# Thread: Trignometric Identities

1. ## Trignometric Identities

$\displaystyle if sinA+cosA=x, Prove that Sin^6A +cos^6A = 4-3(x^2-1)^2 /4$

2. Originally Posted by Sanjana Das
$\displaystyle if sinA+cosA, Prove that Sin^6A +cos^6A = 4-3(x^2-1)^2 /4$
something is wrong here... on the left side you have sin( A ) and on the right side x ???

Anyway... if A = x...

sin^6( x ) + cos^6( x ) = ( sin^2( x ) + cos^2( x ) )^3 - 3sin^4( x )*cos^2( x ) - 3sin^2( x )*cos^4( x ) =

= 1^3 - 3*( sin^2( 2x )/4 )*sin^2( x ) - 3( sin^2( 2x )/4 )*cos^2( x ) =

= 1 - (3/4)*sin^2( 2x )*( sin^2( x ) + cos^2( x ) ) =

= 1 - (3/4)*sin^2( 2x )

3. Originally Posted by josipive
something is wrong here... on the left side you have sin( A ) and on the right side x ???

Anyway... if A = x...

sin^6( x ) + cos^6( x ) = ( sin^2( x ) + cos^2( x ) )^3 - 3sin^4( x )*cos^2( x ) - 3sin^2( x )*cos^4( x ) =

= 1^3 - 3*( sin^2( 2x )/4 )*sin^2( x ) - 3( sin^2( 2x )/4 )*cos^2( x ) =

= 1 - (3/4)*sin^2( 2x )*( sin^2( x ) + cos^2( x ) ) =

= 1 - (3/4)*sin^2( 2x )
no no the answer is not tht plsss check again ...it is sinA+cosA=x
and also A=thetha

4. Originally Posted by sanjana das
no no the answer is not tht plsss check again ...it is sina+cosa=x
and also a=thetha
plsssssss someone solve this question plssssssssssssssss

5. ok.... than

you see this part that i solved:

sin^6( A ) + cos^6( A ) = ... = 1 - (3/4)*sin^2( 2A )

this is equal to:

1 - ( 3/4 )*sin^2( 2A ) = 4 - 3*( x^2 - 1 )^2 / 4

(3/4)*sin^2( 2A ) = ( 3/4 )*( x^2 - 1 )^2 - 3

sin^2( 2A ) = ( x^2 - 1 )^2 - 4