How do I find the maximum and minimum value of any trigonometric expression

Question

Is there any proper general method to find the maximum and the minimum value of any trigonometric expression (for example trigonometric expressions of the form $a \sin x +b \cos x$ or $a\sin x\times\cos x$ or any other such expression) elegantly (without using calculus)? I do not think that my question is broad. I am asking for a technique that works in most of the cases.

Answer 1

By C-S $$a\sin{x}+b\cos{x}\leq\sqrt{(a^2+b^2)(\sin^2x+\cos^2x)}=\sqrt{a^2+b^2}.$$ The equality occurs for $(a,b)||(\sin{x},\cos{x})$.

From here $$\max(a\sin{x}+b\cos{x})=\sqrt{a^2+b^2}$$ and $$\min(a\sin{x}+b\cos{x})=-\sqrt{a^2+b^2}.$$ $a\sin{x}\cos{x}=\frac{1}{2}a\sin2x$ and from here $$\max(a\sin{x}\cos{x})=\frac{|a|}{2}$$ and

$$\min(a\sin{x}\cos{x})=-\frac{|a|}{2}.$$

Answer 2

note that we have the following inequalities for your first question its that $$-\sqrt{a^2+b^2}\leq a\sin(x)+b\cos(x)\leq\sqrt{a^2+b^2}$$ and for your second question we have $\sin(x)\cos(x)=\frac{\sin(2x)}{2}$.You can prove first one by multiplying and dividing by $\sqrt{a^2+b^2}$ and using $\frac{a}{\sqrt{a^2+b^2}}=\cos(a)$ so $\frac{b}{\sqrt{a^2+b^2}}=\sin(a)$ hence we have $\sqrt{a^2+b^2}(\sin(a+x))$ now $-1\leq \sin \leq 1$ hence the proof.

Answer 3

for your first example you can write $$\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2}}\cos(x)\right)=\sqrt{a^2+b^2}(\sin(x)\cos(\phi)+\sin(\phi)\cos(x))$$ $$=\sqrt{a^2+b^2}\sin(x+\phi)$$ where $$\\cos(\phi)=\frac{a}{\sqrt{a^2+b^2}}$$ and $$\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$ your second term $$a\sin(x)\cos(x)=\frac{a}{2}\sin(2x)$$