# 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.

• I don’t understand. Why do you write $\frac{72}2$ instead of $36$, for instance? – Lubin Jun 20 '17 at 20:13
• So was written on that sheet of paper, but I know that it is 36 :). – user36339 Jun 20 '17 at 20:15
• Do you know these formulea ? \begin{eqnarray*} \cos(a+b)+\cos(a-b) = 2 \cos(a) \cos(b) \\ -\cos(a+b)+\cos(a-b) = 2 \sin(a) \sin(b) \end{eqnarray*} – Donald Splutterwit Jun 20 '17 at 20:15
• No, I didn't.Thanks! – user36339 Jun 20 '17 at 20:50
• Note that if these are degrees, then the degree sign is obligatory; otherwise the unit is taken to be radians. For instance, $\sin 90^\circ = 1$, while $\sin 90 = 0.89399\ldots$. The degree sign can be typeset using ^\circ. – Théophile Jun 20 '17 at 21:05

## 2 Answers

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.

• In fact, one can say that the statement with angles in radians must be false on general principles: it would imply that $\cos 12$ is algebraic. (The left side is a polynomial in $\cos 12$ using angle-sum identities.) Then Henning Makholm's answer here provides a contradiction. – Will Orrick Jun 21 '17 at 15:13
• Thanks, Will Orrick! – user36339 Jun 21 '17 at 15:45

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 :

$$2\cos \frac{72^\circ}{2} \cos \frac{24^\circ}{2}+2\sin \frac{96^\circ}{2} \sin \frac{72^\circ}{2}= \cos\bigg(\frac{72^\circ}{2} + \frac{24^\circ}{2}\bigg) + \cos\bigg(\frac{72^\circ}{2} - \frac{24^\circ}{2}\bigg) + \cos\bigg(\frac{96^\circ}{2} - \frac{72^\circ}{2}\bigg) - \cos\bigg(\frac{96^\circ}{2} + \frac{72^\circ}{2}\bigg)$$

$$=$$

$$\cos(48^\circ) + \cos(24^\circ) + \sin(12^\circ) - \sin(84^\circ) = \frac{1}{2}$$

• So how do you prove the last equality, given that none of these angles are mirrors of each other? – Steven Stadnicki Jun 20 '17 at 21:36