Let f(x) = sin^2 x + cos^4 x + 2 and g(x) = c | vedclass

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(C) $f(x) = \sin^2 x + \cos^4 x + 2$$= \sin^2 x + (1 - \sin^2 x)^2 + 2$$= \sin^2 x + 1 - 2\sin^2 x + \sin^4 x + 2$$= \sin^4 x - \sin^2 x + 3$$= (\sin^

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$T_1 = T_2$

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(C) $f(x) = \sin^2 x + \cos^4 x + 2$$= \sin^2 x + (1 - \sin^2 x)^2 + 2$$= \sin^2 x + 1 - 2\sin^2 x + \sin^4 x + 2$$= \sin^4 x - \sin^2 x + 3$$= (\sin^2 x - \frac{1}{2})^2 + 3 - \frac{1}{4} = (\sin^2 x - \frac{1}{2})^2 + \frac{11}{4}$Using $\sin^2 x = \frac{1 - \cos 2x}{2}$,we get $f(x) = (\frac{1 - \cos 2x}{2} - \frac{1}{2})^2 + \frac{11}{4} = (-\frac{\cos 2x}{2})^2 + \frac{11}{4} = \frac{1}{4} \cos^2 2x + \frac{11}{4} = \frac{1}{4} (\frac{1 + \cos 4x}{2}) + \frac{11}{4} = \frac{1}{8} \cos 4x + \frac{23}{8}$.The period $T_1$ of $\cos 4x$ is $\frac{2\pi}{4} = \frac{\pi}{2}$.For $g(x) = \cos(\cos x) + \cos(\sin x)$,$g(x + \frac{\pi}{2}) = \cos(\cos(x + \frac{\pi}{2})) + \cos(\sin(x + \frac{\pi}{2})) = \cos(-\sin x) + \cos(\cos x) = \cos(\sin x) + \cos(\cos x) = g(x)$.Thus,the period $T_2 = \frac{\pi}{2}$.Therefore,$T_1 = T_2$.

Let $f(x) = \sin^2 x + \cos^4 x + 2$ and $g(x) = \cos(\cos x) + \cos(\sin x)$. Also,let the period of $f(x)$ and $g(x)$ be $T_1$ and $T_2$ respectively,then:

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