# Thread: Deceptively difficult trig problem

1. ## Deceptively difficult trig problem

I'm not sure what I'm missing here:
Solve $a\sin(2\theta)-b\cos(2\theta)=\sin\theta$ for θ in terms of a and b.
Every attempt I've made has gotten rather messy.

--Kevin C.

2. Originally Posted by TwistedOne151
I'm not sure what I'm missing here:
Solve $a\sin(2\theta)-b\cos(2\theta)=\sin\theta$ for θ in terms of a and b.
Every attempt I've made has gotten rather messy.

--Kevin C.
Hi Kevin,

3. I get to $4(a^2+b^2)\cos^4\theta-4a\cos^3\theta+(1-4(a^2+b^2))\cos^2\theta+4a\cos\theta+(b^2-1)=0$, a quartic equation for $\cos\theta$. Attempting to solve that via the standard "double completing the square" method led to a horrible cubic.

--Kevin C.

4. Originally Posted by TwistedOne151
I'm not sure what I'm missing here:
Solve $a\sin(2\theta)-b\cos(2\theta)=\sin\theta$ for θ in terms of a and b.
Every attempt I've made has gotten rather messy.

--Kevin C.
Do you have reason to believe a 'nice' general solution exists .....?

5. I think a lot of times as math students we want a nice or "pretty" if you will solution to problems. Sometimes though, we get solutions that look messy and automatically think they must be wrong.

I do this all the time.

Eventually, you will learn to trust and not doubt yourself so much!