# Solving $\sin (5\phi)-\sin \phi=\sin (2\phi)$

The question is : Find the general solution of this equation $$\sin (5\phi)-\sin \phi=\sin (2\phi)$$ I tried to expand $\sin (5\phi)$ and $\sin (2\phi)$, so the equation only contains $\cos\phi$ and $\sin \phi$.

But I can't make it into a form like $\sin(\phi+a)=n$ to get the general solution.

There's another question like this :

Find the general solution of this equation $$\sin 2x+\sin 4x=\cos 2x+\cos 4x$$

## 3 Answers

Just for your request from avtar:

$\sin(n\theta)=\binom{n}{1}\cos^{n-1}(\theta)\sin(\theta)-\binom{n}{3}\cos^{n-3}(\theta)\sin^3(\theta)+...$

Now put $n=5$.

• Thank you for this equation (Haven't learn it before, I'm gonna write it down later:D)I can expand sin5ϕ−sinϕ and get sin(5A) =sin(2A + 3A) = sin(2A)cos(3A) + cos(2A)sin(3A) = 10sin^5(A) - 8 sin^3(A) - 4sin A = 2sin A [ 5sin^4(A) - 4sin^2(A) - 2 ] But still have no idea why sin5ϕ−sinϕ=2cos3ϕsin2ϕ..Could you give some more detail? – Vic. Jun 23 '12 at 13:05
• @Vic.: See above second comment. – mrs Jun 23 '12 at 13:18

$\sin 5\phi-\sin\phi = 2\cos 3\phi \sin 2\phi$, therefore, equation becomes, $2\cos 3\phi \sin2\phi = \sin 2\phi$ giving possibilities, $\sin2\phi=0$ or $\cos3\phi = 1/2$. For first equation, $2\phi=n\pi \implies \phi=n\pi/2$. For second case, $3\phi = 2n\pi + \pi/3$ or $2n\pi - \pi/3$ giving $\phi = \frac{2n\pi}{3} + \pi/9$ or $\frac{2n\pi}{3} - \pi/9$ . therefore solutions are, $\phi=n\pi/2, \frac{2n\pi}{3} + \pi/9,\frac{2n\pi}{3} - \pi/9$

• The answer is correct, but I don't know why sin5ϕ−sinϕ=2cos3ϕsin2ϕ. How to get this? – Vic. Jun 23 '12 at 12:45
• Vic.: It's trigonometric identity $\sin \alpha - \sin \beta = 2\sin\frac{\alpha-\beta}{2} \cos \frac{\alpha + \beta}{2}$. See Wikipedia – sdcvvc Jun 23 '12 at 13:09
• @Vic:you can visit analyzemath.com/calculus/table/table_math_formulas.html – Aang Jun 23 '12 at 13:10

I think I see a method of solving the second equation. First rewrite the equation as

$$\sin2x-\cos2x=\cos4x-\sin4x$$

Now square both sides

$$\sin^22x-2\sin2x\cos2x+\cos^22x=\cos^24x-2\cos4x\sin4x+\sin^24x$$

$$1-\sin4x=1-2\cos4x\sin4x$$

$$\sin4x(1-2\cos4x)=0$$

This form should be a lot easier to solve, though new roots may have been introduced by squaring both sides.