Question

Show that cos(3theta) = 4cos3(theta) − 3cos (theta).

Answer 1

Step-by-step explanation:

To prove:

$\cos 3\theta=3\cos^3 \theta-3\sin^2 \theta\cos \theta$

Identities used:

$\cos(A+B)=\cos A\cos B+\sin A\sin B$      ......(1)

$\sin 2A=2\sin A\cos A$        ........(2)

$\cos 2A=\cos^2A-\sin^2 A$    .......(3)

$\sin^2A+\cos^2A=1$       .......(4)

Taking the LHS:

$\Rightarrow \cos 3\theta=\cos (\theta +2\theta )$

Using identity 1:

$\Rightarrow \cos (\theta +2\theta )=\cos \theta \cos 2\theta-\sin \theta \sin 2\theta$

Using identities 2 and 3:

$\Rightarrow \cos \theta (\cos ^2\theta-\sin^2 \theta )-\sin \theta (2\sin \theta \cos \theta )\\\\\Rightarrow \cos^3\theta -\sin^2\theta \cos \theta -2\sin^2 \theta \cos \theta \\\\\Rightarrow \cos^3\theta -3\sin^2\theta \cos \theta$

Using identity 4:

$\Rightarrow \cos^3\theta -3(1-\cos^2\theta) \cos \theta\\\\\Rightarrow \cos^3\theta-3\cos \theta+3\cos^3\theta\\\\\Rightarrow 4\cos^3\theta -3\cos \theta$

As, LHS = RHS

Hence proved