How to express sin 3theta in terms of trigonometric functions of theta ?

gelepikb165

gelepikb165

Answered question

2023-02-22

How to express sin 3 θ in terms of trigonometric functions of θ ?

Answer & Explanation

daneetuhxxtj

daneetuhxxtj

Beginner2023-02-23Added 8 answers

sin 3 x = sin ( 2 x + x ) = sin 2 x cos x + cos 2 x sin x
We possess dual identities:
sin 2 x = 2 sin x cos x
cos 2 x = cos 2 x - sin 2 x
sin 3 x = 2 sin x cos x cos x + ( cos 2 x - sin 2 x ) sin x
sin 3 x = 2 sin x cos 2 x + sin x cos 2 x - sin 3 x
sin 3 x = 3 sin x cos 2 x - sin 3 x
Vincent Burke

Vincent Burke

Beginner2023-02-24Added 9 answers

We can apply de Moivre's equation:
( cos θ + i sin θ ) n = cos n θ + i sin n θ
as follows:
cos 3 θ + i sin 3 θ
= ( cos θ + i sin θ ) 3
= cos 3 θ + 3 i cos 2 θ sin θ - 3 cos θ sin 2 θ - i sin 3 θ
= ( cos 3 θ - 3 cos θ sin 2 θ ) + i ( 3 cos 2 θ sin θ - sin 3 θ )
Then, when we compare the real and imaginary parts, we discover:
{ cos 3 θ = cos 3 θ - 3 cos θ sin 2 θ sin 3 θ = 3 cos 2 θ sin θ - sin 3 θ
We can also use the trigonometric version of Pythagoras' theorem:
cos 2 θ + sin 2 θ = 1
to refactor these formulae as follows:
cos 3 θ = cos 3 θ - 3 cos θ sin 2 θ
cos 3 θ = cos 3 θ - 3 cos θ ( 1 - cos 2 θ )
cos 3 θ = 4 cos 3 θ - 3 cos θ
sin 3 θ = 3 cos 2 θ sin θ - sin 3 θ
sin 3 θ = 3 ( 1 - sin 2 θ ) sin θ - sin 3 θ
sin 3 θ = 3 sin θ - 4 sin 3 θ

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