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Trigonometrical Equations
normal
$sin^{2n}x + cos^{2n}x$ lies between
A
$-1$ and $1$
B
$0$ and $1$
C
$1$ and $2$
D
None of these
Solution
Since $0 \leq \sin ^{2 \pi} x \leq \sin ^{2} x$
$0 \leq \cos ^{2 \pi} x \leq \cos ^{2} x$
$\left[\because \sin ^{4} x=\sin ^{2} x, \sin ^{2} x \leq \sin ^{2} x \cdot 1\right.$
$\left.\therefore \sin ^{4} x \leq \sin ^{2} x \text { etc. }\right]$
$\Rightarrow \quad 0<\sin ^{2 \pi} x+\cos ^{2 n} x \leq \sin ^{2} x+\cos ^{2} x=1$
$\Rightarrow \quad 0<\sin ^{2 n} x+\cos ^{2 n} x \leq 1$
Standard 11
Mathematics