"How to go from 2cos⁡2x+1 → sin⁡3xsin⁡x" was answered by our community

High School
Geometry
Trigonometry
Trigonometric equations and identities

deiteresfp

Answered question

2021-12-29

How to go from

2 \cos 2 x + 1 \to \frac{\sin 3 x}{\sin x}

Answer & Explanation

alexandrebaud43

Beginner2021-12-30Added 36 answers

2 \cos 2 x + 1 = 2 (1 - 2 \sin^{2} x) + 1 = 3 - 4 \sin^{2} x = \frac{3 \sin x - 4 \sin^{3} x}{\sin x} = \frac{\sin 3 x}{\sin x}

Note:

\sin x \neq 0

Timothy Wolff

Beginner2021-12-31Added 26 answers

Hint for LHS to RHS: Convert

\cos 2 x

to one of its three identities, then multiply and divide by

\sin x

Hint for RHS to LHS: Convert

\sin 3 x

to a known identity, then divide by

\sin x

. Once you have that, add and subtract 1 to get a common factor, then you will have a term in

\cos^{2} x

that can be converted into a double angle formula.

karton

Expert2022-01-08Added 613 answers

We have
$2 \sin α \cos β = \sin (α + β) + \sin (α - β)$
so
$(2 \cos 2 x + 1) \sin x = 2 \sin x \cos 2 x + \sin x = \sin 3 x + \sin (- x) + \sin x = \sin 3 x$
so assuming that $\sin x \neq 0$
$2 \cos 2 x + 1 = \frac{\sin 3 x}{\sin x}$

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