Express $\frac{\sin 3a}{\sin a}$ with only $\cos a$

Question

How can I express $\dfrac{\sin 3a}{\sin a}$ while using only $\cos a$?

Thanks in advance

Answer 1

We know that $$\sin 3a=\sin(2a+a)= \ldots = 3\sin a - 4\sin^3 a$$

So $$\frac{\sin 3a}{\sin a}=3-4\sin^2a=3-4+4\cos^2 a=4\cos ^2a-1$$

Answer 2

$$ \sin3a=\sin(2a+a)=\sin2a\cos a+\cos2a\sin a =2\sin a\cos^2a+(\cos^2a-\sin^2a)\sin a $$

Can you go on?

Answer 3

$$\frac{\sin3a}{\sin a}=\frac{3\sin a-4\sin^3a}{\sin a}$$ $$=3-4\sin^2a$$ $$=3-4(1-\cos^2a)$$ $$=3-4+4\cos^2a$$ $$=4\cos^2a-1$$