Question

The period of 2sinxcosx is

$ {\text{A}}{\text{. 4}}{\pi ^2} \\
  {\text{B}}{\text{. 2}}\pi \\
  {\text{C}}{\text{. 4}}\pi \\
  {\text{D}}{\text{. }}\pi \\ $

Answer 1

Hint: -To solve this question first we have to shorten the question using standard trigonometric results then using property of finding period of trigonometric function we have to find its period.
We have given,
2sinxcosx
We know the formula ( sin2x = 2 sinx.cosx) using this formulae we get,
Sin2x
And we know that period of sinkx is $\dfrac{{2\pi }}{{\left| k \right|}}$
So here period of sin2x will be $\dfrac{{2\pi }}{{\left| 2 \right|}} = \pi $
Hence $\pi $ is the period of 2sinxcosx
Hence option D is the correct option.
Note:- Whenever we get this type of question the key concept of solving is we have to have knowledge of property that period of sinkx is $\dfrac{{2\pi }}{{\left| k \right|}}$ and many more properties like this should be remembered to solve questions of period easily.