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Question

The period of sin2θ is


A

π2

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B

π

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C

2π

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D

π2

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Solution

The correct option is B

π


Step 1: Form a function for given trigonometric expression

Given, sin2θ

We know that,

cos2θ=1-2sin2θ

2sin2θ=1-cos2θ

sin2θ=1-cos2θ2

sin2θ=12-cos2θ2 ...i

Step 2: Calculate the period

Now, as we know that the period of cosθ is 2π.

i.e., Tcosθ=2π

Tcos2θ=2π2

Tcos2θ=π

T12-cos2θ2=π

Tsin2θ=π [using equation i]

i.e., the period of sin2θ is π.

Hence, option (B) is the correct option.


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