Proving 2sin⁡x+sin⁡2x2sin⁡x−sin⁡2x=csc2x+2csc⁡xcot⁡x+cot2x Proving right hand side to left hand side: csc2x+2csc⁡xcot⁡x+cot2x=1sin2x+2cos⁡xsin2x+cos2xsin2x (1) =cos2x+2cos⁡x+1sin2x (2) =1−sin2x+1+sin⁡2xsin⁡xsin2x (3) =2sin⁡x−sin3x+sin⁡2xsin⁡xsin2x (4) =2sin⁡x−sin3x+sin⁡2xsin3x (5) I could not prove further to the left hand side from here.

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Proving 2sin⁡x+sin⁡2x2sin⁡x−sin⁡2x=csc2x+2csc⁡xcot⁡x+cot2x  Proving right hand side to left hand side:  csc2x+2csc⁡xcot⁡x+cot2x=1sin2x+2cos⁡xsin2x+cos2xsin2x     (1)  =cos2x+2cos⁡x+1sin2x    (2)  =1−sin2x+1+sin⁡2xsin⁡xsin2x       (3)  =2sin⁡x−sin3x+sin⁡2xsin⁡xsin2x         (4)  =2sin⁡x−sin3x+sin⁡2xsin3x        (5)  I could not prove further to the left hand side from here.

hysgubwyri3 · Accepted Answer

We have that, using sin⁡2x=2sin⁡xcos⁡x and cancelling out 2sin⁡x≠0  2sin⁡x+sin⁡2x2sin⁡x−sin⁡2x=2sin⁡x+2sin⁡xcos⁡x2sin⁡x−2sin⁡xcos⁡x=1+cos⁡x1−cos⁡x  then, assuming cos⁡x≠−1 ( which holds since we need sin⁡x≠0 for the original expression), multiply by 1+cos⁡x1+cos⁡x and simplify to obtain the result using that 1−cos2x=sin2x

Cheryl King · Accepted Answer

The following relations would be used in the answer:  1−cos2x=sin2x  sin⁡2x=2sin⁡xcos⁡x  Taking LHS,  2sin⁡x+sin⁡2x2sin⁡x−sin⁡2x=2sin⁡x(1+cos⁡x)2sin⁡x(1−cos⁡x)  Cancelling sin⁡2x  =1+cos⁡x1−cos⁡x  Multiplying 1+cos⁡x to numerator and denominator  =(1+cos⁡x)21−cos2x  which is equivalent to  =(1+cos2x+2cos⁡x)sin2x  Divinding each term by denominator, we get LHS as  =csc2x+2csc⁡xcot⁡x+cot2x  which is the RHS.  Hence, Proved.

nick1337 · Accepted Answer

2sin⁡x+sin⁡2x2sin⁡x−sin⁡2x=2sin⁡x+2sin⁡xcos⁡x2sin⁡x−2sin⁡xcos⁡x=1+cos⁡x1−cos⁡x=(1+cos⁡x)21−cos2⁡x=1+2cos⁡x+cos2⁡xsin2⁡x=csc2⁡x+2csc⁡xcot⁡x+cot2⁡x

Proving \frac{2 \sin x+\sin 2x}{2 \sin x-\sin 2x}=\csc^2

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