Test Index

AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Section: Mathematics

© examsnet.com

Question : 21 of 160

Marks: +1, -0

If

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

are in a harmonic progression, then

\cos x=

$\frac{3}{2}$
$1$
$\frac{\sqrt{3}}{2}$
$\sqrt{\frac{3}{2}}$

Solution:

It is given that,

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

are in

\text{H.P.}

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

\cos x = \frac{2 \left(\cos^2 x-\sin^2 \frac{\pi}{3}\right)}{2 \cos x \times \cos \frac{\pi}{3}}

\cos^2 x \left(1-\cos \frac{\pi}{3}\right) = \sin^2 \frac{\pi}{3}

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

Expanding the above we get.

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

\cos^2 x = 1+\cos \frac{\pi}{3}

\cos^2 x = 2 \cos^2 \frac{\pi}{6} = 2 \times \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{2}

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

© examsnet.com

Go to Question:

Prev Question

Next Question