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...

Arthur Pratt

Arthur Pratt

Answered

2021-12-30

Proving 2sinx+sin2x2sinxsin2x=csc2x+2cscxcotx+cot2x
Proving right hand side to left hand side:
csc2x+2cscxcotx+cot2x=1sin2x+2cosxsin2x+cos2xsin2x (1)
=cos2x+2cosx+1sin2x (2)
=1sin2x+1+sin2xsinxsin2x (3)
=2sinxsin3x+sin2xsinxsin2x (4)
=2sinxsin3x+sin2xsin3x (5)
I could not prove further to the left hand side from here.

Answer & Explanation

hysgubwyri3

hysgubwyri3

Expert

2021-12-31Added 43 answers

We have that, using sin2x=2sinxcosx and cancelling out 2sinx0
2sinx+sin2x2sinxsin2x=2sinx+2sinxcosx2sinx2sinxcosx=1+cosx1cosx
then, assuming cosx1 ( which holds since we need sinx0 for the original expression), multiply by 1+cosx1+cosx and simplify to obtain the result using that 1cos2x=sin2x
Cheryl King

Cheryl King

Expert

2022-01-01Added 36 answers

The following relations would be used in the answer:
1cos2x=sin2x
sin2x=2sinxcosx
Taking LHS,
2sinx+sin2x2sinxsin2x=2sinx(1+cosx)2sinx(1cosx)
Cancelling sin2x
=1+cosx1cosx
Multiplying 1+cosx to numerator and denominator
=(1+cosx)21cos2x
which is equivalent to
=(1+cos2x+2cosx)sin2x
Divinding each term by denominator, we get LHS as
=csc2x+2cscxcotx+cot2x
which is the RHS.
Hence, Proved.
nick1337

nick1337

Expert

2022-01-08Added 573 answers

2sinx+sin2x2sinxsin2x=2sinx+2sinxcosx2sinx2sinxcosx=1+cosx1cosx=(1+cosx)21cos2x=1+2cosx+cos2xsin2x=csc2x+2cscxcotx+cot2x

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?