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PLEASE HELP ME WITH THIS TRIG!!!! THANK YOU IF YOU DO!!!!
Solution
Assuming you are trying to reduce this to simplest terms…the easiest way to reduce something like this is to break it down into parts and simplify then put back together. Sometimes it takes some trial and error to figure out the best way to break up the parts.
First lets consider some trig identities.
(1) sin (2π – x) = – sin x
(2) cos 2x = 1 – 2 (sin x)2
(3) cos 2x = 2 (cos x)2
Using the identities 2 and 3 repeatedly will allow you to get expressions for higher powers of sin and cos. However, to simplify, I will use some identities collected in Gradshteyn and Ryzhik Table of Integrals, Series, and Products (If you are going into a field that uses a lot of math, I recommend you buy this book)
(4) sin6 x = (1/32)*[ – cos(6x) + 6 cos(4x) – 15 cos(2x) + 10]
(5) cos6 x = (1/32)*[ cos(6x) + 6 cos(4x) + 15 cos(2x) + 10]
(6) sin4 x = (1/8) [cos(4x) – 4 cos(2x) + 3]
(7) cos4 x = (1/8) [cos(4x) + 4 cos(2x) + 3]
The third term in your expression is easiest to reduce using identity (1): sin (2π – a) = – sin a
Using identities (4) and (5), the first term in your expression reduces to:
2*[ sin6 (a) + cos6 (a) ] = 2*(1/32)[ 12 cos(4a) + 20] = [(3/4) cos(4a)] + (5/4)
Using identities (6) and (7) the second term reduces to:
3*[ sin4 (a) + cos4 (a) ] = 3*(1/8)[ 2 cos(4a) + 6] = [(3/4) cos(4a) + (9/4)
combining these terms and applying the correct signs from the problem:
2*[ sin6 (a) + cos6 (a) ] – 3*[ sin4 (a) + cos4 (a) ] – sin (2π – a) + 1
= {[(3/4) cos(4a)] + (5/4)} – { [(3/4) cos(4a) + (9/4)} – (- sin a) + 1
= -1 + sin(a) + 1 = sin(a)