Solve. 1. sin x cos^3 x-cosx sin^3 x=1/4sin^4x (cos x+cos 4x+cos 7x)/(sin x+sin 4x+sin 7x)=cot 4x

Darian Hubbard 2022-08-01 Answered
Solve.
1. sin x cos 3 x cos x sin 3 x = 1 / 4 sin 4 x
2. cos x + cos 4 x + cos 7 x sin x + sin 4 x + sin 7 x = cot 4 x
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Answers (1)

Monastro3n
Answered 2022-08-02 Author has 15 answers
1) Take out common factors:
sin x cos x [ cos 2 x sin 2 x ] = ( 1 / 4 ) sin ( 4 x )
double angle identity:
sin ( 2 x ) = 2 sin ( x ) cos ( x )
So,
sin ( 2 x ) 2 [ cos ( 2 x ) ] = sin ( 4 x ) 4
sin ( 2 x ) cos ( 2 x ) 2 = sin ( 4 x ) 4
We already know that:
sin ( 2 u ) = sin ( u ) cos ( u )
if we let u = 2x:
sin [ 2 ( 2 x ) ] = sin ( 2 x ) cos ( 2 x )
sin ( 4 x ) = sin ( 2 x ) cos ( 2 x )
sin ( 4 x ) 4 = sin ( 4 x ) 4
2) we know that cos x + cos 7 x = 2 cos ( 7 x + x ) / 2 c o s ( 7 x x ) / 2 = 2 cos 4 x cos 3 x
and sin x + sin 7 x = 2 sin ( 7 x + x ) / 2 cos ( 7 x x ) / 2 = 2 sin 4 x cos 3 x
cos x + cos 4 x + cos 7 x sin x + sin 4 x + sin 7 x = 2 cos 4 x cos 3 x + cos 4 x 2 sin 4 x cos 3 x + sin 4 x = cos 4 x ( 2 cos 3 x + 1 ) sin 4 x ( 2 cos 3 x + 1 ) = cos 4 x sin 4 x = cot x
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