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Question Number 40277 by Raj Singh last updated on 18/Jul/18

Answered by MJS last updated on 18/Jul/18

$$\frac{d}{d\theta }\left[{\sin }^p\theta {\cos }^q\theta \right]={\sin }^{p-1}\theta {\cos }^{q-1}\theta (p{\cos }^2\theta -q{\sin }^2\theta )$$$$\mathrm{zeros}\mathrm{at}$$$$\sin \theta =0\Rightarrow \theta =2n\pi$$$$\cos \theta =0\Rightarrow \theta =(2n+1)\frac{\pi }{2}$$$$$$$$p{\cos }^2\theta -q{\sin }^2\theta =0$$$${pc}^2-{qs}^2=0$$$$p(1-s^2)-{qs}^2=0$$$$p-{ps}^2-{qs}^2=0$$$${\sin }^2\theta =\frac{p}{p+q}$$$$\Rightarrow {\cos }^2\theta =1-\frac{p}{p+q}=\frac{q}{p+q}$$$$\Rightarrow {\tan }^2\theta =\frac{p}{q}$$$$\Rightarrow \theta =\arctan \sqrt{\frac{p}{q}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jul/18

$$y={sin}^p\theta {cos}^q\theta$$$$lny=plnsin\theta +qlncos\theta$$$$\frac{1}{y}\frac{dy}{d\theta }=pcot\theta -qtan\theta$$$$formax/min\frac{dy}{d\theta }=0$$$$pcot\theta -qtan\theta =0$$$$qtan\theta =pcot\theta$$$${tan}^2\theta =\frac{p}{q}sotan\theta =\sqrt{\frac{p}{q}}$$$$\frac{dy}{d\theta }={sin}^p\theta {cos}^q\theta (pcot\theta -qtan\theta )$$$$\frac{dy}{d\theta }=y(pcot\theta -qtan\theta )$$$$\frac{d^{2}y}{d{\theta }^2}=\frac{dy}{d\theta }(pcot\theta -qtan\theta )+y(-{pcosec}^2\theta -{qsec}^2\theta )$$$$\frac{d^{2}y}{d{\theta }^2}=\frac{dy}{d\theta }(p.\sqrt{\frac{q}{p}}-q.\sqrt{\frac{p}{q}})-y\{p(1+\frac{q}{p})+q(1+\frac{p}{q}$$$$\frac{d^{2}y}{d{\theta }^2}=0-2y(p+q)=-2y(p+q)$$$$attan\theta =\sqrt{\frac{p}{q}}\frac{d^{2}y}{d{\theta }^2}is-ve$$$$soattan\theta =\sqrt{\frac{p}{q}}{sin}^p\theta {cos}^q\theta ismaximum$$$$$$

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