The value of \(3[sin^4({3\pi \over 2}-a)+sin^4(3\pi+a)] - 2[sin^6({\pi \over 2}+a)+sin^6(5\pi-a)]\) is
Thank you.
sin (3pi/2 - a) = sin(3pi/2)cosa + cos(3pi/2)sina = -cosa
So
sin4 ( 3pi/2 -a) = (-cos a)^4 = (cos a)^4
sin ( 3pi + a) = sin (3pi) cos a - cos(3pi)sina = sina
So
sin 4 ( 3pi + a) = ( sin a)^4
sin ( pi/2 + a) = sin (pi/2)cos a - sin a cos (pi/2) = cos a
So
sin 6 (pi/2 + a) = (cos a)^6
sin (5pi -a) = sin (5pi)cosa + sin a cos (5pi) = -sina
So
sin 6 ( 5pi -a) = (-sin a)^6 = (sin a)^6
So
3 [ (cos a)^4 + (sin a)^4 ] - 2 [ (cos a)^6 + (sin a)^6 ] =
Special identites
cos ^4 a = (3 + 4cos (2a) + cos (4a)) / 8
sin^4 a = (3 - 4cos (2a) + cos (4a) ) / 8
So.....the first expression evaluates to
(3/8) ( 6 + 2cos (4a) ) =
(9/4) + (3/4)cos(4a)
Also
cos^6 a = (1/32) ( 15cos (2a) + 6cos (4a) + cos (6a) + 10 )
sin ^6 a = (1/32) (-15cos (2a) + 6cos (4a) - cos (6a) + 10 )
So......the second expression evaluates to
-(2/32) ( 12cos (4a) + 20) =
-(1/16) ( 12cos (4a) + 20) =
-(3/4)cos (4a) - 5/4
So
[ (9/4) + (3/4)cos (4a) ] - [ (3/4)cos (4a) + 5/4) ] =
9/4 - 5/4 =
4/4 =
1