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Question Number 16740 by Tinkutara last updated on 26/Jun/17

$$\mathrm{The}\mathrm{maximum}\mathrm{value}\mathrm{of}$$$${\cos }^2\left(\cos (33\pi +\theta )\right)+{\sin }^2\left(\sin (45\pi +\theta )\right)$$$$\mathrm{is}$$$$\left(1\right)1+{\sin }^{2}1$$$$\left(2\right)2$$$$\left(3\right)1+{\cos }^{2}1$$$$\left(4\right){\cos }^{2}2$$

Commented by ajfour last updated on 26/Jun/17

$$\mathrm{mrW}1\mathrm{Sir},\mathrm{please}\mathrm{see}\mathrm{to}\mathrm{this}\mathrm{question}..$$

Commented by Tinkutara last updated on 26/Jun/17

$$\mathrm{I}\mathrm{simplified}\mathrm{the}\mathrm{question}\mathrm{to}$$$${\cos }^2\left(\cos \theta \right)+{\sin }^2\left(\sin \theta \right).\mathrm{I}\mathrm{put}\theta =\frac{\pi }{2}$$$$\mathrm{and}\mathrm{got}1+{\sin }^{2}1.\mathrm{But}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}$$$$\mathrm{why}\mathrm{it}\mathrm{has}\mathrm{maximum}\mathrm{value}\mathrm{at}\frac{\pi }{2}.$$

Commented by prakash jain last updated on 26/Jun/17

$$u={\cos }^2\left(\cos \theta \right)+{\sin }^2\left(\sin \theta \right)$$$$=1+\frac{1}{2}(\cos \left(2\cos \theta \right)-\cos \left(2\sin \theta \right))$$$$u^{\prime }=\sin \theta \sin \left(2\cos \theta \right)+\cos \theta \sin \left(2\sin \theta \right)$$$$=\sin \theta \{\left(2\cos \theta \right)-\frac{{\left(2\cos \theta \right)}^3}{3!}+...\}+$$$$+\cos \theta \{\left(2\sin \theta \right)-\frac{{\left(2\sin \theta \right)}^3}{3!}+-..\}$$$$=\sin 2\theta [\{1-\frac{{\left(2\cos \theta \right)}^2}{3!}+..\}+\{1-\frac{{\left(2\sin \theta \right)}^2}{3!}+..\}]$$$$u^{\prime }=0\mathrm{when}\sin 2\theta =0\Rightarrow \theta =0,\frac{\pi }{2}$$$$\mathrm{double}\mathrm{derivate}\mathrm{will}\mathrm{give}\mathrm{us}$$$$\mathrm{whether}\mathrm{it}\mathrm{is}\mathrm{maxima}\mathrm{or}\mathrm{minima}.$$$$0\mathrm{minima}$$$$\frac{\pi }{2}\mathrm{maxima}$$

Answered by mrW1 last updated on 26/Jun/17

$$\cos (33\pi +\theta )=-\cos \theta$$$$\sin (45\pi +\theta )=-\sin \theta$$$${\cos }^2\left(\cos (33\pi +\theta )\right)+{\sin }^2\left(\sin (45\pi +\theta )\right)$$$$={\cos }^2(-\cos \theta )+{\sin }^2(-\sin \theta )$$$$={\cos }^2\left(\cos \theta \right)+{\sin }^2\left(\sin \theta \right)$$$$\max .\mathrm{is}\mathrm{when}\cos \theta =0\mathrm{and}\sin \theta =\pm 1$$$$\mathrm{or}\theta =\pm \frac{\pi }{2}$$$$\max .=1+{\sin }^{2}1$$$$$$$$\Rightarrow \mathrm{answer}\left(1\right)\mathrm{is}\mathrm{right}.$$

Commented by Tinkutara last updated on 26/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Commented by ajfour last updated on 26/Jun/17

$$\mathrm{yes}\mathrm{Sir},\mathrm{obviously}!$$

Commented by Tinkutara last updated on 26/Jun/17

$$\mathrm{Can}\mathrm{it}\mathrm{be}\mathrm{said}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{an}\mathrm{increasing}$$$$\mathrm{functiom}?$$

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