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PANKAJ

· started a discussion

· 1 Months ago

Trick (Question:111088.0)

Any other method?

Question:
The value of cos3\(\theta\) + cos3 (120° + \(\theta\)) + cos3 (\(\theta\)-120°) is :
Options:
A) \(\cfrac{\sqrt[]{3}}{2}cos\ \theta\)
B) \(\cfrac{3}{4}\sec^3 \theta\)
C) \(\cfrac{3}{2} \ ten^3 \theta\)
D) \(\cfrac{3}{4}\cos3 \theta\)
Solution:
Ans: (d)

cos3 θ + cos3 (120° +θ) + cos3 (θ-120°)

\(=\cfrac{1}{4}\)[(cos3θ+ 3cosθ)] + \(=\cfrac{1}{4}\)[cos (360° + 3θ) + 3cos (120° +θ)] + \(\cfrac{1}{4}\)[cos (3θ- 360°) + 3 cos (θ-120°)]

\(=\cfrac{1}{4}\)[cos 3θ + 3cosθ + cos3θ + 3 cos (120° + θ) + cos 3θ+ 3 cos (θ-120°)]

[∵ cos (360 + 3θ) = cos (360 - 3θ) = cos 3θ, cos (-θ) = cosθ]

\(=\cfrac{1}{4}\)[3 cos3θ+ 3cosθ + 3 cos (120° +θ) + 3 cos (θ - 120°)]

\(=\cfrac{1}{4}\)[3 cos 3θ + 3cosθ +\(\cfrac{3}{4}\){cos 120° cosθ - sin 120° sinθ} +\(\cfrac{3}{4}\) (cos 120° cosθ+ sin 120° sinθ )}]

\(=\cfrac{3}{4}\) [cos3θ + cos θ] +\(\cfrac{3}{4}\) (cosθ. 2cos120°)]

= \(\cfrac{3}{4}\)[cos 3θ+ cos θ + 2 × \(\cfrac{-1}{2}\) cosθ]

= \(\cfrac{3}{4}\)[cos 3θ + cos θ - cosθ]

[∵ cos 120°= \(\cfrac{-1}{2}]\) 

=\(\cfrac{3}{4}\)cos 3θ

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