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Evaluation of a Determinant

The value of ...

Question

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

(b) 0

(c) $- \frac{1}{2} \cos 2 θ$

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

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Solution

(a) $\frac{1}{2} \cos 2 θ$

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

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Similar questions

Q. The value of

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

\frac{1}{2} \cos 2 x

- \frac{1}{2} \cos 2 x

\frac{1}{2}

Q. Question 3
The value of the expression

c o s e c (75^{\circ} + θ) - s e c (15^{\circ} - θ) - t a n (55^{\circ} + θ) + c o t (35^{\circ} - θ)

is

(A) -1
(B) 0
(C) 1
(D)

\frac{3}{2}

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

then k = _____________.

(θ) = s i n θ (s i n θ + s i n 3 θ) . T h e n f (θ)

Q. The value of

\sin (\frac{π}{4} + θ) - \cos (\frac{π}{4} - θ)

is
(a) 2 cosθ
(b) 2 sinθ
(c) 1
(d) 0

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