question_answer

If \[y=a(1+\cos \theta )\] and \[x=a(\theta -\sin \theta ),\] find \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\,at\,\,\theta =\frac{\pi }{2}.\]

OR

If \[\cos \frac{x}{2}\cdot \cos \frac{x}{4}\cdot \cos \frac{x}{8}...=\frac{\sin x}{x},\] prove that  \[\frac{1}{{{2}^{2}}}{{\sec }^{2}}\frac{x}{2}+\frac{1}{{{2}^{4}}}{{\sec }^{2}}\frac{x}{4}+...=cose{{c}^{2}}x-\frac{1}{{{x}^{2}}}.\]

Answer:

Given \[y=a(1+\cos \theta )\]  and \[x=a(\theta -\sin \theta )\]

On differentiating both sides w.r.t.\[\theta \], we get

\[\frac{dy}{d\theta }=-\,a\sin \theta \] and \[\frac{dx}{d\theta }=a(1-cos\theta )\]

\[\therefore \]   \[\frac{dy}{dx}=\frac{\left( \frac{dy}{d\theta } \right)}{\left( \frac{dx}{d\theta } \right)}=\frac{-\,a\sin \theta }{a(1-cos\theta )}\]

\[=\frac{-\sin \theta }{1-cos\theta }\]

\[\Rightarrow \]   \[\frac{dy}{dx}=\frac{-\,2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\sin }^{2}}\frac{\theta }{2}}\]    \[\left[ \begin{align}   & \because \,\,\sin \theta =2\sin \frac{\theta }{2}\,\cos \frac{\theta }{2} \\  & \text{and}\,\,1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2} \\ \end{align} \right]\]

\[=-\cot \frac{\theta }{2}\]

Now,     \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)\]

\[=\frac{d}{dx}\left( -\cot \frac{\theta }{2} \right)\]

\[=\frac{d}{d\theta }\left( -\cot \frac{\theta }{2} \right)\times \frac{d\theta }{dx}\] \[\left[ \because \frac{d}{dx}\left[ f(\theta ) \right]=\frac{d}{d\theta }f(\theta )\times \frac{d\theta }{dx} \right]\]

\[=\frac{1}{2}\text{cose}{{\text{c}}^{2}}\frac{\theta }{2}\times \frac{1}{a(1-cos\theta )}\]

\[\left[ \because \,\frac{dx}{d\theta }=a(1-cos\theta )\Rightarrow \frac{d\theta }{dx}=\frac{1}{a(1-cos\theta )} \right]\]

\[=\frac{1}{2a}\frac{\cos e{{c}^{2}}\frac{\theta }{2}}{2{{\sin }^{2}}\frac{\theta }{2}}\]          \[\left[ \because \,\,1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2} \right]\]

\[=\frac{1}{4a}\cos e{{c}^{4}}\frac{\theta }{2}\]

\[{{\left[ \frac{{{d}^{2}}y}{d{{x}^{2}}} \right]}_{at\,\,\theta \,\,=\,\,\frac{\pi }{2}}}=\frac{1}{4a}\cos e{{c}^{4}}\left( \frac{\pi }{4} \right)\]

\[=\frac{1}{4a}\cdot {{(\sqrt{2})}^{4}}\]            \[\left[ \because \,\,\,\cos ec\frac{\pi }{4}=\sqrt{2} \right]\]            \[=\frac{1}{4a}\times 4=\frac{1}{a}\]

OR

Given, \[\cos \frac{x}{2}\cdot \cos \frac{x}{4}\cdot \cos \frac{x}{8}....=\frac{\sin x}{x}\]

Taking log on both sides, we get

\[\log \left[ \cos \frac{x}{2}\cdot \cos \frac{x}{4}\cdot \cos \frac{x}{8}.... \right]=\log \left[ \frac{\sin x}{x} \right]\]

\[\Rightarrow \] \[\log \cos \frac{x}{2}+\log \cos \frac{x}{4}+\log \cos \frac{x}{8}....\]

\[=\log \sin x-\log x\]

\[\left[ \begin{align}   & \because \log (m\cdot n\cdot p)=\log m+\log n \\  & +\log p\,\,\text{and}\,\,\log \frac{m}{n}=\log m-\log n \\ \end{align} \right]\]

On differentiating both sides w.r.t. x, we get

\[-\frac{1}{2}\cdot \frac{\sin x/2}{\cos x/2}-\frac{1}{4}\cdot \frac{\sin x/4}{\cos x/4}-\frac{1}{8}\cdot \frac{\sin x/8}{\cos x/8}-...\]

\[=\frac{\cos x}{\sin x}-\frac{1}{x}\]

\[\Rightarrow \] \[-\frac{1}{2}\tan \frac{x}{2}-\frac{1}{4}\tan \frac{x}{4}-\frac{1}{8}\tan \frac{x}{8}-...=\cot x-\frac{1}{x}\]

Again differentiating both sides w.r.t. x, we get

\[-\frac{1}{{{2}^{2}}}{{\sec }^{2}}\frac{x}{2}-\frac{1}{{{4}^{2}}}{{\sec }^{2}}\frac{x}{4}-\frac{1}{{{8}^{2}}}{{\sec }^{2}}\frac{x}{8}-...\]

\[=-\,\text{cose}{{\text{c}}^{2}}x+\frac{1}{{{x}^{2}}}\]

\[\Rightarrow \] \[-\frac{1}{{{2}^{2}}}{{\sec }^{2}}\frac{x}{2}+\frac{1}{{{2}^{4}}}{{\sec }^{2}}\frac{x}{4}+\frac{1}{{{2}^{6}}}{{\sec }^{2}}\frac{x}{8}+...\]

\[=\text{cose}{{\text{c}}^{2}}x-\frac{1}{{{x}^{2}}}\]

Hence proved.