Proving a Trigonometrical Identity: $2\sin2x\cos6x=\sin8x-\sin4x$

Question

How can i prove
$$2\sin2x\cos6x=\sin(8 x)-\sin(4 x)$$

As per this identity $\sin2x=2\sin x\cos x$ but how to proceed from here

Answer 1

Hint:you can use the well-known prosthaphaeresis formulas $$\sin x \cos y=\frac{\sin(x+y) - \sin(y-x)}{2}$$ and get $$2\sin2x\cos6x=2\frac{\sin(8x) - \sin(4x)}{2}=\sin(8 x)-\sin(4 x)$$

Answer 2

Hint: use this identity; $$sin (A) cos (B) = 1/2 (sin (A + B) + sin (A - B) )$$

Answer 3

Write $8x=6x+2x$ and $4x=6x-2x$ and expand.

Answer 4

The easiest way is to start by using your product-to-sum formulas. http://www.sosmath.com/trig/prodform/prodform.html So, for example, on the RHS, sin(8x) - sin(4x) = sin(6x+2x) - sin(6x-2x) on the LHS, letting a = 2x and b = 6x, you use the identity for sin(a)cos(b). (If you haven't had these formulas, you can derive them easily enough with your angle addition formulas.) The rest should now be clear.