The maximum value of c o s 2 ( π 3 − x ) − c o s 2 ( π 3 + x ) is Options: A . &nbsp - √ 3 2 B . &nbsp 1 2 C . &nbsp √ 3 2 D . &nbsp 3 2

Answer: Option C : C c o s 2 ( π 3 − x ) - c o s 2 ( π 3 + x ) = { c o s ( π 3 − x ) + c o s ( π 3 + x ) } { c o s ( π 3 − x ) − c o s ( π 3 + x ) } = { 2 c o s π 3 c o s x } { 2 s i n π 3 s i n x } = sin 2 π 3 sin 2x = √ 3 2 sin 2x Its maximum value is √ 3 2 , {- 1 ≤ sin 2x ≤ 1}.

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

Discussion Forum : Trigonometric Functions

Question -

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

Options:

A . -

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

B .

\frac{1}{2}

C .

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

D .

\frac{3}{2}

Answer: Option C
:
C

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

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

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

= 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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Discussion Forum : Trigonometric Functions

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