Question

solve with explanations sin^6 +cos^6 = 1-3sin^2cos^2

Answer 1

Identity Formulas :

  • sinθ * cosecθ = 1

  • cosθ * secθ = 1

  • tanθ * cotθ = 1

  • sin²θ + cos²θ = 1

  • sec²θ - tan²θ = 1

  • cosec²θ - cot²θ = 1

Proof :

Now, LHS = sin⁶θ + cos⁶θ

= (sin²θ)³ + (cos²θ)³

  [ ∵ xᵃᵇ = (xᵃ)ᵇ ]

= (sin²θ + cos²θ)

{(sin²θ)² + (cos²θ)² - sin²θ cos²θ}

 [ ∵ a³ + b³ = (a + b) (a² + b² - ab) ]

= 1 * {(sin²θ + cos²θ)² - 2sin²θ cos²θ - sin²θ cos²θ}

 [ ∵ a² + b² = (a + b)² - 2ab ]

= 1 - 3 sin²θ cos²θ

= RHS

↠ sin⁶θ + cos⁶θ = 1 - 3 sin²θ cos²θ

Hence, proved.

Answer 2

$= > { (sin(2) }^{3} + { \cos(2) }^{3}$

$= > { \sin }^{2} + {cos}^{2} ( {sin}^{4} - {sin}^{2} \times {cos}^{2} \\ + {cos}^{4} )$
$= > { \sin(2) }^{2} + { \cos(2) }^{2} - {cos}^{2} \times {sin}^{2}$
$= > {sin}^{2} + {cos}^{2} - 2 {sin}^{2} \times {cos}^{2} - \\ {sin}^{2} \times {cos}^{2}$
$= > 1 - 3 {sin}^{2} \times {cos}^{2}$
$= > lhs = rhs \: hence \: proof$
hope this answer is helpful plz mark it as brainliest if u think I can help u just follow me