Question

prove that : sin⁶0 + cos⁶0 =1 - 3 sin²0 cos² 0 ( 0 is thetha)


prove LHS = RHS​

Answer 1

Answer:

Here I am using A instead of theta.

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We know the algebraic identities:

1 ) a³ + b³ = ( a + b )³ - 3ab( a + b )

2 ) a² + b² = ( a + b )² - 2ab

and

Trigonometric identity :

1 ) sin² A + cos² A = 1

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Now ,

i ) sin^6 A + cos^6A

= ( sin² A)³ + ( cos² A )³

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

= 1 - 3sin²A cos²A --- ( 1 )

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ii ) sin⁴A + cos⁴A

= ( sin² A )² + ( cos²A )²

= ( sin²A + cos²A )² - 2sin²Acos²A

= 1 - 2sin²Acos²A -----( ii )

__________________________

Now ,

LHS

= 2(sin^6A+cos^6A)-3(sin⁴A+cos⁴A)+1

{ From ( i ) & ( ii ) , we get }

=2(1 -3sin²Acos²A)-3(1 - 2sin²Acos²A)+1

= 2 -6sin²Acos²A-3+6sin²Acos²A + 1

= 2 - 3 + 1

= 3 - 3

= 0

= RHS

•••••

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Answer 2

Step-by-step explanation:

taking LHS,

0 +1

1

taking RHS

1-3×0×1

1

LHS = RHS