Option A
[tex]\cos 3 x=\cos x-4 \cos x \sin ^{2} x[/tex]
Solution:
Given that we have to rewrite with only sin x and cos x
Given is cos 3x
[tex]cos 3x = cos(x + 2x)[/tex]
We know that,
[tex]\cos (a+b)=\cos a \cos b-\sin a \sin b[/tex]
Therefore,
[tex]\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x[/tex] ---- eqn 1
We know that,
[tex]\sin 2 x=2 \sin x \cos x[/tex]
[tex]\cos 2 x=\cos ^{2} x-\sin ^{2} x[/tex]
Substituting these values in eqn 1
[tex]\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)[/tex] -------- eqn 2
We know that,
[tex]\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x[/tex]
Applying this in above eqn 2, we get
[tex]\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)[/tex]
[tex]\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}[/tex]
[tex]\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x[/tex]
Therefore,
[tex]\cos 3 x=\cos x-4 \cos x \sin ^{2} x[/tex]
Option A is correct
