Prove that $\frac{\sin^2 (2\alpha)}{\sin^2 (\alpha)}=4-4\sin^2 (\alpha)$.

Question

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

I have to solve the left hand side to equal the right hand side.

Answer 1

We have $\sin(2\alpha) = 2 \sin \alpha \cos \alpha$, hence $$ \frac{\sin^2(2\alpha)}{\sin^2(\alpha)} = \frac{4 \sin^2 (\alpha) \cos ^2 (\alpha)}{\sin^2(\alpha)} = 4 \cos^2 \alpha = 4 - 4\sin^2 \alpha $$ as $\cos^2 \alpha + \sin^2 \alpha = 1$.

Answer 2

Left-hand side: $\sin (2t) = 2\sin(t)\cos(t)$, so $$\sin^2(2t) = (\sin (2t))^2 = (2\sin t\cos t)^2 = 4\sin^2 t \cos^2 t$$

That gives us $$\dfrac{\sin^2(2t)}{\sin^2 t} = \require{cancel}\dfrac{4\cancel{\sin^2 t} \cos^2 t}{\cancel{\sin^2 t}} = 4\cos^2 t$$

On the right-hand side, note that $$4-4\sin^2 t = 4(1-\sin^2 t) = 4\cos^2 t$$

Answer 3

HINT:

Use Double angle formula $$\sin2x=2\sin x\cos x$$

Then, Fundamental Theorem of Trigonometry

Answer 4

$$\frac{\sin^2(2\alpha)}{\sin^2(\alpha)} = \left(\frac{\sin(2\alpha)}{\sin(\alpha)}\right)^2 = \left(\require{cancel}\frac{2\cancel{\sin(\alpha)}\cos(\alpha)}{\cancel{\sin(\alpha)}}\right)^2= 4\cos^2(\alpha) = 4 - 4\sin^2(\alpha)$$

List of idenitities used:

• $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$
• $\sin^2(\alpha)+\cos^2(\alpha) = 1$

Answer 5

Using $\sin x =\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos x =\frac{e^{ix}+e^{-ix}}{2}$, and $\sin^2 x + \cos^2 x = 1$, write

$$\boxed{\dfrac{\sin^2 2\alpha}{\sin^2 \alpha}} = \left(\frac{\sin 2\alpha}{\sin \alpha}\right)^2 = \left(\frac{(e^{2i\alpha}-e^{-2i\alpha})/2i}{(e^{i\alpha}-e^{-i\alpha})/2i}\right)^2$$

$$ = \left(\frac{(e^{i\alpha}+e^{-i\alpha})(e^{i\alpha}-e^{-i\alpha})}{(e^{i\alpha}-e^{-i\alpha})}\right)^2 = \left(e^{i\alpha}+e^{-i\alpha}\right)^2=\left(2\cdot\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}\right)\right)^2$$

$$ = 4\cos^2 \alpha = 4(1-\sin^2\alpha) = \boxed{4 - 4\sin^2 \alpha}$$