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

The correct option is A 1/2 cos 2 θ 1/2 cos 2 θ = cos^2 ( π/6+ θ ) sin^2 ( π/6 θ ) = cos ( π/6 + θ + π/6 θ ) cos ( π/6+6 π/6 + θ ) = cos 2 π/6/cos 2θ = ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

The value of ...

Question

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

\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 $\frac{1}{2} c o s 2 θ$

$\frac{1}{2} c o s 2 θ = c o s^{2} (\frac{π}{6} + θ) - s i n^{2} (\frac{π}{6} - θ)$
$= c o s (\frac{π}{6} + θ + \frac{π}{6} - θ) c o s (\frac{π}{6} + 6 - \frac{π}{6} + θ)$
$= \frac{c o s \frac{2 π}{6}}{c o s 2 θ}$
$= \frac{1}{2} c o s 2 θ$

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Similar questions

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

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The value of cos2(π6+θ)−sin2(π6−θ)

The correct option is A 12cos2θ 12cos2θ=cos2(π6+θ)−sin2(π6−θ) =cos(π6+θ+π6−θ)cos(π6+6−π6+θ) =cos2π6cos2θ =12cos2θ

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