Prove the sin 2x + sin 4x + cos 2x + cos 4x = 2 sqrt.2 cos x sin (3x + pi/4)
Hence find all the angles x between 0 and pi inclusive for which
sin 2x + sin 4x + cos 2x + cos 4x = 0
so i used:
2sqrt.2 cos x sin(3x + pi/4) = 0
cos x = 0 sin(3x + pi/4) = 0, pi, 2 pi
x = pi/2, 3 pi/2 (rej) x = -pi/4,(rej) pi/12, 5pi/12
however the answers are pi/4, pi/2, 7pi/12 and 11pi/12...
I subbed my answers into 2sqrt.2 cos x sin(3x + pi/4) =0 and it is correct. however, when substituted into sin 2x + sin 4x + cos 2x + cos 4x, i got another answer. but if you use the book's answer and sub it into sin 2x + sin 4x + cos 2x + cos 4x, you get 0. but then again, isnt sin 2x + sin 4x + cos 2x + cos 4x = 2 sqrt.2 cos x sin (3x + pi/4), as stated ? if that is the case, there shouldn't be anything amiss with my answers.
it would be of great help if someone could screen through this. thanks a million times.
sin(3x + pi/4) = 0
0<x<pi (inclusive)
0<3x<3pi
pi/4<(3x+pi/4)< 3.25pi
Basic angle = 0
3x+pi/4 = 0(NA), pi, 2pi or 3pi
3x = 3pi/4, 7pi/4 or 11pi/4
x = pi/4, 7pi/12 or 11pi/12
Add in the answer from the cos x =0 and everything is all right.
2sqrt.2 cos x sin(3x + pi/4) = 0
cos x = 0 sin(3x + pi/4) = 0
x = pi/2, 3 pi/2 (rej) 3x + pi/4 = 0, pi, 2pi, 3pi, 4pi
x = -pi/12 (rej), pi/4, 7pi/12, 11pi/12, 5pi/4 (rej)
combining, x = pi/4, pi/2, 7pi/12 and 11pi/12
I think you were careless :D
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