Question

evluate sin^6 Θ - cos^6 Θ
${ \sin }^{6} a \: - { \cos }^{6} a$
plz solve this
expecting expert answer plz

Answer 1

HELLO DEAR,

we know that;-

(a³-b³)=(a²-b²)(a⁴+b⁴+ab)----------(1)

(a⁴+b⁴)=(a²+b²)² - 2a²b²---------------(2)

sin²Θ - cos²Θ => (2sin²θ-1)--------------(3)

(sin²Θ + cos²Θ )=1---------------(1)

[ If we know that cos²θ=1−sin²θ, we can replace cos²θ with 1−sin²θ in the expression :-

 (sin²Θ - cos²Θ) = sin²θ -(1−sin²θ)

 (sin²Θ - cos²Θ) = (2sin²θ-1) ]

now Solve

$( \sin ^{6} \alpha - { \cos}^{2} \alpha ) = ( { \sin }^{2} \alpha)^{3} - { (\cos ^{2} \alpha ) }^{ 3} \\ = >( { \sin }^{2} \alpha - \cos^{2} \alpha )( \sin^{4} \alpha + \cos^{4} \alpha + \sin\alpha \times \cos \alpha ) ======= using...(1) \\ = > (2 { \sin }^{2} \alpha - 1)[ ({\sin }^{2} \alpha + \cos^{2} \alpha )^{2} - 2 { \sin}^{2} \alpha { \cos}^{2} \alpha +\sin\alpha \times \cos \alpha )] === using.(2)..and..(3) \\ = > (2 { \sin }^{2} \alpha - 1)[ {1} - 2 { \sin}^{2} \alpha { \cos}^{2} \alpha +\sin\alpha \times \cos \alpha )] using--(4)$

I HOPE ITS HELP YOU DEAR,
THANKS