Prove $\cos 2α = 2 \sin^2β + 4\cos (α + β) \sin α \sin β + \cos 2(α + β)$

Question

We have to prove

Prove that $\cos 2α = 2 \sin^2β + 4\cos (α + β) \sin α \sin β + \cos 2(α + β)$.

It is difficult for me to start off. I request someone to provide me a hint.

Answer 1

Hint:

$$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$

From that identity you can substitute the three $\cos$ you have.

Answer 2

$$2\sin\alpha\sin\beta=\cos(\alpha-\beta)+\cos(\alpha+\beta)$$

Now like Help with trigonometric equation, apply $$\cos(\alpha-\beta)\cos(\alpha+\beta)=\cos^2\alpha-\sin^2\beta$$

Finally use $\cos2A=2\cos^2A-1$ for $\cos2(\alpha+\beta)$

Answer 3

$$P=2 \sin^2β + 4\cos (α + β) \sin α \sin β + \cos 2(α + β)$$

We have that:

$$2\cos (\alpha + \beta)\sin (\beta)=\sin (\alpha + 2\beta)-\sin (\alpha)$$

so,

$$4\cos (\alpha + \beta)\sin (\beta)\sin(\alpha)=2\sin (\alpha + 2\beta)\sin(\alpha)-2\sin^2(\alpha)$$

but,

$$2\sin (\alpha + 2\beta)\sin(\alpha)=\cos(2\beta)-\cos[2(\alpha + \beta)]$$

then

$$P=2 \sin^2β +\cos(2\beta)-\cos[2(\alpha + \beta)]-2\sin^2(\alpha)+\cos[2(\alpha + \beta)]\\ P=1-\cos(2\beta)+\cos(2\beta)-2\sin^2(\alpha)=\cos(2\alpha)$$