Maximum value of $\sin(3x) + 2 \cos(3x)$ without calculus

Question

This is a follow up to the question I posted here in which I was seeking assistance in finding the maximum value of a trig expression. I realize now that using the first and second derivative tests is one way to solve the problem but a few users hinted at another way that didn’t require calculus.

I’m not sure how to solve this without Calculus and am curious to see how this would be done. Again the problem is to find the maximum value of the expression $$\sin(3x) + 2 \cos(3x) \text{ where } - \infty < x < \infty$$ without the aid of Calculus.

Answer 1

You can look to write $\sin (3x)+2\cos(3x)=a\sin (3x+\phi)$. This will be possible when the frequency of the two waves is the same, as here. Now use the angle-sum formula to get $$\sin (3x)+2\cos(3x)=a \sin(3x) \cos (\phi) + a \cos (3x) \sin(\phi)$$

Then $a \cos(\phi)=1,\\ a \sin(\phi)=2$

so $\tan (\phi)=2, \\ \sin(\phi)=\frac 2{\sqrt 5},\\ \cos(\phi)=\frac 1{\sqrt 5},\\ a=\sqrt 5$

Now we can see that the maximum is $\sqrt 5$ and the minimum is $-\sqrt 5$

Answer 2

Hint: $a \sin(t) + b \cos(t) = r \sin(t+s)$ for some $r$ and $s$.

Answer 3

Let $y=A\sin\theta+B\cos\theta$

So, $y=A\frac{2t}{1+t^2}+B\frac{1-t^2}{1+t^2}$ where $t=\tan \frac{\theta}2$

So, $$(y+B)t^2-2At+(y-B)=0$$

As $t$ is real, the discriminant $(2A)^2-4(y-B)(y+B)\ge0$ $\implies y^2\le A^2+B^2\implies -\sqrt{A^2+B^2}\le y\le \sqrt{A^2+B^2}$

$\implies -\sqrt{A^2+B^2}\le A\sin\theta+B\cos\theta\le \sqrt{A^2+B^2}$

Here $\theta=3x,A=1,B=2\implies A^2+B^2=1^2+2^2=5$

$\implies -\sqrt5\le \sin3x+2\cos2x\le \sqrt5$