Question

If,
$\sin\alpha + { \sin }^{2} \alpha + { \sin }^{3} \alpha = 1$
Prove that;
${ \cos }^{6} \alpha - 4 { \cos}^{4} \alpha + 8 { \cos }^{2 } \alpha = 4$

Answer 1

Heya !!

Given :- sinA + sin²A + sin³A = 1

To prove :- cos^6A - 4cos⁴A + 8cos²A = 4

Proof :- sinA + sin²A + sin³A = 1

=> sinA + sin³A = 1 - sin²A

=> sinA + sin³A = cos²A

Squaring on both sides

(sinA + sin³A)² = (cos²A)²

=> sin²A + 2sin⁴A + sin^6A = cos⁴A

=> (1 - cos²A) + 2(sin²A)² + (sin²A)³ = cos⁴A

=> (1 - cos²A) + 2(1 - cos²A)² + (1 - cos²A)³ = cos⁴A

=> (1 - cos²A) + (2 - 4cos²A + 2cos⁴A) + (1 - 3cos²A + 3cosA -cos^6A) = cos⁴A

=> 4 - 8cos²A + 5cos⁴A - cos^6A = cos⁴A

=> 4 - 8cos²A + 4cos⁴A - cos^6A = 0

=> cos^6A - 4cos⁴A + 8cos²A = 4

Hence, proved.