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

The maximum value of cos^2 (cos (33π + θ)) + sin^2 (sin (45π + θ)) is (1) 1 + sin^2 1 (2) 2 (3) 1 + cos^2 1 (4) cos^2 2

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

Commented by ajfour last updated on 26/Jun/17

mrW1 Sir, please see to this question..

$$\mathrm{mrW1}\:\mathrm{Sir},\:\mathrm{please}\:\mathrm{see}\:\mathrm{to}\:\mathrm{this}\:\mathrm{question}.. \\ $$

Commented by Tinkutara last updated on 26/Jun/17

I simplified the question to cos^2 (cos θ) + sin^2 (sin θ). I put θ = (π/2) and got 1 + sin^2 1. But I don′t know why it has maximum value at (π/2).

$$\mathrm{I}\:\mathrm{simplified}\:\mathrm{the}\:\mathrm{question}\:\mathrm{to} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{cos}\:\theta\right)\:+\:\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:\theta\right).\:\mathrm{I}\:\mathrm{put}\:\theta\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{got}\:\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{1}.\:\mathrm{But}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know} \\ $$$$\mathrm{why}\:\mathrm{it}\:\mathrm{has}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{at}\:\frac{\pi}{\mathrm{2}}. \\ $$

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

u=cos^2 (cos θ)+sin^2 (sin θ) =1+(1/2)(cos (2cos θ)−cos (2sin θ)) u′=sin θsin (2cos θ)+cos θsin (2sin θ) =sin θ{(2cos θ)−(((2cosθ)^3 )/(3!))+...}+ +cos θ{(2sin θ)−(((2sin θ)^3 )/(3!))+−..} =sin 2θ[{1−(((2cos θ)^2 )/(3!))+..}+{1−(((2sin θ)^2 )/(3!))+..}] u′=0 when sin 2θ=0⇒ θ=0, (π/2) double derivate will give us whether it is maxima or minima. 0 minima (π/2) maxima

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

Answered by mrW1 last updated on 26/Jun/17

cos (33π + θ)=−cos θ sin (45π + θ)=−sin θ cos^2 (cos (33π + θ)) + sin^2 (sin (45π + θ)) =cos^2 (−cos θ) + sin^2 (−sin θ) =cos^2 (cos θ) + sin^2 (sin θ) max. is when cos θ=0 and sin θ=±1 or θ=±(π/2) max.=1+sin^2 1 ⇒answer (1) is right.

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

Commented by Tinkutara last updated on 26/Jun/17

Thanks Sir!

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

Commented by ajfour last updated on 26/Jun/17

yes Sir, obviously!

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

Commented by Tinkutara last updated on 26/Jun/17

Can it be said that it is an increasing functiom?

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

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