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

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.

• This is not an equation, it is a function, or an expression if you want. An equation involves an equality. Feb 8, 2013 at 2:22
• If it bothers you that much your free to edit the question, but I believe most people understand what I'm trying to say here. Feb 8, 2013 at 2:27
• I does not bother me that much. But it never hurts to use the appropriate words. Most people would still understand the question if you replaced equation by turtle. Feb 8, 2013 at 2:30
• touche, haha that made me laugh, I'll go ahead and change it then Feb 8, 2013 at 2:32

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 $-\sqrt 5$

• How do you know the frequency of the two waves will ever be the same? Feb 8, 2013 at 2:25
• @AmateurMathGuy: In this case they are. In some problems they are, in others not. If they are integer multiples of some frequency you can use the en.wikipedia.org/wiki/… to get something like $a\sin ax \sin bx$ and the max/min is again $\pm a$. If the frequencies don't match up, at some time they will be almost at peak together and you can take the sum of the maxima as the amplitude. Feb 8, 2013 at 2:32

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

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

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