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

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(A) Given $f(x) = \sin^2 x + \cos^4 x + 2$.Using $\cos^2 x = 1 - \sin^2 x$,we have $f(x) = \sin^2 x + (1 - \sin^2 x)^2 + 2 = \sin^2 x + 1 - 2\sin^2 x

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$T_1 = 2T_2$

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(A) Given $f(x) = \sin^2 x + \cos^4 x + 2$.Using $\cos^2 x = 1 - \sin^2 x$,we have $f(x) = \sin^2 x + (1 - \sin^2 x)^2 + 2 = \sin^2 x + 1 - 2\sin^2 x + \sin^4 x + 2 = \sin^4 x - \sin^2 x + 3$.Using $\sin^2 x = \frac{1 - \cos 2x}{2}$,we can express $f(x)$ in terms of $\cos 2x$ or $\cos 4x$. Alternatively,$f(x) = \sin^2 x + (1 - \sin^2 x)^2 + 2 = \sin^2 x + 1 - 2\sin^2 x + \sin^4 x + 2 = \sin^4 x - \sin^2 x + 3$.Since $\sin^2 x = \frac{1 - \cos 2x}{2}$,the period of $\sin^2 x$ is $\pi$. The period of $\sin^4 x$ is also $\pi$. Thus,$T_1 = \pi$.Wait,let's re-evaluate: $f(x) = \sin^2 x + \cos^4 x + 2$. $f(x+\pi) = \sin^2(x+\pi) + \cos^4(x+\pi) + 2 = (-\sin x)^2 + (-\cos x)^4 + 2 = \sin^2 x + \cos^4 x + 2 = f(x)$.Checking for $\pi/2$: $f(x+\pi/2) = \sin^2(x+\pi/2) + \cos^4(x+\pi/2) + 2 = \cos^2 x + \sin^4 x + 2 
eq f(x)$. So $T_1 = \pi$.Now for $g(x) = \cos(\cos x) + \cos(\sin x)$.$g(x+\pi/2) = \cos(\cos(x+\pi/2)) + \cos(\sin(x+\pi/2)) = \cos(-\sin x) + \cos(\cos x) = \cos(\sin x) + \cos(\cos x) = g(x)$.Thus,$T_2 = \pi/2$.Therefore,$T_1 = 2T_2$.

Let $f(x) = \sin^2 x + \cos^4 x + 2$ and $g(x) = \cos(\cos x) + \cos(\sin x)$. If the periods of $f(x)$ and $g(x)$ are $T_1$ and $T_2$ respectively,then:

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