ate a trigonometric identity that is useful in evaluating ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}2si{n}^{2}xdx$

hoose the correct answer below.

A$.sinx=cos(\frac{}{2}-x)$

B$.sin2x=2sinxcosx$

C$.si{n}^{2}x+co{s}^{2}x=1$

D$.si{n}^{2}x=\frac{1-cos2x}{2}$