The maximum value of cos2 -3 - x - cos2 -3 - x is

Cos^2 ( π/3 x ) cos^2 ( π/3 + x ) = {cos(π/3 x) + cos(π/3 + x )} {cos( π/3 x ) cos(π/3 + x)} = {2 cos π/3 cos x} {2 sin π/3 sin x} = sin 2 π/3 sin ...

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Byju's Answer

Standard XIII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

The maximum v...

Question

The maximum value of $c o s^{2} (\frac{π}{3} - x) - c o s^{2} (\frac{π}{3} + x)$ is

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

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$\frac{1}{2}$

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

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$\frac{3}{2}$

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Solution

The correct option is C
$\frac{\sqrt{3}}{2}$

$c o s^{2} (\frac{π}{3} - x)$ - $c o s^{2} (\frac{π}{3} + x)$

= ${c o s (\frac{π}{3} - x) + c o s (\frac{π}{3} + x)}$ ${c o s (\frac{π}{3} - x) - c o s (\frac{π}{3} + x)}$

= ${2 c o s \frac{π}{3} c o s x}$ ${2 s i n \frac{π}{3} s i n x}$

= sin $\frac{2 π}{3}$ sin 2x = $\frac{\sqrt{3}}{2}$ sin 2x

Its maximum value is $\frac{\sqrt{3}}{2}$ , {- 1 $\leq$ sin 2x $\leq$ 1}.

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The maximum value of cos2(π3−x)−cos2(π3+x) is

The correct option is C √32 cos2(π3−x) - cos2(π3+x) = {cos(π3−x)+cos(π3+x)} {cos(π3−x)−cos(π3+x)} = {2cosπ3cosx} {2sinπ3sinx} = sin 2π3 sin 2x = √32 sin 2x Its maximum value is √32, {- 1 ≤ sin 2x ≤ 1}.

The maximum value of $c o s^{2} (\frac{π}{3} - x) - c o s^{2} (\frac{π}{3} + x)$ is