# deriving sine half angle formula from sine double angle formula

I can derive the sin, cos and tan half angle formulas from the cosine double angle formula. But I'm having trouble deriving the sine half angle formula from the sine double angle formula

Below is my attempt at deriving sine half angle formula from sine double angle formula

And I could go no further.

Could someone provide me with a hint?

Edit 1:

below is the sine half identity I want to derive from sine double angle identity

• Which half-angle formula are you attempting to derive? Also note that when taking square roots you have to consider the $\,\pm\,$ sign choice. – dxiv Sep 30 '18 at 6:06
• Easier to start with: $$\cos(2y) = \cos^2(y)-\sin^2(y)=1-2\sin^2(y)$$ – David Peterson Sep 30 '18 at 6:09
• @Thor There is no single sine half angle formula. You just showed how to derive one such formula (though the derivation is not complete, and apparently not what you were after, either). So, to repeat the question from my first comment, which half-angle formula are you attempting to derive? – dxiv Sep 30 '18 at 6:10
• The problem is, that you introduced $\cos(y/2)$, so you need the half-angle identity for $\cos$ to continue – David Peterson Sep 30 '18 at 6:16
• @Thor Then replace the denominator with the similar $\,\cos\,$ half-angle formula and simplify. – dxiv Sep 30 '18 at 6:19

Square everything: $$\sin^2 y =4\cos^2(y/2)\sin^2(y/2)=4(1-\sin^2 (y/2))\sin^2(y/2)=4\sin^2(y/2)-4\sin^4(y/2)$$
So $$1-\sin^2 y = 1 - 4 \sin^2(y/2)+4\sin^4(y/2)=(1-2 \sin^2 (y/2))^2$$ That is, $$\cos^2 y = (1-2 \sin^2 (y/2))^2$$
Unfortunately this only gives you $$\pm \cos y = 1 - 2 \sin^2(y/2)$$ (the cost of squaring in the first place). But since both sides are smooth, we have to make the same choice of sign everywhere — and at $$0$$ it clearly is the positive one. So we have showed that: $$\cos y = 1 - 2 \sin^2(y/2)$$ and so $$\sin (y/2)=\pm \sqrt{\frac{1-\cos y}{2}}$$