Differentiation (Part-38) | S N De

 

$\,6.\,$ If $\,\,\sin\theta=k\sin(\theta+\phi),\,\,$ show that, $\,\,\tan(\theta+\phi)=\frac{\sin\phi}{\cos\phi-k}.$

Sol. $\,\,\sin\theta=k\sin(\theta+\phi),\,\,\text{(given)} \\ \Rightarrow \sin(\theta+\phi-\phi)=k\sin(\theta+\phi)\\ \Rightarrow \sin(\theta+\phi)\cos\phi-\cos(\theta+\phi)\sin\phi\\ =k\sin(\theta+\phi)\\ \Rightarrow \sin(\theta+\phi)(\cos\phi-k)=\cos(\theta+\phi)\sin\phi \\ \Rightarrow \frac{\sin(\theta+\phi)}{\cos(\theta+\phi)}=\frac{\sin\phi}{\cos\phi-k}\\ \Rightarrow \tan(\theta+\phi)=\frac{\sin\phi}{\cos\phi-k}\quad\text{(showed)}$ 

$\,7.\,$ If $\,\,\frac{\cot(\alpha-\beta)}{\cot\alpha}+\frac{\cos^2\gamma}{\cos^2\alpha}=1,\,\,$ show that, $\,\,\tan^2\gamma+\tan\alpha\cot\beta=0.$

Sol. $\,\,\frac{\cot(\alpha-\beta)}{\cot\alpha}+\frac{\cos^2\gamma}{\cos^2\alpha}=1\\ \Rightarrow \frac{\cos^2\gamma}{\cos^2\alpha}\\=1-\frac{\cot(\alpha-\beta)}{\cot\alpha}\\=1-\frac{\cos(\alpha-\beta)\sin\alpha}{\sin(\alpha-\beta)\cos\alpha}\\=\frac{\sin(\alpha-\beta)\cos\alpha-\cos(\alpha-\beta)\sin\alpha}{\sin(\alpha-\beta)\cos\alpha}\\=\frac{\sin(\alpha-\beta-\alpha)}{\sin(\alpha-\beta)\cos\alpha}\\ \Rightarrow \frac{\cos^2\gamma}{\cos\alpha}=\frac{-\sin\beta}{\sin(\alpha-\beta)}\\ \Rightarrow \cos^2\gamma=\frac{-\sin\beta\cos\alpha}{\sin(\alpha-\beta)} \\ \Rightarrow\sec^2\gamma=\frac{\sin(\alpha-\beta)}{-\cos\alpha\sin\beta} \\ \Rightarrow 1+\tan^2\gamma=\frac{\sin(\alpha-\beta)}{-\cos\alpha\sin\beta} \\ \Rightarrow \tan^2\gamma\\=-\left[1+\frac{\sin(\alpha-\beta)}{\cos\alpha\sin\beta}\right]\\=-\left[1+\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\sin\beta}\right]\\=-\frac{\sin\alpha\cos\beta}{\cos\alpha\sin\beta}\\=-\tan\alpha\cot\beta. \\ \therefore \tan^2\gamma+\tan\alpha\cot\beta=0.\quad\text{(showed)}$ 

$\,8.\,$ If $\,\,\tan\theta=\frac{x\sin\alpha+y\sin\beta}{x\cos\alpha+y\cos\beta},\,\,$ prove that, $\,\,x\sin(\theta-\alpha)+y\sin(\theta-\beta)=0.$

Sol. $\,\,\tan\theta=\frac{x\sin\alpha+y\sin\beta}{x\cos\alpha+y\cos\beta}\\ \Rightarrow \frac{\sin\theta}{\cos\theta}=\frac{x\sin\alpha+y\sin\beta}{x\cos\alpha+y\cos\beta} \\ \Rightarrow x(\sin\theta\cos\alpha-\cos\theta\sin\alpha)\\=-y(\sin\theta\cos\beta-\cos\theta\sin\beta) \\ \Rightarrow x\sin(\theta-\alpha)=-y\sin(\theta-\beta) \\ \Rightarrow x\sin(\theta-\alpha)+y\sin(\theta-\beta)=0\quad\text{(proved)}$

$\,9(i).\,$ Show that for all real value of $\,x,\,\,\,c-\sqrt{a^2+b^2}\leq a\cos x +b\sin x+c \leq c+\sqrt{a^2+b^2}.$

Sol. $\,\,a\cos x+b\sin x+c\\=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\right)+c\\=\sqrt{a^2+b^2}(\sin\alpha\cos x+\cos\alpha\sin x)+c\,\,[**]\\=\sqrt{a^2+b^2}\sin(\alpha+x)+c$

Note[**] : $\,\,\sin\alpha=\frac{a}{\sqrt{a^2+b^2}},\,\,\cos\alpha=\frac{b}{\sqrt{a^2+b^2}}.$ 

Now, $\,\, -1\leq \sin(\alpha+x)\leq 1 \\ \Rightarrow -\sqrt{a^2+b^2}+c\leq \sqrt{a^2+b^2}\sin(\alpha+x)+c\\ \leq \sqrt{a^2+b^2}+c \\ \Rightarrow c-\sqrt{a^2+b^2}\leq a\cos x +b\sin x+c \\ \leq c+\sqrt{a^2+b^2}\quad\text{(showed)} $  

$\,9(ii).\,$ Express $\,(\cos\theta-\sin\theta)\,$ in the form $\,\,r\cos(\theta+\alpha)\,\,$ and $\,\,\,(\sqrt3\sin\theta+\cos\theta)\,\,$ in the form $\,\,r\sin(\theta+\beta).$

Sol. $\,\,\,\cos\theta-\sin\theta\\=\sqrt2\left(\frac{1}{\sqrt2}\cos\theta-\frac{1}{\sqrt2}\sin\theta\right)\\=\sqrt2\left(\cos\theta\cos\frac{\pi}{4}-\sin\theta\sin\frac{\pi}{4}\right)\\=\sqrt2\cos\left(\theta+\frac{\pi}{4}\right)$

Again,$ \, \sqrt3\sin\theta+\cos\theta \\=2\left(\frac{\sqrt3}{2}\sin\theta+\frac 12\cos\theta\right)\\=2\left(\sin\theta\cos\frac{\pi}{6}+\cos\theta\sin\frac{\pi}{6}\right)\\=2\sin\left(\theta+\frac{\pi}{6}\right)$

$\,9(iii).\,$ Find the maximum and minimum values of $\,\,5\cos\theta+12\sin\theta+12.$

Sol.  $\,\,5\cos\theta+12\sin\theta+12\\ =13\left(\frac{5}{13}\cos\theta+\frac{12}{13}\sin\theta\right)+12\\=13(\sin\alpha\cos\theta+\cos\alpha\sin\theta)+12\,\,[**]\\=13\sin(\theta+\alpha)+12 \rightarrow(1)$

Note[**] : $\,\,\sin\alpha=\frac{5}{13},\,\,\,\cos\alpha=\frac{12}{13}.$

Now,$\,\,\,-1\leq \sin(\theta+\alpha)\leq 1 \\ \Rightarrow -13 \leq 13 \sin(\theta+\alpha)\leq 13 \\ \Rightarrow -13+12 \leq 13\sin(\theta+\alpha)+12\leq 13+12 \\ \Rightarrow -1 \leq 5\cos\theta+12\sin\theta+12\leq 25\quad [\text{By (1)}]$

Hence,the maximum and minimum values of $\,\,5\cos\theta+12\sin\theta+12\,\,$ is $\,\,25\,\,$ and $\,\,-1\,\,$ respectively.