CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Property 7

The value of ...

Question

The value of $\cos^{2} (\frac{π}{6} + x) - \sin^{2} (\frac{π}{6} - x)$ is
(a) $\frac{1}{2} \cos 2 x$

(b) 0

(c) $- \frac{1}{2} \cos 2 x$

(d) $\frac{1}{2}$

Open in App

Solution

(a) $\frac{1}{2} \cos 2 x$

$We have, \cos^{2} (\frac{π}{6} + x) - \sin^{2} (\frac{π}{6} - x) = \cos^{2} (\frac{π}{6} + x) - \cos^{2} [\frac{π}{2} - (\frac{π}{6} - x)] = \cos^{2} (\frac{π}{6} + x) - \cos^{2} (\frac{π}{3} + x) = [\cos (\frac{π}{6} + x) + \cos (\frac{π}{3} + x)] [\cos (\frac{π}{6} + x) - \cos (\frac{π}{3} + x)] = 2 \cos (\frac{\frac{π}{6} + x + \frac{π}{3} + x}{2}) \cos (\frac{\frac{π}{6} + x - \frac{π}{3} - x}{2}) 2 \sin (\frac{\frac{π}{6} + x + \frac{π}{3} + x}{2}) \sin (\frac{\frac{π}{3} + x - \frac{π}{6} - x}{2}) = 4 \cos (\frac{π}{4} + x) \cos (- \frac{π}{12}) \sin (\frac{π}{4} + x) \sin (\frac{π}{12}) = 4 \cos (\frac{π}{4} + x) \cos (\frac{π}{12}) \sin (\frac{π}{4} + x) \sin (\frac{π}{12}) = [2 \sin (\frac{π}{4} + x) \cos (\frac{π}{4} + x)] [2 \sin (\frac{π}{12}) \cos (\frac{π}{12})] = \sin (\frac{π}{2} + 2 x) \sin \frac{π}{6} = \cos 2 x \times \frac{1}{2} = \frac{1}{2} \cos 2 x$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The value of

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

\frac{1}{2} \cos 2 θ

- \frac{1}{2} \cos 2 θ

\frac{1}{2}

\cos^{2} (\frac{π}{6} + x) - \sin^{2} (\frac{π}{6} - x) = k \cos 2 x

then k = _____________.

The value of $\int_{- 1}^{2} \frac{| x |}{x}$ dx is

\frac{1}{x}

, x > 0, then its greatest value is

(a)

-

2

(b) 0

(c) 3

(d) none of these

Q. The value of

\int_{0}^{1} \tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x,

is

(a) 1
(b) 0
(c) −1
(d) π/4

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Property 7

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon