Question

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$2( { \sin }^{6} \alpha + { \cos }^{6} \alpha ) - 3( { \sin }^{4} \alpha - { \cos }^{4} \alpha ) + 1 = 0$

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Answer 1

Hope it helps you ❣️☑️☑️☑️

Answer 2

Answer:

2(sin 6 α+cos 6 α)−3(sin 4 α−cos 4 α)+1=0

Rewrite the expression by applying the distributive rule

a³+b³=(a+b)(a²−ab+b²)

2(sin² a+ cos² a)(sin⁴ a − sin² acos²b+ cos² a)−3(sin⁴ a +cos⁴ a)+1=0

Put the value of pythagorean identity

sin²(x)+cos²(x)=1

, to get:

2(1)(sin⁴ a−sin² a cos² b+cos² a)−3(sin⁴ a+cos⁴ a)+1=0

2 sin⁴ a − 2sin² a cos² b+2cos⁴ a−3 sin⁴ a−3cos⁴ a+1=0

Simplify:

−sin⁴ (a) − 2sin² a cos² b−cos⁴ a+1=0

−(sin⁴ a+2sin² a cos² b+cos⁴ a)+1=0

−(sin² a+cos² a)2+1=0

Put the value of pythagorean identity

sin² (x)+cos² (x)=1 , to get:−1+1=0

0 = 0

HENCE, PROVED