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
1 Answer
$\begingroup$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$$ \sin (y/2)=\pm \sqrt{\frac{1-\cos y}{2}} $$
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