sin^{2n}x + cos^{2n}x lies between | vedclass

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Question

Since $0 \leq \sin ^{2 \pi} x \leq \sin ^{2} x$

$0 \leq \cos ^{2 \pi} x \leq \cos ^{2} x$

$\left[\because \sin ^{4} x=\sin ^{2} x, \sin ^{2} x \

Vedclass · Accepted Answer

$0$ and $1$

Vedclass · Answer

Since $0 \leq \sin ^{2 \pi} x \leq \sin ^{2} x$

$0 \leq \cos ^{2 \pi} x \leq \cos ^{2} x$

$\left[\because \sin ^{4} x=\sin ^{2} x, \sin ^{2} x \leq \sin ^{2} x \cdot 1ight.$

$\left.	herefore \sin ^{4} x \leq \sin ^{2} x 	ext { etc. }ight]$

$\Rightarrow \quad 0<\sin ^{2 \pi} x+\cos ^{2 n} x \leq \sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \quad 0<\sin ^{2 n} x+\cos ^{2 n} x \leq 1$

$sin^{2n}x + cos^{2n}x$ lies between

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