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 $$ \sin 2x+\sin 4x=\cos 2x+\cos 4x $$


Just for your request from avtar:


Now put $n=5$.

  • $\begingroup$ 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? $\endgroup$ – Vic. Jun 23 '12 at 13:05
  • 1
    $\begingroup$ @Vic.: See above second comment. $\endgroup$ – 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, $\phi=n\pi/2, \frac{2n\pi}{3} + \pi/9,\frac{2n\pi}{3} - \pi/9$

  • $\begingroup$ The answer is correct, but I don't know why sin5ϕ−sinϕ=2cos3ϕsin2ϕ. How to get this? $\endgroup$ – Vic. Jun 23 '12 at 12:45
  • 2
    $\begingroup$ Vic.: It's trigonometric identity $\sin \alpha - \sin \beta = 2\sin\frac{\alpha-\beta}{2} \cos \frac{\alpha + \beta}{2}$. See Wikipedia $\endgroup$ – sdcvvc Jun 23 '12 at 13:09
  • 1
    $\begingroup$ @Vic:you can visit analyzemath.com/calculus/table/table_math_formulas.html $\endgroup$ – Aang Jun 23 '12 at 13:10

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


Now square both sides




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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.