Question

If sin theta + sin^2 theta + sin^3 theta= 1, prove that cos^6 theta -4cos^4 + 8cos^2=4

Answer 1

Answer:

4 = $Cos^{6}\theta$ - $4Cos^{4}\theta$ + $8Cos^{2}\theta$

Step-by-step explanation:

In the question  

We have  

$Sin\theta$ + $Sin^{2}\theta$+ $Sin^{3}\theta$ = 1

$Sin\theta$ + $Sin^{3}\theta$ = $1 - Sin^{2}\theta$

$Sin\theta$ ($1 +Sin^{2}\theta$) = $Cos^{2}\theta$

$Sin\theta$($1 + 1 - Cos^{2}\theta$) = $Cos^{2}\theta$

$Sin\theta$ ($2 - Cos^{2}\theta$) = $Cos^{2}\theta$

Squaring on both sides

$[Sin\theta(2 - Cos^{2}\theta)]^2$ =$[Cos^2}\theta]^2$

$Sin^{2}\theta$ $(2 - Cos^{2}\theta)^2$ = $Cos^{4}\theta$

($1 - Cos^{2}\theta$) ($4 + Cos^{4}\theta$ - $4Cos^{2}\theta$) = $Cos^{4}\theta$

$4 + Cos^{4}\theta$  $- 4Cos^{2}\theta$  - 4$Cos^{2}\theta$  - $Cos^{6}\theta$  + $4Cos^{4}\theta$ = $Cos^{4}\theta$

4 = $Cos^{6}\theta$ + $8Cos^{2}\theta$ - $4Cos^{4}\theta$

4 = $Cos^{6}\theta$ - $4Cos^{4}\theta$ + $8Cos^{2}\theta$

  Hence the result.

Answer 2

Step-by-step explanation:

step by step explanation

one of the most important question of trigonometry RDSHARMA CLASS 10