Get the solution to "Prove thatcos2(θ)+cos2(θ+120∘)+cos2(θ−120∘)=32"

Alfredo Holmes

Answered question

2022-03-31

Prove that

\cos^{2} (θ) + \cos^{2} (θ + 120^{\circ}) + \cos^{2} (θ - 120^{\circ}) = \frac{3}{2}

pastuh7vka

Beginner2022-04-01Added 13 answers

I am assuming that

θ

is in degrees. Then using the following formulas:

\cos (x + y) = \cos x \cos y - \sin x \sin y

\cos (x - y) = \cos x \cos y + \sin x \sin y

{(a - b)}^{2} + {(a + b)}^{2} = 2 (a^{2} + b^{2})

\cos (120) = - \frac{1}{2}

\sin (120) = \frac{\sqrt{3}}{2}

\cos^{2} (x) + \sin^{2} (x) = 1

we have:

\cos^{2} (θ) + \cos^{2} (θ + 120) + \cos^{2} (θ - 120)

= \cos^{2} θ + {[\cos (θ) \cos (120) - \sin (θ) \sin (120)]}^{2} + {[\cos (θ) \cos (120) + \sin (θ) \sin (120)]}^{2}

= \cos^{2} (θ) + 2 [\cos^{2} (θ) \cos^{2} (120) + \sin^{2} (θ) \sin^{2} (120)]

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

= \cos^{2} (θ) + 2 [\frac{1}{4} \cos^{2} (θ) + \frac{3}{4} \sin^{2} (θ)]

= \cos^{2} (θ) + \frac{1}{2} \cos^{2} (θ) + \frac{3}{2} \sin^{2} (θ)

= (1 + \frac{1}{2}) \cos^{2} (θ) + \frac{3}{2} \sin^{2} (θ)

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

= \frac{3}{2} [\cos^{2} (θ) + \sin^{2} (θ)]

= \frac{3}{2} \cdot 1 = \frac{3}{2}

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