Question

pls, answer properly........

If sin θ + sin² θ + sin³ θ = 1, prove that cos⁶ θ - 4cos⁴ θ + 8cos² θ = 4

Answer 1

Let : sin θ + sin² θ + sin³ θ = 1................ (1)

Then : sin θ + sin³ θ = 1 - sin² θ

......... sin θ ( 1 + sin² θ ) = cos² θ

......... sin θ ( 1 + 1 - cos² θ ) = cos² θ

......... sin θ ( 2 - cos² θ ) = cos² θ ...........(2)

Squaring both sides of (2),

sin² θ ( 2 - cos² θ )² = cos^4 θ

( 1 - cos² θ )( 4 - 4 cos² θ + cos^4 θ ) = cos^4 θ

( 4 - 4 cos² θ + cos^4 θ ) - 4 cos² θ + 4 cos^4 θ - cos^6 θ = cos^4 θ

4 - 8 cos² θ + 5 cos^4 θ - cos^6 θ = cos^4 θ

4 = cos^4 θ + 8 cos² θ - 5 cos^4 θ + cos^6 θ

Rearranging the above equation,

cos^6 θ - 4 cos^4 θ + 8 cos² θ = 4 .............................. Ans.

Hope it's Helpful for you.......

Answer 2

Step-by-step explanation:

$We \: have \\ = sinx + {sin}^{2}x + {sin}^{3} x = 1 \\ = sinx + {sin}^{3} x = 1 - {sin}^{2} x \\ = sinx + {sin}^{3} x = {cos}^{2} x \\ = sinx(1 + {sin}^{2} x) = {cos}^{2} x \\ = sinx(2 - {cos}^{2} x) = {cos}^{2} x \\ = {sin}^{2}x( {2 - {cos}^{2}x) }^{2} = {cos}^{4} x \\ = (1 - {cos}^{2} x)(4 + {cos}^{4} x - 4 {cos}^{2} x) = {cos}^{4} x \\ = 4 - 4 {cos}^{2} x + {cos}^{4} x - 4 {cos}^{2} x + 4 {cos}^{4} x + {cos}^{6} x = {cos}^{4} x \\ = {cos}^{6} x - 4 {cos}^{4} x + 8 {cos}^{2} x = 4 \\ \\ Hence \: proved \: \\ Hope \: it \: will \: help \: you.$