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How do you even approach this question. Is there something fundamental that i'm missing here.

Renascence_5.
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Julian Lopez
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  • What you’re missing is that this is not one equation for $f$, this is a family of equations, an infinite number of them — one for each $x$ (for each $x \in [0,2\pi]$ actually, because once you have those, the others don’t say anything new). They determine $f(t)$ for $t \in [-1,1]$ (the range of $\sin$) uniquely. You can solve them for $f$ explicitly, or use tricks like $\sin x = \cos(x-\pi/2)$ or $\sin^2 x + \cos^2 x = 1$ to obtain the answer without doing that. – Alex Shpilkin Jan 14 '17 at 07:29

3 Answers3

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Hint: use that $\cos(\alpha-90^\circ)=\sin(\alpha)$ for any $\alpha$.

Arthur
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Hint: $f(\sin x) = 1 -2\sin^2x \implies f(x) = 1 - 2x^2 \implies f(\cos x) = 1-....$

DeepSea
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$$f(\sin(x))=\cos(2x)=1-2\sin^2(x)$$ HINT : Change of variable : $\quad X=\sin(x)$ $$f(X)=1-2X^2$$ Now, the function $f$ is known. You can put $\cos(x)$ into it : $$f(\cos(x))= ...?$$

JJacquelin
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