Prove the trigonometric identity?

Question

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

1519 views around the world

You can reuse this answer
Creative Commons License

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.!