Proof that sin &#x2061;<!-- ⁡ --> x + 3 sin &#x2061;<!-- ⁡ --> 2 x +

letumsnemesislh

letumsnemesislh

Answered question

2022-07-04

Proof that sin x + 3 sin 2 x + 3 sin 3 x + sin 4 x > 0 for 0 < x < 2 π

Answer & Explanation

Kathryn Moody

Kathryn Moody

Beginner2022-07-05Added 10 answers

Since
sin x + 3 sin 2 x + 3 sin 3 x + sin 4 x = 2 sin 5 x 2 cos 3 x 2 + 6 sin 5 x 2 cos x 2 =
= 2 sin 5 x 2 ( cos 3 x 2 + 3 cos x 2 ) = 8 sin 5 x 2 cos 3 x 2 ,
we get the answer:
2 π 5 < x < 4 π 5 or π < x < 6 π 5 or 8 π 5 < x < 2 π.
Jaydan Aguirre

Jaydan Aguirre

Beginner2022-07-06Added 2 answers

hint
use the formula
sin ( a ) + sin ( b ) =
2 sin ( a + b 2 ) cos ( a b 2 )
for sin ( x ) + sin ( 4 x ) and for
3 ( sin ( 2 x ) + sin ( 3 x ) )

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