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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.

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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

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

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  • $\begingroup$ 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? $\endgroup$ – KBC Jul 20 '17 at 17:22
  • $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Jul 20 '17 at 17:26
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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.

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

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

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