6
$\begingroup$

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?

  • 1
    Where is this problem from?2011-05-12

3 Answers 3

7

$ \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 $).

1

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 $\cos{5A}\cdot\sin(A)$.

0

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