Find a function $f:\mathbb{R}^*\longrightarrow\mathbb{R}$, such as $f>0$ and it satisfies the following equation: $$f(\sin(x))+f(\cos(x))=\left(\frac{2}{\sin(2x)}\right)^2,\quad \forall x\in\mathbb{R}\setminus\left\{\frac{k\pi}{2},k\pi\;k\in\mathbb{Z}\right\} $$ I don't know how to start this, any idea would be helpful.
Find a function that satisfies $f(\sin(x))+f(\cos(x))=\left(\frac{2}{\sin(2x)}\right)^2$
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calculus
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1I'd start with some identities; see what you can get out of $\sin(2x)$. – 2017-01-25
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0If one wants to find more than one solution then it's useful to note that if $f$ is a solution then so is $f+g$ if $g(\sin(x)) + g(\cos(x)) = 0$. For example $g(x) = C\left[\frac{1}{2} - x^2\right]$ has this property (one also have to make sure that $f+g>0$ is satisfied) – 2017-01-25
3 Answers
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Hint:
$$\left(\frac{2}{\sin(2x)}\right)^2=\frac{1}{\sin^2x}+\frac{1}{\cos^2 x}$$
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Hint
$$\left(\frac{2}{\sin(2x)}\right)^2 = \left(\frac{1}{\sin x \cos x}\right)^2 = \frac{\sin^2x+\cos^2x}{\sin^2 x \cos^2 x} = \frac{1}{\cos^2 x}+\frac{1}{\sin^2 x}$$
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3That's a pretty big hint :) – 2017-01-25
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0You're right, so I put it in bold ;-D. – 2017-01-25
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$$f(\sin x)+f(\cos x)=\frac{1}{\sin^2x\cos^2x}$$ one solution is the set of all function with property $$f(x)+f(y)=\frac{1}{x^2y^2}$$ where $x^2+y^2=1$.
Another, with $$f(\sin x)+f(\cos x)=\frac{1}{\sin^2x}+\frac{1}{\cos^2x}$$ is the set of all function with property $$f(x)+f(y)=\frac{1}{x^2}+\frac{1}{y^2}$$ where $x^2+y^2=1$.
Better is that take $\sin x=\dfrac{2t}{1+t^2}$ and $\cos x=\dfrac{1-t^2}{1+t^2}$ thus $$f(\dfrac{2t}{1+t^2})+f(\dfrac{1-t^2}{1+t^2})=\frac{(1+t^2)^2}{2t(1-t^2)}$$