Differentiation (Part-38) | S N De

 

$\,6.\,$ If $\,\alpha\,$ and $\,\beta\,$ are positive acute angles and $\,\,\cos 2\alpha=\frac{n\cos2\beta-1}{n-\cos2\beta},\,(n >1),\,\,$ show that, $\,\sqrt{n-1}\tan\alpha=\sqrt{n+1}\tan\beta$

Sol. We have,$\,\,\cos2\alpha=\frac{n\cos2\beta-1}{n-\cos2\beta} \\ \Rightarrow \frac{1-\cos2\alpha}{1+\cos2\alpha}=\frac{n-\cos2\beta-n\cos2\beta+1}{n-\cos2\beta+n\cos2\beta-1} \\ \Rightarrow \tan^2\alpha=\frac{(n+1)-(n+1)\cos 2\beta}{(n-1)+(n-1)\cos2\beta} \\ \Rightarrow\tan^2\alpha=\frac{n+1}{n-1}. \frac{1-\cos2\beta}{1+\cos2\beta} \\~~~~~~~~~~~~~~~~ =\frac{n+1}{n-1}. \tan^2\beta \,\,[**]\\ \Rightarrow \tan \alpha=\sqrt{\frac{n+1}{n-1}}. \tan\beta \\ \Rightarrow \sqrt{n-1}.\tan\alpha=\sqrt{n+1}.\tan\beta \\ [**]=>\alpha\,,\beta\,\,\, \text{are positive acute angles}$

$\,7.\,$ If $\,\,\frac{\tan x}{\tan y}=\frac{1+\cos^2x}{1+\sin^2x},\,\,$ show that $\,\,\sin(3x+y)=7\sin(x-y).$

Sol. $\,\,\frac{\tan x}{\tan y}=\frac{1+\cos^2x}{1+\sin^2x} \\ \Rightarrow \frac{\sin x\cos y}{\sin y\cos x}=\frac{2(1+\cos^2x)}{2(1+\sin^2x)} \\ \Rightarrow \frac{\sin x\cos y}{\sin y\cos x}= \frac{2+1+\cos 2x}{2+1-\cos 2x} \\ \Rightarrow \frac{\sin x\cos y+\cos x \sin y}{\sin x \cos y-\cos x \sin y}=\frac{6}{2\cos 2x} \\ \Rightarrow \frac{\sin(x+y)}{\sin(x-y)}=\frac{6}{2\cos 2x} \\ \Rightarrow 2\sin(x+y)\cos 2x=6\sin(x-y) \\ \Rightarrow \sin(3x+y)-\sin(x-y)=6\sin(x-y) \\ \Rightarrow \sin(3x+y)=7\sin(x-y)$

$\,8.\,$ If $\,\,\sin\theta+\sin2\theta=a,\,\text{and}\,\,\cos\theta+\cos2\theta=b\,$ show that $\,\,(a^2+b^2)(a^2+b^2-3)=2b$

Sol. $\,\,a^2+b^2\\=(\sin\theta+\sin2\theta)^2+(\cos\theta+\cos2\theta)^2\\=\sin^2\theta+\sin^22\theta+2\sin\theta.\sin2\theta+\cos^2\theta\\+\cos^22\theta+2\cos\theta\cos2\theta\\=(\sin^2\theta+\cos^2\theta)+(\sin^22\theta+\cos^22\theta)\\+2(\cos2\theta\cos\theta+\sin2\theta\sin\theta)\\=1+1+2\cos(2\theta-\theta)\\=2+2\cos\theta \cdots(1)\\ \therefore a^2+b^2-3\\=2+2\cos\theta-3$

Hence, $\,\,(a^2+b^2)(a^2+b^2-3)\\=(2+2\cos\theta)(2+2\cos\theta-3)\,\,\,[\text{By (1)}]\\=4\cos^2\theta-2\cos\theta+4\cos\theta-2\\=2(1+\cos2\theta)+2\cos\theta-2\\=2(\cos\theta+\cos2\theta)\\=2b$

$\,9.\,$  If $\,\,\theta=\frac{\pi}{2^n+1},\,\,$ show that, $\,\,2^n\cos\theta\cos2\theta\cos2^2\theta\cos2^3\theta\cdots\cos2^{n-1}\theta=1$

Sol. Let $\,\,P=2^n\cos\theta\cos2\theta\cos2^2\theta\cos2^3\theta \\\cdots\cos2^{n-1}\theta \\ \Rightarrow P.\sin\theta=2^{n-1}(2\sin\theta\cos\theta)\cos2\theta\cos2^2 \theta \\ \cos2^3\theta\cdots\cos2^{n-1} \\ \Rightarrow P.\sin\theta=2^{n-2}(2\sin2\theta\cos2\theta)\cos2^2  \theta \\ \cos2^3 \theta  \cdots\cos2^{n-1}  \\ \Rightarrow P.\sin\theta=2^{n-2}.\sin2^2\theta\cos2^2\theta\cos2^3\theta \\ \cdots\cos2^{n-1}\\ \Rightarrow \cdots \cdots \\ \Rightarrow P.\sin\theta=\sin2^n\theta \\ \Rightarrow P=\frac{\sin2^n\theta}{\sin\theta}\\~~~~~~~~=\frac{\sin(\pi-\theta)}{\sin\theta}\,\,\,\,[***]\\~~~~~~~~=\frac{\sin\theta}{\sin\theta}\\~~~~~~~~=1 \\ [***]=>\theta=\frac{\pi}{2^n+1} \Rightarrow 2^n\theta=\pi-\theta$

$\,10.\,$ Prove that , $\,\,(2\cos\theta-1)(2\cos2\theta-1)(2\cos2^2\theta-1)\cdots (2\cos2^{n-1}\theta-1)=\frac{2\cos2^n\theta+1}{2\cos\theta+1}$

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Sol. Let $\,\,P=(2\cos\theta-1)(2\cos2\theta-1)(2\cos2^2\theta-1)\cdots (2\cos2^{n-1}\theta-1) \\ \Rightarrow P(2\cos\theta+1)=(2\cos\theta+1)(2\cos\theta-1)(2\cos2\theta-1)(2\cos2^2\theta-1)\cdots (2\cos2^{n-1}\theta-1)\\ \Rightarrow P(2\cos\theta+1)=(4\cos^2\theta-1)(2\cos2\theta-1)(2\cos2^2\theta-1)\cdots (2\cos2^{n-1}\theta-1)\\ \Rightarrow P(2\cos\theta+1)=(2.2\cos^2\theta-1)(2\cos2\theta-1)(2\cos2^2\theta-1)\cdots (2\cos2^{n-1}\theta-1) \\ \Rightarrow P(2\cos\theta+1)=(2(1+\cos2\theta)-1)(2\cos2\theta-1)(2\cos2^2\theta-1)\cdots (2\cos2^{n-1}\theta-1)\\ \Rightarrow P(2\cos\theta+1)=(2\cos2\theta+1)(2\cos2\theta-1)(2\cos2^2\theta-1)\cdots (2\cos2^{n-1}\theta-1)\\ \Rightarrow \cdots\cdots \cdots\cdots \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ \Rightarrow P(2\cos\theta+1)=(2\cos 2^{n-1}\theta+1)(2\cos 2^{n-1}\theta-1) \\ \Rightarrow P(2\cos\theta+1)=(2\cos2^n\theta+1) \\ \Rightarrow P=\frac{2\cos2^n\theta+1}{2\cos\theta+1}$