Step-by-step explanation:
To prove:
[tex]\sin 3x=3\sin x\cos^2 x-\sin^3x[/tex]
Identities used:
[tex]\sin(A+B)=\sin A\cos B+\cos A\sin B[/tex] ......(1)
[tex]\sin 2A=2\sin A\cos A[/tex] ........(2)
[tex]\cos 2A=\cos^2A-\sin^2 A[/tex] .......(3)
Taking the LHS:
[tex]\Rightarrow \sin 3x=\sin (x+2x)[/tex]
Using identity 1:
[tex]\Rightarrow \sin (x+2x)=\sin x\cos 2x+\cos x\sin 2x[/tex]
Using identities 2 and 3:
[tex]\Rightarrow \sin x(\cos ^2x-\sin^2 x)+\cos x(2\sin x\cos x)\\\\\Rightarrow \sin x\cos^2x-\sin^3x+2\sin x\cos^2x\\\\\Rightarrow 3\sin x\cos^2x-\sin^3x[/tex]
As, LHS = RHS
Hence proved
