Simplifying a cosine triple angle expansion

Question

The problem is express $\dfrac{1+\cos 3\theta}{1+\cos \theta }$ in terms of increasing powers of $\sin \theta.$

I have derived upto $1-8\sin^2 \dfrac{\theta}{2}+16\sin^4 \dfrac{\theta}{2}$, but am unaware how to proceed further. Thanks in advance.

Answer 1

Your equation can be transformed first into:

$$(2\cos(\theta)-1)^2$$

which we further can transform using the relationship $\cos(\theta)=1-2\sin^2\frac{\theta}{2}$ to:

$$(2(1-2\sin^2\frac{\theta}{2})-1)^2=(1-4\sin\frac{\theta}{2})^2$$

The half-angle identity for Sine is $sin\frac{\theta}{2}=\sqrt{\frac{1-\cos{\theta}}{2}}$, which unfortunatelly does not allow us for reducing the above-given term further.