Proving $1-\cos2a+2\sin a \sin3a = 2 \sin^22a$ and $\sin^22a - \sin^2a = \sin3a\sin a$

Question

Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.

1) Prove: $$1 - \cos2a + 2 \sin a\, \sin3a = 2 \sin^22a$$

I started off with rewriting $1 - \cos2a$ to $2 \sin^2a$. That gets me to: $$2 \sin^2a + 2\sin a\, \sin3a = 2 \sin^22a$$ but I am lost after this point.

2) Prove: $$\sin^22a - \sin^2a = \sin3a \,\sin a$$

I have begun with replacing $\sin^22a$ with $4 \sin^2a \,\cos^2a$.

That brings me to: $$4 \sin^2a\,\cos^2a - \sin^2a = \sin3a\, \sin a$$ or $$\sin^2a(4 \cos^2a -1) = \sin3a\sin a$$ but I cannot find the solution.

So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?

Thanks for your attention. I’m looking forward to your reply.

Answer 1

Using http://mathworld.wolfram.com/WernerFormulas.html,

$$2\sin a\sin3a=\cos(3-1)a-\cos(3+1)a$$ Then use $\cos2y=1-2\sin^2y$

For the second, use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

See also: Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

Answer 2

Hint:

Linearise the r.h.s.: $$2\sin^22a=1-\cos 4a,$$

and, in the l.h.s., $\;2\sin a\sin 3a$.

Answer 3

1. Hint:

Consider the trigonometric identities $$\sin^2\alpha+\cos^2\alpha=1$$ $$\sin(2\alpha)=2\sin\alpha \cos\alpha\quad \cos(2\alpha)=\cos^2\alpha-\sin^2\alpha$$ $$\sin(3\alpha)=3\sin\alpha-4\sin^3\alpha$$

SOLUTION

$$\sin^2\alpha(1-\sin^2\alpha)=\sin^2\alpha\cos^2\alpha$$ $$= \sin\alpha(\sin\alpha-\sin^3\alpha)\quad \mid·4$$ $$\Rightarrow \sin\alpha(\sin\alpha+3\sin\alpha-4\sin^3\alpha)=4\sin^2\alpha ·\cos^2\alpha$$ $$\sin\alpha(\sin\alpha+\sin(3\alpha))=\sin^2\alpha+\sin\alpha \sin(3\alpha)=4\sin^2\alpha ·\cos^2\alpha=\sin^2(2\alpha)$$ $$\Rightarrow 2\sin^2\alpha+2\sin\alpha\sin(3\alpha)=\bigl(1-\cos^2\alpha+\sin^2\alpha\bigr)+2\sin\alpha\sin(3\alpha)=2\sin^2\alpha$$ Hence $$1-\cos(2\alpha)+2\sin\alpha\sin(3\alpha)=2\sin^2\alpha$$

2. SOLUTION

   $$\sin\alpha(3-4\sin^2\alpha)=\sin(3\alpha)$$ Thus $$\sin\alpha(4-4\sin^2\alpha-1)=\sin\alpha(4\cos^2\alpha-1)=4\sin\alpha\cos^2\alpha-\sin\alpha=\sin(3\alpha)$$ $$\Rightarrow4\sin^2\alpha\cos^2\alpha-\sin^2\alpha=\sin^2(2\alpha)-\sin^2\alpha=\sin(3\alpha)\sin(\alpha)$$