Max and min values of the sum of cosines?

Question

I have a function $m(t) = cos 2 \pi (2000)t + 2 cos2 \pi (4000)t$

I am confused on how to determine the max and min values of m(t). I know that if they had the same frequency the max and min would be 3 and -3 respectively.

For the max I plugged in $0$ for $t$.

This gave me a max of $3$. How can you determine the min without using graphing software?

Answer 1

Note that

$$\cos x + 2 \cos 2x = \cos x+4\cos^2 x-2$$

thus

$$m(t) = \cos (2 \pi (2000)t) + 2 \cos(2 \pi (4000)t)= 4\cos^2(2 \pi (2000)t)+\cos(2 \pi (2000)t)-2$$

and for $y=\cos (2 \pi (2000)t)\in[-1,1]$

$$f(y)=4y^2+y-2\implies f'(y)=8y+1=0\implies f_{min}=f\left(-\frac18\right)=-\frac{33}{16}$$

and maximum on the boundary for $$f_{max}=f(1)=3$$

Answer 2

Rewrite like: $m(t)=\cos(\omega t)+2cos(2\omega t)$ where $\omega=4000\pi$

Then: $\cos(\omega t)+2cos(2\omega t)=\cos(\omega t)+2(2cos^2(\omega t)-1)$

$\to m(t)=4 \cos^2(\omega t)+\cos(\omega t)-2$

$\to m(t)/4=\cos^2(\omega t)+1/4\cos(\omega t)-1/2=(cos(\omega t)+1/8)^2-1/16-1/2$

$\to m(t)=4(cos(\omega t)+1/8)^2-9/4$

Can you take it from here?