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

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.

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

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

• I appreciate your reply but my question was about a general method which can be used to find maximum and minimum values of all trigonometric expressions including this one. BTW what do you mean by C-S? – KBC Jul 20 '17 at 17:22
• @KBC It's Cauchy-Schwartz inequality. I think we have no a general method to solve math problems. Give me a problem and I'll try to solve it. – Michael Rozenberg Jul 20 '17 at 17:26

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.

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 $$a\sin(x)\cos(x)=\frac{a}{2}\sin(2x)$$

• What happens if $a^2+b^2=0$? – Michael Rozenberg Jul 20 '17 at 17:28
• dear Michael, then is $$a=b=0$$ and the equation $$a\sin(x)+b\cos(x)$$ makes no sence – Dr. Sonnhard Graubner Jul 20 '17 at 17:33