Question

2(sin⁶θ + cos⁶θ)-3(sin⁴θ + cos⁴θ)+1=0,Prove it by using trigonometric identities.Prove it by using trigonometric identities.Prove it by using trigonometric identities.

Answer 1

Solution of this question

Identity used

sin²A+cos²A=1

a³+b³=(a+b)(a²+b²-ab)

1 = sin²A + cos²A

(a+b)²= a²+b²+2ab

theta = A

=2[ (sin²A)³+(cos²A)³]-3sin⁴A-3cos⁴A +sin²A+cos²A

=2[(sin²A+cos²A)(sin⁴A+cos⁴A-sin²A.cos²A)]-3sin⁴A-3cos⁴A +sin²A+cos²A

=2[(1)(sin⁴A+cos⁴A-sin²A.cos²A)]-3sin⁴A-3cos⁴A +sin²A+cos²A

=2(sin⁴A+cos⁴A-sin²A.cos²A)]-3sin⁴A-3cos⁴A +sin²A+cos²A

=2sin⁴A+2sin⁴A-2sin²A.cos²A-3sin⁴A-3cos⁴A +sin²A+cos²A

=-sin⁴A-cos⁴A-2sin²Acos²A+1

=-(sin⁴A+cos⁴A+2sin²A.cos²A)+1

=-{(sin²A)²+(cos²A)²+2sin²A.cos²A}+1

=-{sin²A+Cos²A}²+1

=-1+1

=0

RHS

LHS=RHS

HP

Answer 2

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 )
_____________________________

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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