Solve Trigo Prob: Prove \sin{4\alpha} = a

[ron_jay]

Homework Statement

if: $$(1+\sqrt{1+a})\tan{\alpha} = (1+\sqrt{1-a})$$

Prove that: $$\sin{4 \alpha} = a$$

Homework Equations

$$\cos{2\alpha} = 1-2\sin^2{\alpha}$$

$$\tan{\alpha} = \sqrt{\frac{1-\cos{2\alpha}}{1+\cos{2\alpha}}}$$

The Attempt at a Solution

We probably have to simplify to express $$\tan{\alpha}$$ in terms of $$\sin{4\alpha}$$ and hence need to use the equations given above. Trying componendo and dividendo did no good as it got too complicated.

$$\sin{\alpha}+\sin{\alpha}\sqrt{1+a} = \cos{\alpha}+\cos{\alpha}\sqrt{1-a}$$

$$(\sin{\alpha}-\cos{\alpha})^2 = (\cos{\alpha}\sqrt{1-a}-\sin{\alpha}\sqrt{1+a})^2$$

$$1-\sin{2\alpha}=\cos^2{\alpha}(1-a)+\sin^2{\alpha}(1+a)-\sin{2\alpha}\sqrt{1-a^2}$$

Though I could simplify after this, I cannot understand how to get $$\sin{4\alpha}$$ out of it. What is looking even more difficult is how to remove the $$\sqrt{1-a^2}$$ term.

I would be grateful if somebody could guide me in solving this.

 

[varygoode]

Wait, is the $$\sqrt{1-a^2}$$ inside of your last sine term or is it a scalar multiplying your last sine term?

 

[ron_jay]

No, it is a scalar multiplying the last sine term.

 

[ron_jay]

Looks like I got it; beginning from where I left,

$$1-\sin{2\alpha} = \cos^2{\alpha-a\cos^2{\alpha}+\sin^2{\alpha}+a\sin^2{\alpha}-(\sin{2\alpha})(\sqrt{1-a^2})$$

-After taking 'a' common...

$$Since, \cos^2{\alpha}+\sin^2{\alpha}=1$$

$$\Rightarrow\sin{2\alpha}=a\cos{2\alpha}+(\sin{2\alpha})(\sqrt{1-a^2})$$

$$\Rightarrow(\sin{2\alpha}-a\cos{2\alpha})^2=(\sqrt{1-a^2}.\sin{2\alpha})^2$$

$$\Rightarrow\sin^2{2\alpha}+a^2\cos^2{2\alpha}-a\sin{4\alpha}=\sin^2{2\alpha}-a^2\sin^2{2\alpha}$$

the term $$sin^2{2\alpha}$$ gets canceled on both sides and taking 'a' common we get>>

$$\Rightarrow \sin{4\alpha}=a$$

VOILA!