Is \displaystyle{\sin ^2}a = \frac{{1 - \cos 2a}}{2} True?

[Hernaner28]

Is this true?:

\displaystyle{\sin ^2}a = \frac{{1 - \cos 2a}}{2}


Because I've seen it in a book but isn't that

\displaystyle\sin a = \sqrt {1 - {{\cos }^2}a}
then:
\displaystyle{\sin ^2}a = 1 - {\cos ^2}a

and NOT:
\displaystyle{\sin ^2}a = \frac{{1 - \cos 2a}}{2}

thanks!

 

[Mark44]

Is this true?:

\displaystyle{\sin ^2}a = \frac{{1 - \cos 2a}}{2}Because I've seen it in a book but isn't that

\displaystyle\sin a = \sqrt {1 - {{\cos }^2}a}

That would be
$$ sin(a) = \pm \sqrt{1 - cos^2(a)}$$

then:
\displaystyle{\sin ^2}a = 1 - {\cos ^2}a

and NOT:
\displaystyle{\sin ^2}a = \frac{{1 - \cos 2a}}{2}

thanks!

They are both true.

Start with cos(2a) = cos²(a) - sin²(a)
= 1 - sin²(a) - sin²(a) = 1 - 2sin²(a)

So cos(2a) = 1 - 2sin²(a)

A little rearrangenment yields the identity you're asking about.