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

The maximum value of

\cos^{2} (\cos (33 π + θ)) + \sin^{2} (\sin (45 π + θ))

is

(1) 1 + \sin^{2} 1

(2) 2

(3) 1 + \cos^{2} 1

(4) \cos^{2} 2

Commented by ajfour last updated on 26/Jun/17

mrW 1 Sir, please see to this 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 = \frac{π}{2}

and got 1 + \sin^{2} 1 . But I {don}^{'} t know

why it has maximum value at \frac{π}{2} .

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

u = \cos^{2} (\cos θ) + \sin^{2} (\sin θ)

= 1 + \frac{1}{2} (\cos (2 \cos θ) - \cos (2 \sin θ))

u^{'} = \sin θ \sin (2 \cos θ) + \cos θ \sin (2 \sin θ)

= \sin θ {(2 \cos θ) - \frac{{(2 \cos θ)}^{3}}{3!} + \dots} +

+ \cos θ {(2 \sin θ) - \frac{{(2 \sin θ)}^{3}}{3!} + - . .}

= \sin 2 θ [{1 - \frac{{(2 \cos θ)}^{2}}{3!} + . .} + {1 - \frac{{(2 \sin θ)}^{2}}{3!} + . .}]

u^{'} = 0 when \sin 2 θ = 0 \Rightarrow θ = 0, \frac{π}{2}

double derivate will give us

whether it is maxima or minima .

0 minima

\frac{π}{2} maxima

Answered by mrW1 last updated on 26/Jun/17

\cos (33 π + θ) = - \cos θ

\sin (45 π + θ) = - \sin θ

\cos^{2} (\cos (33 π + θ)) + \sin^{2} (\sin (45 π + 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 = \pm 1

or θ = \pm \frac{π}{2}

\max . = 1 + \sin^{2} 1

\Rightarrow answer (1) is right .

Commented by Tinkutara last updated on 26/Jun/17

Thanks Sir!

Commented by ajfour last updated on 26/Jun/17

yes Sir, obviously!

Commented by Tinkutara last updated on 26/Jun/17

Can it be said that it is an increasing

functiom ?