Prove that $2\cdot \cos \frac{72}{2}\cdot \cos \frac{24}{2}+2\cdot \sin \frac{96}{2}\cdot \sin \frac{72}{2}=0.5$. Additional data added To solve it I have tried some options, where in one of them I  applied product to sum formulas, which seemed to be very helpful, but didn't get the answer.
 Used these formulae: 
$$\cos(a+b)+ \cos(a-b)=2\cdot \cos(a)\cdot \cos(b)$$
$$−\cos(a+b)+\cos(a−b)=2\sin(a)\sin(b)$$
And now, I am stuck at this:
$$\cos(48)+\cos(24)-\cos(84)+\cos(12)=0.5$$
Then tried to make the arguments similar using a double angle and a half angle formulae $$2\cos^2(24)-1+\cos(24)+\sqrt{\frac{1+\cos(24)}{2}}+\cos(84)=0.5$$ Anyway, I can't show that expression above really is 0.5.
 A: Using the formulae and supposing that you work on degrees :
$$ \cos(a+b)+\cos(a−b)=2\cos(a)\cos(b)$$
$$−\cos(a+b)+\cos(a−b)=2\sin(a)\sin(b) $$
we apply them and transform your given equation :
$$2\cos \frac{72^\circ}{2} \cos \frac{24^\circ}{2}+2\sin \frac{96^\circ}{2} \sin \frac{72^\circ}{2}= \cos\bigg(\frac{72^\circ}{2} + \frac{24^\circ}{2}\bigg)  + \cos\bigg(\frac{72^\circ}{2} - \frac{24^\circ}{2}\bigg) + \cos\bigg(\frac{96^\circ}{2} - \frac{72^\circ}{2}\bigg) - \cos\bigg(\frac{96^\circ}{2} + \frac{72^\circ}{2}\bigg)  $$
$$=$$
$$\cos(48^\circ) + \cos(24^\circ) +  
\sin(12^\circ) - \sin(84^\circ) = \frac{1}{2}$$
A: If the angles are in degrees, the equation clearly cannot hold: all of the sines and cosines in the expression are positive.  Since sine is increasing between $0^\circ$ and $90^\circ$, we have
$$
2\sin48^\circ\sin36^\circ>2\sin45^\circ\sin30^\circ=\frac{1}{\sqrt{2}}>\frac{1}{2}.
$$
The entire left side is therefore greater than $0.5.$  Indeed, the value of the left side is approximately $2.4563$.
If the angles are in radians, the statement seems incredibly unlikely, and would be spectacular if true.  Alas, a numerical check shows that it isn't.
