Derivation of the tangent half angle identityI'm having trouble proceeding fromsin⁡(θ)1+cos⁡(θ)totan⁡(θ2)Context:Consider the function f defined for all (x,y) such that y≠0 with the rulef(x,y)=yx2+y2+xShow thatf(rcos⁡(θ),rsin⁡(θ))=tan⁡(θ2)So far I've done:f(rcos⁡(θ),rsin⁡(θ))=rsin⁡(θ)r2cos2(θ)+r2sin2(θ)+rcos⁡(θ)=rsin⁡(θ)r2+rcos⁡(θ)=sin⁡(θ)1+cos⁡(θ)Usingcos⁡(θ)=2cos2(θ2)−1⇒2cos2(θ2)=cos⁡(θ)+1We getf(rcos⁡(θ),rsin⁡(θ))=sin⁡(θ)2cos2(θ2)

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Derivation of the tangent half angle identityI&#039;m having trouble proceeding fromsin⁡(θ)1+cos⁡(θ)totan⁡(θ2)Context:Consider the function f defined for all (x,y) such that y≠0 with the rulef(x,y)=yx2+y2+xShow thatf(rcos⁡(θ),rsin⁡(θ))=tan⁡(θ2)So far I&#039;ve done:f(rcos⁡(θ),rsin⁡(θ))=rsin⁡(θ)r2cos2(θ)+r2sin2(θ)+rcos⁡(θ)=rsin⁡(θ)r2+rcos⁡(θ)=sin⁡(θ)1+cos⁡(θ)Usingcos⁡(θ)=2cos2(θ2)−1⇒2cos2(θ2)=cos⁡(θ)+1We getf(rcos⁡(θ),rsin⁡(θ))=sin⁡(θ)2cos2(θ2)

Drahthaare89c · Accepted Answer

Hint: The numerator can be written assin⁡θ=sin⁡(2⋅θ2)=2sin⁡θ2cos⁡θ2

Ruben Gibson · Accepted Answer

It is a lot easier to use some trig. identities:sin⁡(θ)=2sin⁡(θ2)cos⁡(θ2) and 1+cos⁡θ=2cos2(θ2). You will immediately get the result from these two formulas.

Derivation of the tangent half angle identity I'm having

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