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Maths-

General

Easy

Question

Period of $sin space x sin space open parentheses 120 to the power of ring operator minus x close parentheses sin space open parentheses 120 to the power of ring operator plus x close parentheses$ is

$\frac{π}{3}$
$\frac{2 π}{3}$
$π$
$\frac{π}{2}$

Hint:

In the question we have to find the fundamental period of the given equation. For this we will first simplify the given equation using some trigonometric identities then by using periodic functions we can find the required result.

The correct answer is: $fraction numerator 2 pi over denominator 3 end fraction$

$sin space x sin space open parentheses 120 to the power of ring operator minus x close parentheses sin space open parentheses 120 to the power of ring operator plus x close parentheses equals sin x space left parenthesis sin squared 120 degree minus sin squared x right parenthesis equals sin x space open square brackets open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses squared minus sin squared x close square brackets equals sin x space open square brackets 3 over 4 minus sin squared x close square brackets equals 1 fourth open square brackets 3 space sin x space minus 4 sin squared x close square brackets equals 1 fourth sin 3 x f u n d a m e n t a l space p e r i o d space o f space sin space a x space i s space fraction numerator 2 straight pi over denominator a end fraction. S o comma space p e r i o d space o f space 1 fourth sin 3 x space i s fraction numerator 2 straight pi over denominator 3 end fraction.$

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