Question

sinx+ sin²x + sin³x = 1

find cos⁶x-4cos⁴x+8cos²x
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Answer 1

Answer:

sin X + sin 2 X + sin 3 X = 1

sin X + sin 3 X = 1 - sin 2 X

( sin X + sin 3 X ) 2 = ( 1 - sin 2 X ) 2

sin 2 X + sin 6 X + 2 sin 4 X = cos 4 X

1 - cos 2 x + ( 1 cos 2 X ) 3 + 2 ( 1 - code 2 X ) 2

= cos 4 X

1 - cos 2 X + 1 - 3 cos 2 X + 3 cos 4 x - cos 6 X + 2

- 4 cos 2 X + 2 cos 4 X = cos for X

cos 6 X - 4 cos 4 X + 8 cos 2 X = 4