I encountered this problem in Khan Academy's differential calculus program.
I made an equation to show the distance between the functions
$$a=2\cos(x)-\sin(2x)$$
and took its derivative, which turned out to be:
$$a'=-2\sin(x)-2\cos(2x)$$
To find the minimum, I set the derivative to be equal to zero to find a critical point:
$$0=-2\sin(x)-2\cos(2x)$$
I then spent a very frustrating time trying to find a trig identity that would help me find the value of x. I rearranged things using $\cos(2x)=1-2\sin^2(x)=2\cos^2(x)-1$ but got nowhere. Finally I tested the multiple choice answers and selected the correct one to be able to move on, hoping that Khan would offer to let me see the solution (it sometimes does this, sometimes not). Unfortunately, it didn't! Can anyone give me a nudge in the right direction for how to find $x$ when there is a sine and cosine with a different argument (one has $x$, one has $2x$). As an aside, I also thought of using tangent, but would have needed $\sin(x)$ and $\cos(x)$ for that, right?
Best Answer
You are on the right path, however I am confused as to where you got "stuck" after substituting $\cos{2x} = 1 - 2\sin^2{x}$. That leaves you with only one unary function, $\sin{x}$. Consider the polynomial pattern that appears in the new equation.
Hint