Question

Someone help me pliisss

Answer 1

The prove of the given identity is shown below:

Explanation:

It seems that the question involves the proving of a trigonometric identity.

Therefore, we are going to work independently on each side of the equal sign.

We start by considering
`cos(\alpha)=cos((\alpha)/(2) +(\alpha)/(2))` since the angle alpha is the same as the addition of two halves of it. Then we apply the identity for the cosine of an addition of angles:
`cos ((\alpha)/(2) +(\alpha)/(2))=cos((\alpha)/(2))*cos((\alpha)/(2))-sin((\alpha)/(2))*sin((\alpha)/(2))= cos^2((\alpha)/(2))-sin^2((\alpha)/(2))`

and now we use the Pythagorean identity to replace the cosine square expression:

`cos^2((\alpha)/(2))=1-sin^2((\alpha)/(2))`

in the difference of square trig functions we had arrived to:

`cos^2((\alpha)/(2))-sin^2((\alpha)/(2))=1-sin^2((\alpha)/(2))--sin^2((\alpha)/(2))=1-2\,sin^2((\alpha)/(2))`

We have arrived at the exact same expression the original identity shown on the righthand side of the equal sign, so we have proved the identity.