Find the value of ² x + ²≤ft(x + π/6) + ²≤ft(x - π/6).

Correct Answer is : rac3+2cos⁡2x2rac{3+2 2x}{2}rac3+2cos2x2

Correct Answer is : rac3+2cos⁡2x2rac{3+2 2x}{2}rac3+2cos2x2

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UPSC NDA Math Model Paper 5

Question : 117 of 120

Marks: +1, -0

\cos^{2} x + \cos^{2}\left(x + \frac{\pi}{6}\right) + \cos^{2}\left(x - \frac{\pi}{6}\right)

Solution:

We have to find the value of

\cos^{2} x + \cos^{2}\left(x + \frac{\pi}{6}\right) + \cos^{2}\left(x - \frac{\pi}{6}\right)

Rewrite the given expression by using double angle formula as follows:

\cos^{2} x + \cos^{2}\left(x + \frac{\pi}{6}\right) + \cos^{2}\left(x - \frac{\pi}{6}\right)

= \frac{1+\cos 2x}{2} + \frac{1+\cos\left(2x+\frac{\pi}{3}\right)}{2}

+ \frac{1+\cos\left(2x-\frac{\pi}{3}\right)}{2}

= \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}[\cos 2x + \cos\left(2x+\frac{\pi}{3}\right)]

+ \cos\left(2x-\frac{\pi}{3}\right)

= \frac{1}{2}[3 + \cos 2x + 2 \cos 2x \cos\left(\frac{\pi}{3}\right)]

[\because \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)]

= \frac{3+2\cos 2x}{2}

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