[Math] Show that $\sin x(\sin 2x+ \sin 4x +\sin 6x) = \sin 3x \sin 4x$

geometrytrigonometry

I got a trigonometric question in an exam and it was confused me to solve.The question contains 3 parts.

  1. Show that $\sin 2x + \sin 4x + \sin 6x = (1 + 2\cos 2x) \sin 4x$.
  2. By using the above show that $\sin x(\sin 2x + \sin 4x + \sin 6x) = \sin 3x \sin 4x$.
  3. Derive the values for $\sin (\pi/12)$

I solved the 1st part as bellow. But 2nd and 3rd was unable to solve. So I am Looking for a help to solve this. Thanx šŸ™‚

sin2x+sin4x+sin6x=(1+2cos2x)sin4x
Using sum to product identities
LHS=sin6x+sin2x+sin4x
sinĪ±+sinĪ²=2 sinā”((Ī±+Ī²)/2)  cosā”((Ī±-Ī²)/2)
LHS=sin4x+2 sinā”((6x+2x)/2)  cosā”((6x-2x)/2)
=sin4x+(2 sinā”4xcos2x)
=sin4x(1+2cos2x)

Best Answer

Once you are able to write down $\sin(3x)$ in the following way, you are good to go. $$ \begin{align} \sin(3x) &= \sin(x+2x)\\ & = \sin(x)\cos(2x) + \cos(x)\sin(2x)\\ &=\sin(x) \cos(2x) + \cos(x)2\sin(x)\cos(x)\\ &=\sin(x) (\cos(2x) + 2\cos^2(x)) \\ &=\sin(x) (\cos(2x) + 2\cos^2(x) -1 +1) \\ &=\sin(x) (\cos(2x) + \cos(2x) +1)\\ &=\sin(x) (1 + 2\cos(2x) ) \end{align} $$

Related Question