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.
 A: what i have found is this expression
$$32\,\sin \left( x \right)  \left( \cos \left( x \right)  \right) ^{4}-
24\,\sin \left( x \right)  \left( \cos \left( x \right)  \right) ^{2}+
2\,\sin \left( x \right) -3\,\cos \left( x \right) 
$$
and the first derivative with respect to $x$ is given by
$$f'(x)=32\sin(x)\cos(x)^4-24\sin(x)\cos(x)^2+2\sin(x)-3\cos(x)$$
A: 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).
A: I suppose that you are allowed to plot the function for $0\leq x \leq 2\pi$. Doing it, you notice that the maximum is located close to $3$; let us say $\pi$.
So, use Taylor expansions around $x=\pi$ to get $$f(x)=2\sin (5x)-3\cos( x)=3-10 (x-\pi )-\frac{3}{2} (x-\pi )^2+\frac{125}{3} (x-\pi )^3+O\left((x-\pi
   )^4\right)$$ Ignoring all terms of degree higher than $3$, consider, as an approximation, that you need to find the maximum of $$g(x)=3-10 (x-\pi )-\frac{3}{2} (x-\pi )^2+\frac{125}{3} (x-\pi )^3$$ The derivative is $$g'(x)=125 (x-\pi )^2-3 (x-\pi )-10$$ which is just a quadratic. Solve it as usual and keep the root which is the closest to $\pi$; it will be $$x_*=\frac{1}{250} \left(3-\sqrt{5009}+250 \pi \right)\approx 2.8705$$ Now $$g(x_*)=\frac{539973+5009 \sqrt{5009}}{187500}\approx 4.7706$$
Using calculus (not simple), you would find that the maximum value of the function is $\approx 4.8613$ corresponding to $x\approx 2.8450$.
