Differentiation (Part-38) | S N De

 $\,10(i).\,$ Express $\, \sin5\theta\,$ in terms of $\,\sin\theta.$

Sol. $\,\,\sin5\theta\\=\sin(2\theta+3\theta)\\=\sin2\theta\cos3\theta+\cos2\theta\sin3\theta\\=2\sin\theta\cos\theta(4\cos^3\theta-3\cos\theta)\\+(1-2\sin^2\theta)(3\sin\theta-4\sin^3\theta)\\=8\sin\theta\cos^4\theta-6\sin\theta\cos^2\theta\\+3\sin\theta-4\sin^3\theta-6\sin^3\theta+8\sin^5\theta\\=8\sin\theta(1-\sin^2\theta)^2-6\sin\theta(1-\sin^2\theta)\\+3\sin\theta-10\sin^3\theta+8\sin^5\theta\\=8\sin\theta(1-2\sin^2\theta+\sin^4\theta)-6\sin\theta\\+6\sin^3\theta+3\sin\theta-10\sin^3\theta+8\sin^5\theta\\=8\sin\theta-16\sin^3\theta+8\sin^5\theta-3\sin\theta \\-4\sin^3\theta+8\sin^5\theta\\=16\sin^5\theta-20\sin^3\theta+5\sin\theta$ 

$\,10(ii).\,$ Express $\, \cos6\theta\,$ in terms of $\,\cos\theta.$

Sol. $\,\cos6\theta\\=\cos(2\times 3\theta)\\=2\cos^23\theta-1\\=2(4\cos^3\theta-3\cos\theta)^2-1\\=32\cos^6\theta-48\cos^4\theta+18\cos^2\theta-1$

$\,10(iii).\,$ Express $\, \tan4\theta\,$ in terms of $\,\tan\theta.$

Sol. $\,\,\tan4\theta\\=\tan(2\times 2\theta)\\=\frac{2\tan2\theta}{1-\tan^22\theta}\\=\frac{2\times \frac{2\tan\theta}{1-\tan^2\theta}}{1-\left(\frac{2\tan\theta}{1-\tan^2\theta}\right)^2}\\=\frac{4\tan\theta(1-\tan^2\theta)}{(1-\tan^2\theta)^2-4\tan^2\theta}\\=\frac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}$

$\,10(iv).\,$ Express $\, \cos4A\,$ in terms of $\,\tan A.$

Sol. $\,\,\cos4A\\=\cos(2\times 2A)\\=\frac{1-\tan^22A}{1+\tan^22A}\\=\frac{1-\left(\frac{2\tan A}{1-\tan^2A}\right)^2}{1+\left(\frac{2\tan A}{1-\tan^2A}\right)^2}\\=\frac{(1-\tan^2A)^2-4\tan^2A}{(1-\tan^2A)^2+4\tan^2A}\\=\frac{1-6\tan^2A+\tan^4A}{(1+\tan^2A)^2}$

$\,11.\,$ Show that, $\,\sec x=\frac{2}{\sqrt{2+\sqrt{2+2\cos4x}}}$

Sol. $\,\,\frac{2}{\sqrt{2+\sqrt{2+2\cos4x}}}\\=\frac{2}{\sqrt{2+\sqrt{2(1+\cos4x)}}}\\=\frac{2}{\sqrt{2+\sqrt{4\cos^22x}}}\\=\frac{2}{\sqrt{2+2\cos2x}}\\=\frac{2}{\sqrt{2(1+\cos2x)}}\\=\frac{2}{\sqrt{2.2\cos^2x}}\\=\frac{2}{2\cos x}\\=\frac{1}{\cos x}\\=\sec x$

$\,12.\,$ If $\,\,\tan x,\tan y,\,\tan z\,$ are in G.P., show that , $\,\cos2y=\frac{\cos(x+z)}{\cos(x-z)}$

Sol. If $\,\,\tan x,\tan y,\,\tan z\,$ are in G.P., $\,\tan^2y=\tan x\,\,\tan z \\ \Rightarrow \frac{\sin^2y}{\cos^2y}=\frac{\sin x\sin z}{\cos x\cos z} \\ \Rightarrow \frac{\sin^2y+\cos^2y}{\sin^2y-\cos^2y}=\frac{\sin x\sin z+\cos x\cos z}{\sin x\sin z-\cos x\cos z} \\ \Rightarrow \frac{1}{-\cos 2y}=\frac{\cos(x-z)}{-\cos(x+z)} \\ \Rightarrow \cos 2y=\frac{\cos(x+z)}{\cos(x-z)}$

$\,13.\,$ Using the formula $\,\,\tan A=\frac{\sin 2A}{1+\cos 2A},\,$ find the values of $\,\tan75^{\circ}\,\,$ and $\,\cot22^{\circ}30'.$

Sol. We have, $\,\,\tan A=\frac{\sin 2A}{1+\cos 2A} \cdots(1).$ 

Putting $\,A=75^{\circ} \,$ in (1), we get $\,\tan75^{\circ}=\frac{\sin(2\times 75^{\circ})}{1+\cos{(2\times 75^{\circ})}}\\~~~~~~~~~~~~~=\frac{\sin150^{\circ}}{1+\cos150^{\circ}}\\~~~~~~~~~~~~~=\frac{\sin30^{\circ}}{1-\cos30^{\circ}}\\~~~~~~~~~~~~~=\frac{1/2}{1-\sqrt3/2}\\~~~~~~~~~~~~~=\frac{1}{2-\sqrt3}\\~~~~~~~~~~~~~=2+\sqrt3$

Again, putting $\,A=22^{\circ}30'\,\,$ in (1), we get $\,\,\tan22^{\circ}30'=\frac{\sin(2\times 22^{\circ}30')}{1+\cos(2\times 22^{\circ}30')} \\~~~~~~~~~~~~~~~~~~~ =\frac{\sin45^{\circ}}{1+\cos45^{\circ}}\\~~~~~~~~~~~~~~~~~~~=\frac{1/\sqrt2}{1+1/\sqrt2}\\~~~~~~~~~~~~~~~~~~~=\frac{1}{1+\sqrt2} \\ \text{So,}\,\,\cot22^{\circ}30'=1+\sqrt2.$

$\,14.\,$ If $\,\,32\sin^6\theta=10-15\cos2\theta+b\cos4\theta\\+a\cos6\theta,\,\,$ show that $\,\,\tan\theta\tan5\theta=\frac{a+b-5}{a-b+7}$

Sol. $\,\,32\sin^6\theta=10-15\cos2\theta+b\cos4\theta\\+a\cos6\theta \\ \Rightarrow 4(2\sin^2\theta)^3=10-15\cos2\theta\\+b\cos4\theta+a\cos6\theta\\ \Rightarrow4(1-\cos2\theta)^3=10-15\cos2\theta+b\cos4\theta\\+a\cos6\theta\\ \Rightarrow 4-12\cos2\theta+12\cos^22\theta-4\cos^32\theta=10\\-15\cos2\theta+b\cos4\theta+a\cos6\theta \\ \Rightarrow-6+12\cos^22\theta-b\cos4\theta=-3\cos2\theta\\+4\cos^32\theta+a\cos6\theta \\ \Rightarrow 6(2\cos^22\theta-1)-b\cos4\theta=\cos6\theta\\+a\cos6\theta\\ \Rightarrow6\cos4\theta-b\cos4\theta=\cos6\theta+a\cos6\theta \\ \Rightarrow(6-b)\cos4\theta=(1+a)\cos6\theta \\ \Rightarrow \frac{\cos4\theta}{\cos6\theta}=\frac{1+a}{6-b}\\ \Rightarrow \frac{\cos4\theta+\cos6\theta}{\cos4\theta-\cos6\theta}=\frac{1+a+6-b}{1+a-6+b}\\ \Rightarrow \frac{2\cos\frac{4\theta+6\theta}{2}\cos\frac{4\theta-6\theta}{2}}{2\sin\frac{4\theta+6\theta}{2}\sin\frac{6\theta-4\theta}{2}}=\frac{a-b+7}{a+b-5} \\ \Rightarrow \frac{\cos5\theta\cos\theta}{\sin5\theta\sin\theta}=\frac{a-b+7}{a+b-5} \\ \Rightarrow \tan5\theta\tan\theta=\frac{a+b-5}{a-b+7} $

$\,15.\,$ If $\,\tan(\alpha+\beta)=a+b\,$ and $\,\tan(\alpha-\beta)=a-b,\,\,$ then show that $\,\,a\tan \alpha-b\tan\beta=a^2-b^2.$

Sol. $\,\tan(\alpha+\beta)=a+b \\ \Rightarrow \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=a+b \\ \Rightarrow 1-\tan\alpha\tan\beta=\frac{\tan\alpha+\tan\beta}{a+b} \\ \Rightarrow \tan\alpha\tan\beta=1-\frac{\tan\alpha+\tan\beta}{a+b}  \cdots(1)$

Again, $\,\,\tan(\alpha-\beta)=a-b \\ \Rightarrow \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=a-b \\ \Rightarrow 1+\tan\alpha\tan\beta=\frac{\tan\alpha-\tan\beta}{a-b}\\ \Rightarrow 1+1-\frac{\tan\alpha+\tan\beta}{a+b}=\frac{\tan\alpha-\tan\beta}{a-b}\,\,[\text{By (1)}] \\ \Rightarrow \frac{\tan\alpha-\tan\beta}{a-b}+\frac{\tan\alpha+\tan\beta}{a+b}=2 \\ \Rightarrow \frac{a\tan\alpha+b\tan\alpha-a\tan\beta-b\tan\beta+a\tan\alpha\\-b\tan\alpha+a\tan\beta-b\tan\beta}{a^2-b^2}=2 \\ \Rightarrow \frac{2a\tan\alpha-2b\tan\beta}{a^2-b^2}=2 \\ \Rightarrow a\tan\alpha-b\tan\beta=a^2-b^2.$