If sin⁡x+sin⁡y=aandcos⁡x+cos⁡y=b then find tan⁡(x−y2)

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Benedict · Accepted Answer

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bandsin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B. cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡Bandcos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B. These are true for all A and B. So, sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡Bandcos⁡(A+B)+cos⁡(A−B)=2cos⁡Acos⁡B. Therefore relating A and B to x and y we getA+B=xandA−B=y.FromtheseA=x+y2andB=x−y2. We can now write sin⁡x+sin⁡y=2sin⁡12(x+y)cos⁡12(x−y)=aandcos⁡x+cos⁡y=2cos⁡12(x+y)cos⁡12(x−y)=b. Therefore, dividing these two we get tan⁡12(x+y)=ab. If we square each of the original equations we get: sin2x+sin2y+2sin⁡x.sin⁡y=a2andcos2x+cos2y+2cos⁡x.cos⁡y=b2 Adding these two equations we get 2+2cos⁡(x−y)=a2+b2[sin2+cos2=1 for ⊥h xandy] We can expand cos⁡(x−y) into 2cos2((x−y)/2)−1, so we can write

Jeffrey Jordon · Accepted Answer

The value of tan⁡(x−y2)  Step-by-step explanation:  Given: sin⁡x+sin⁡y=a,cos⁡x+cos⁡y=b  To find: The value of tan⁡(x−y2)  Solution:  sin⁡x+sin⁡y=a  We know, sin⁡x+sin⁡y=2sin⁡(x+y2)cos⁡(x−y2)  2sin⁡(x+y2)cos⁡(x−y2)=a....(1)  cos⁡x+cos⁡y=b  We know, cos⁡x+cos⁡y=2cos⁡(x+y2)sin⁡(x−y2)  2cos⁡(x+y2)sin⁡(x−y2)=b...(2)  Divide (1) and (2),  2sin⁡(x+y2)cos⁡(x−y2)2cos⁡(x+y2)sin⁡(x−y2)  Therefore, the value of tan⁡(x+y2)=ab

If sinx+siny=a and cosx+cosy=b then find tan(x-y/2)

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