\[(1-t^{2})^{2}-4t^{2}=1-2t^{2}+t^{4}-4t^{2}=1-6t^{2}+t^{4}. \]Therefore\[\tan 4x=\frac{4\tan x\,(1-\tan^{2}x)}{1-6\tan^{2}x+\tan^{4}x}. \]Hence \(\displaystyle \tan 4x=\dfrac{4\tan x\left(1-\tan^{2}x\right)}{1-6\tan^{2}x+\tan^{4}x}\).

Double angle twice, taking \(\displaystyle 4x=2(2x)\).Use \(\displaystyle \cos 2\theta=1-2\sin^{2}\theta\) with \(\displaystyle \theta=2x\):\[\cos 4x=1-2\sin^{2}2x . \]Now substitute \(\displaystyle \sin 2x=2\sin x\cos x\):\[\cos 4x=1-2(2\sin x\cos x)^{2}=1-2\cdot 4\sin^{2}x\cos^{2}x=1-8\sin^{2}x\cos^{2}x . \]Hence \(\displaystyle \cos 4x=1-8\sin^{2}x\cos^{2}x\).

Write \(\displaystyle 6x=2(3x)\) and use the triple-angle formula for cosine.Two rules are needed:\[\cos 2\theta=2\cos^{2}\theta-1,\qquad \cos 3x=4\cos^{3}x-3\cos x . \]Apply the first with \(\displaystyle \theta=3x\):\[\cos 6x=2\cos^{2}3x-1 . \]Now put \(\displaystyle c=\cos x\) and substitute \(\displaystyle \cos 3x=4c^{3}-3c\):\[\cos^{2}3x=(4c^{3}-3c)^{2}=16c^{6}-24c^{4}+9c^{2}. \]Hence\[\cos 6x=2\left(16c^{6}-24c^{4}+9c^{2}\right)-1=32c^{6}-48c^{4}+18c^{2}-1 . \]Hence \(\displaystyle \cos 6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1\).