Question

show that
sin^6 A + cos^6 A + 3sin^2 A cos^2 A = 1​

Answer 1

Answer:

1

Step-by-step explanation:

two identities

${ \sin(x) }^{2} + { \cos(x) }^{2} = 1$

${(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

I hope this answer is explanatory.

Answer 2

Step-by-step explanation:

sin^6A + cos^6A + 3sin²Acos²A

= (sin²A)³ + (cos²A)³ + 3sin²Acos²A

=(sin²A + cos²A)(sin⁴A - sin²Acos²A + cos⁴A) + 3sin²Acos²A

[ a³ + b³ = (a + b)(a² - ab + b²)]

=1(sin⁴A - sin²Acos²A + cos⁴A) + 3sin²Acos²A

[sin²A + cos²A = 1]

=sin⁴A + 2 sin²Acos²A + cos⁴A

=(sin²A)² + 2 sin²Acos²A + (cos²A)²

=(sin²A + cos²A)²

=1. (Proved)

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