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