Respuesta :
Step-by-step explanation:
To prove:
[tex]\cos 3\theta=3\cos^3 \theta-3\sin^2 \theta\cos \theta[/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)
[tex]\sin^2A+\cos^2A=1[/tex] .......(4)
Taking the LHS:
[tex]\Rightarrow \cos 3\theta=\cos (\theta +2\theta )[/tex]
Using identity 1:
[tex]\Rightarrow \cos (\theta +2\theta )=\cos \theta \cos 2\theta-\sin \theta \sin 2\theta[/tex]
Using identities 2 and 3:
[tex]\Rightarrow \cos \theta (\cos ^2\theta-\sin^2 \theta )-\sin \theta (2\sin \theta \cos \theta )\\\\\Rightarrow \cos^3\theta -\sin^2\theta \cos \theta -2\sin^2 \theta \cos \theta \\\\\Rightarrow \cos^3\theta -3\sin^2\theta \cos \theta[/tex]
Using identity 4:
[tex]\Rightarrow \cos^3\theta -3(1-\cos^2\theta) \cos \theta\\\\\Rightarrow \cos^3\theta-3\cos \theta+3\cos^3\theta\\\\\Rightarrow 4\cos^3\theta -3\cos \theta[/tex]
As, LHS = RHS
Hence proved
