The value of cos 2-6-sin 2-6- is

The correct option is B 1/2 cos 2 θ We have, cos^2 (π/6+θ ) sin^2 (π/6 θ ) = cos^2 (π/6+ θ ) cos^2 [π/2 (π/6 θ ) ] = cos^2 (π/6+θ ) cos^2 (π/3+θ ) = [cos ...

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Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

The value of ...

Question

The value of $c o s^{2} (\frac{π}{6} + θ) - s i n^{2} (\frac{π}{6} - θ)$ is

$\frac{1}{2} c o s 2 θ$

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$- \frac{1}{2} c o s 2 θ$

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$\frac{1}{2}$

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Solution

The correct option is A
$\frac{1}{2} c o s 2 θ$

We have,

$c o s^{2} (\frac{π}{6} + θ) - s i n^{2} (\frac{π}{6} - θ) = c o s^{2} (\frac{π}{6} + θ) - c o s^{2} [\frac{π}{2} - (\frac{π}{6} - θ)] = c o s^{2} (\frac{π}{6} + θ) - c o s^{2} (\frac{π}{3} + θ)$

$= [c o s (\frac{π}{6} + θ) + c o s (\frac{π}{3} + θ)] [c o s (\frac{π}{6} + θ) - c o s (\frac{π}{3} + θ)] = 2 c o s (\frac{\frac{π}{6} + θ + \frac{π}{3} + θ}{2}) c o s (\frac{\frac{π}{6} + θ - \frac{π}{3} - θ}{2}) 2 s i n (\frac{\frac{π}{6} + θ + \frac{π}{3} + θ}{2}) s i n (\frac{\frac{π}{2} + θ - \frac{π}{6} - θ}{6})$

$= 4 c o s (\frac{π}{4} + θ) c o s (- \frac{π}{12}) s i n (\frac{π}{4} + θ) s i n (\frac{π}{12}) = 4 c o s (\frac{π}{4} + θ) c o s (\frac{π}{12}) s i n (\frac{π}{4} + θ) s i n (\frac{π}{12}) = [2 s i n (\frac{π}{4} + θ) c o s (\frac{π}{4} + θ)] [2 s i n (\frac{π}{12}) c o s (\frac{π}{12})] = s i n (\frac{π}{2} + 2 θ) c o s \frac{π}{6} = c o s 2 θ \times \frac{1}{2} = \frac{1}{2} c o s 2 θ$

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The value of cos2 (π6+θ)− sin2 (π6−θ) is

The value of $c o s^{2} (\frac{π}{6} + θ) - s i n^{2} (\frac{π}{6} - θ)$ is