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How do you solve this equation for A: $~~\sin(5A) + \cos(5A)\sin(A) - \cos(3A) = 0$

I've tried expanding it many times, but I can't seem to be able to reduce it to a format I can work with. Is there a simpler method of solution than repeated expansion?

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Sketching the graph of the function suggests it has 10 real roots in $[0,2\pi)$ and is then periodic with period $2\pi$. Unless you enjoy solving order-10 polynomials, you may be restricted to numerical methods. –  Henry Mar 13 '11 at 14:56
    
Multiplying out the brackets gives sin(6A)+2sin(5A)-sin(4A)-2cos(3A)=0. I don't know if that helps or not. –  quanta Apr 12 '11 at 13:29
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Where is this problem from? –  Qiaochu Yuan May 12 '11 at 10:10
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3 Answers

up vote 7 down vote accepted

$$ \sin(5A) + \cos(5A) \sin(A) - \sin(3A) = 0 $$ Let $ x = e^{iA} $ and use De Moivre's, $$ \frac{x^5 - x^{-5}}{2i} + \frac{x^5 + x^{-5}}{2} \frac{x - x^{-1}}{2i} - \frac{x^3-x^{-3}}{2i} = 0 $$ Multiply by $ 4i x^6 $, $$ 2(x^{11} - x) + (x^{10} + 1)(x^2 - 1) - 2(x^9 - x^3) = $$ $$ (x^2 - 1)(x^{10} + 2x^9 + 2x + 1) = 0 $$ The phase of each root to the polynomial above (the ones with $ | x | = 1 $ at least) is a solution $ A $ to your equation (up to an integer addition of $ 2\pi $).

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Try using $\cos{A} = \sin\frac{\pi}{2}-A$ and the $\sin{A} + \sin{B}$ or $\cos{A} + \cos{B}$ formulas. You can also write $2 \cos{A}\sin{B} = \sin(A+B) - \sin(A-B)$, which actually will reduce $\cos{5A}\cdot\sin(A)$.

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Wolfram Alpha gives an explicit $12^{\text{th}}$ order polynomial and finds ten real roots.

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