Differentiation (Part-38) | S N De

 

$\,10.\,$ If $\,\,\cos\alpha+\cos\beta=\cos\frac{3\pi}{7},\,\,\sin\alpha+\sin\beta=\sin\frac{3\pi}{7},\,\,$ find $\,\,\cos^2\frac{\alpha-\beta}{2}.$

Sol. $\,\,(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=\cos^2\frac{3\pi}{7}+\sin^2\frac{3\pi}{7} \\ \Rightarrow \cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta+\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta=1 \\ \Rightarrow (\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=1 \\ \Rightarrow 1+1+2\cos(\alpha-\beta)=1 \\ \Rightarrow 2\left[1+\cos(\alpha-\beta)\right]=1 \\ \Rightarrow 2.2\cos^2\frac{\alpha-\beta}{2}=1 \\ \Rightarrow \cos^2\frac{\alpha-\beta}{2}=\frac 14.$

$11.\,\,$ If $\,\cos A=\frac{x}{y+z},\,\cos B=\frac{y}{z+x},\,\,\cos C=\frac{z}{x+y}\,\,$ show that $\,\,\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2=1$

Sol. $\,\,\tan^2\frac A2\\=\frac{1-\cos A}{1+\cos A}\\=\frac{1-\frac{x}{y+z}}{1+\frac{x}{y+z}}\\=\frac{y+z-x}{y+z+x}$

Similarly, $\,\,\tan^2\frac B2\\=\frac{1-\cos B}{1+\cos B}\\=\frac{x+z-y}{x+z+y}$

and $\,\,\tan^2\frac C2\\=\frac{1-\cos C}{1+\cos C}\\=\frac{x+y-z}{x+y+z}$

$\,\,\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2\\=\frac{y+z-x}{y+z+x}+\frac{x+z-y}{x+z+y}+\frac{x+y-z}{x+y+z}\\=\frac{y+z-x+x+z-y+x+y-z}{x+y+z}\\=\frac{x+y+z}{x+y+z}\\=1$

$\,12.\,$ If $\,\frac{\cos x}{a_1}=\frac{\cos2x}{a_2}=\frac{\cos3x}{a_3},\,\,$ show that, $\,\,\frac{\sin^2x}{2}=\frac{2a_2-a_1-a_3}{4a_2}$

Sol. Let $\,\frac{\cos x}{a_1}=\frac{\cos2x}{a_2}=\frac{\cos3x}{a_3}=\frac 1k\,\,(k \neq 0) \\ \Rightarrow a_1=k\cos x,\,a_2=k\cos2x,\,a_3=k\cos3x.$

Now , $\quad \frac{2a_2-a_1-a_3}{4a_2}\\=\frac{2\cos2x-(\cos x+\cos3x)}{4\cos2x}\\=\frac{2\cos2x-2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}}{4\cos2x}\\=\frac{2\cos2x-2\cos2x\cos x}{4\cos2x}\\=\frac{2\cos2x(1-\cos x)}{4\cos2x}\\=\frac{1-\cos x}{2}\\=\frac{2\sin^2\frac x2}{2}\\=\sin^2\frac x2$

$\,13.\,$ If $\,\cos\alpha=\frac{x}{y+z},\,\,\cos\beta=\frac{y}{z+x},\,\cos\gamma=\frac{z}{x+y},\,\,$ show that,  $\,\,\tan^2\frac{\alpha}{2}+\tan^2\frac{\beta}{2}+\tan^2\frac{\gamma}{2}=1$

Sol. Same as Question No. $11.$

$\,14.\,$ If $\,\cos x+\cos y=a,\,\,\sin x+\sin y=b, \,\,$ show that $\,\,\sin2x+\sin2y=2ab \left(1-\frac{2}{a^2+b^2}\right)$

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Sol. We have, $\,\cos x+\cos y=a,\,\,\sin x+\sin y=b \\ \Rightarrow a^2+b^2=(\cos x+\cos y)^2+(\sin x+\sin y)^2 \\ \Rightarrow a^2+b^2=\cos^2x+\cos^2y+2\cos x\cos y\\+\sin^2x+\sin^2y+2\sin x\sin y \\ \Rightarrow a^2+b^2=(\sin^2x+\cos^2x)+(\sin^2y+\cos^2y)\\+2(\cos x\cos y+\sin x\sin y) \\ \Rightarrow a^2+b^2=1+1+2\cos(x-y) \\ \Rightarrow a^2+b^2=2[1+\cos(x-y)] \\ \Rightarrow \frac{a^2+b^2}{2}=1+\cos(x-y)\\ \Rightarrow \cos(x-y)=\frac{a^2+b^2}{2}-1 \rightarrow(1)$

Again, $\,\,\frac ba=\frac{\sin x+\sin y}{\cos x+\cos y} \\ \Rightarrow \frac ba=\frac{2\sin\frac{x+y}{2}\cos\frac{x+y}{2}}{2\cos\frac{x+y}{2}\cos\frac{x+y}{2}} \\ \Rightarrow \frac ba=\tan\frac{x+y}{2}$

Hence, $\,\,\sin(x+y)\\=\sin\left(2 \times \frac{x+y}{2}\right)\\=\frac{2\tan\frac{x+y}{2}}{1+\tan^2\frac{x+y}{2}}\\=\frac{2.b/a}{1+(b/a)^2}\\=\frac{2ab}{a^2+b^2}\rightarrow(2)$

Now, $\,\,\sin2x+\sin2y\\=2\sin\frac{2x+2y}{2}\cos\frac{2x-2y}{2}\\=2\sin(x+y)\cos(x-y)\\=2.\frac{2ab}{a^2+b^2}.\left(\frac{a^2+b^2}{2}-1 \right)\,\,[\text{By (1),(2)}]\\=2ab-\frac{4ab}{a^2+b^2}\\=2ab\left(1-\frac{2}{a^2+b^2}\right)$

$\,15.\,$ In $\,\,\Delta PQR,\,\,\angle R=90^{\circ},\,$ and if $\,\tan\frac P2,\tan\frac Q2\,\,$ are the roots of $\,\,ax^2+bx+c=0\,\,$ then show that $\,\,a+b=c.$

Sol. In $\,\,\Delta PQR,\,\,\angle R=90^{\circ},\,$ and so $\,\,P+Q=90^{\circ} \\ \Rightarrow\frac 12(P+Q)=45^{\circ} \rightarrow(1)$

Since, $\,\tan\frac P2,\tan\frac Q2\,\,$ are the roots of $\,\,ax^2+bx+c=0\,\,$ so, $\,\,\tan\frac P2+\tan\frac Q2=-\frac ba,\\ \text{and}\,\,\tan\frac P2.\tan\frac Q2=\frac ca.$

Now, from (1), we get $\,\,\tan(\frac{P+Q}{2})=\tan 45^{\circ}\\ \Rightarrow \frac{\tan\frac P2+\tan\frac Q2}{1-\tan\frac P2.\tan\frac Q2}=1 \\ \Rightarrow \tan\frac P2+\tan\frac Q2=1-\tan\frac P2\tan\frac Q2 \\ \Rightarrow -\frac ba=1-\frac ca \\ \Rightarrow -b=a-c \\ \Rightarrow a+b=c.$