<s>[INST] Find maximum $2\sin 5x-3\cos x$. Is it possible to find the maximum of $2\sin 5x-3\cos x $ without using calculus nor numerical methods?
I suspect there is a way to play around with trig identities until the expression is only in terms of either $\cos$ or $\sin$. I have done some trials but have only managed to make the expression messier.
 [/INST] We have
$$2\sin(5x)-3\cos(x)=\\2\sin(3x+2x)-3\cos(x)=\\2\sin(3x)\cos(2x)+2\cos(3x)\sin(2x)-3\cos(x)$$
We make use of the identities: $$\sin(3x)=3\sin(x)-4\sin^3(x)\\ \cos(3x)=4\cos^3(x)-3\cos(x)$$
and $$\sin(2x)=2\cos(x)\sin(x)\\\cos(2x)=1-2\sin^2(x)$$
Thus we obtain
$$2[(3\sin(x)-4\sin^3(x))(1-\sin^2(x))]+2[(4\cos^3(x)-3\cos(x))2\sin(x)\cos(x)]-3\cos(x)=\\16\sin^5(x)-20\sin^3(x)+6\sin(x)+16\cos^4(x)\sin(x)-12\sin(x)\cos^2(x)-3\cos(x)$$
We can now make use of the rather obscure identities $$\sin(x)=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$$ and $$\cos(x)=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$$
to get an expression solely on terms of $u=\tan{\frac{x}{2}}$
Still,I am not sure if this will help though (in terms of not using calculus or numerical methods per request).
</s>