If sin⁡α+sin⁡β=a and cos⁡α−cos⁡β=b, then what is tan⁡(α−β2)? If I square both the equations 2+2sin⁡(α−β)=a2+b2 sin⁡(α−β)=a2+b2−22 Since sin⁡2θ=2tan⁡θ1+tan2θ, then sin⁡(α−β)=2tan⁡α−β21+(tan⁡α−β2)2 It’s obviously the too long to solve, so is there a shorter way to do this?

Question

If sin⁡α+sin⁡β=a  and  cos⁡α−cos⁡β=b, then what is tan⁡(α−β2)?  If I square both the equations  2+2sin⁡(α−β)=a2+b2  sin⁡(α−β)=a2+b2−22  Since sin⁡2θ=2tan⁡θ1+tan2θ, then  sin⁡(α−β)=2tan⁡α−β21+(tan⁡α−β2)2  It’s obviously the too long to solve, so is there a shorter way to do this?

sineurosi0f · Accepted Answer

Since a2=sin⁡α+β2cos⁡α−β2 while −b2=sin⁡α+β2sin⁡α−β2,tan⁡α−β2=−ba

If \sin \alpha+\sin \beta=a \text{ and } \cos \alpha−\cos \beta=b,

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