Deceptively difficult trig problem

**TwistedOne151** · April 28th 2008, 05:44 AM

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.

**Isomorphism** · April 28th 2008, 06:01 AM

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

,

Tell us your answer. I have a nagging suspicion that you have the right answer

**TwistedOne151** · April 28th 2008, 09:13 AM

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.

**mr fantastic** · April 28th 2008, 08:02 PM

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 .....?

**elizsimca** · April 28th 2008, 08:48 PM

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!

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