Expert answers to "Prove thatcos⁡2θ−3sin⁡2θ≡2cos⁡(2θ+π3)≡−2sin⁡(2θ−π6)"

Richie Patterson

Answered question

2022-02-26

Prove that

\cos 2 θ - \sqrt{3} \sin 2 θ \equiv 2 \cos (2 θ + \frac{π}{3}) \equiv - 2 \sin (2 θ - \frac{π}{6})

an2gi2m9gg

Beginner2022-02-27Added 9 answers

Use the identities for the sine and cosine of the sum or difference of two angles:

\sin (2 θ - \frac{π}{6}) = \sin 2 θ \cos \frac{π}{6} - \cos 2 θ \sin \frac{π}{6}

= \frac{\sqrt{3}}{2} \sin 2 θ - \frac{1}{2} \cos 2 θ

and

\cos (θ + \frac{π}{3}) = \cos 2 θ \cos \frac{π}{3} - \sin 2 θ \sin \frac{π}{3}

= \frac{1}{2} \cos 2 θ - \frac{\sqrt{3}}{2} \sin 2 θ

Elsie Tillman

Beginner2022-02-28Added 5 answers

First, let's recap:

\sin \frac{π}{3} = \frac{\sqrt{3}}{2}

\cos \frac{π}{3} = \frac{1}{2}

\sin \frac{π}{6} = \frac{1}{2}

\cos \frac{π}{6} = \frac{\sqrt{3}}{2}

Going from

\cos 2 θ - \sqrt{3} \sin 2 θ

2 \cos (2 θ + \frac{π}{3})

\cos 2 θ - \sqrt{3} \sin 2 θ = \cos^{2} θ - \sin^{2} θ - \sqrt{3} \cdot 2 \sin θ \cos θ

= 2 \cdot \frac{1}{2} (\cos^{2} θ - \sin^{2} θ - \sqrt{3} \cdot 2 \sin θ \cos θ)

= 2 (\frac{1}{2} (\cos^{2} θ - \sin^{2} θ) - \frac{1}{2} \sqrt{3} \cdot 2 \sin θ \cos θ)

= 2 ((\cos^{2} θ - \sin^{2} θ) \cos \frac{π}{3} - 2 \sin θ \cos θ \sin \frac{π}{3})

= 2 (\cos 2 θ \cos \frac{π}{3} - \sin 2 θ \sin \frac{π}{3})

= 2 \cos (2 θ + \frac{π}{3})

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