Question

verify the identity sin3x = 3 sinx - 4 sin^3x

Answer 1

The identity is

$\sin 3x=3\sin x-4\sin ^3x$

Since the angle sum formula of sine is:

$\sin (a+b)=\sin (a)\cos (b)\text{ + cos(a)sin(b)}$

and the double angle formula of cosine is:

$\cos (2a)=cos^2(a)-sin^2(a)\text{ = }1-2\sin ^2(a)$

then:

$\begin{gathered} \sin (3x)\text{ = sin(2x +x) = sin(2x)cos(x) +cos(2x)sin(x)} \\ =(2\sin (x)\cos (x))\cos (x)+(1-2sin^2(x))\sin (x) \\ =\text{ }(2\sin (x)\cos ^2(x))+sin^2(x)-2\sin ^3(x) \\ =\text{ 2sin(x) }\cdot(1-\sin ^2(x))\text{ }+sin^2(x)-2\sin ^3(x) \\ =\text{ }2\sin (x)-2\sin ^3(x)\text{ + }\sin (x)-2\sin ^3(x) \\ =\text{ 3}\sin (x)-4\sin ^3(x)\text{ } \end{gathered}$