Prove the trigonometric identity?

Question

#(1+3sintheta-4sin^3theta)/(1+2sintheta)^2=(1-sintheta)#

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Answer 1 · 2018-06-03T10:17:14

For a Proof, please refer to Explanation.

Let, for ease of writing, #sintheta=s#.

Then, we have, #1+3sintheta-4sin^3theta#,

#=1+3s-4s^3#.

In this polynomial of #s#, the sum of its co-efficients is #0#.

#:. (1-s)# must be a factor.

So, we arrange the terms of #1+3s-4s^3# in such a way that

#(1-s)# can be factored out.

Thus, #1+3s-4s^3=ul(1-s)+ul(4s-4s^2)+ul(4s^2-4s^3)#,

#=1(1-s)+4s(1-s)+4s^2(1-s)#,

#=(1-s)(1+4s+4s^2)#,

#=(1-s)(1+2s)^2#.

#:. (1+3s-4s^3)/(1+2s)^2=(1-s)#.

# rArr (1+3sintheta-4sin^3theta)/(1+2sintheta)^2=(1-sintheta)#, as desired!

Enjoy Maths.!