Answer:
The prove of the given identity is shown below:
Step-by-step 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 [tex]cos(\alpha)=cos(\frac{\alpha}{2} +\frac{\alpha}{2})[/tex] 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: [tex]cos (\frac{\alpha}{2} +\frac{\alpha}{2})=cos(\frac{\alpha}{2})*cos(\frac{\alpha}{2})-sin(\frac{\alpha}{2})*sin(\frac{\alpha}{2})= cos^2(\frac{\alpha}{2})-sin^2(\frac{\alpha}{2})[/tex]
and now we use the Pythagorean identity to replace the cosine square expression:
[tex]cos^2(\frac{\alpha}{2})=1-sin^2(\frac{\alpha}{2})[/tex]
in the difference of square trig functions we had arrived to:
[tex]cos^2(\frac{\alpha}{2})-sin^2(\frac{\alpha}{2})=1-sin^2(\frac{\alpha}{2})--sin^2(\frac{\alpha}{2})=1-2\,sin^2(\frac{\alpha}{2})[/tex]
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.
