# How to change the appearence of the correct answer of $\cos55^\circ\cdot\cos65^\circ\cdot\cos175^\circ$

I represented the problem in the following view and solved it: \begin{align}-\sin35^\circ\cdot\sin25^\circ\cdot\sin85^\circ\cdot\sin45^\circ&=A\cdot\sin45^\circ\\ -\frac{1}{2}(\cos20^\circ-\cos70^\circ)\cdot\frac{1}{2}(\cos50^\circ-\cos120^\circ)&=A\cdot\sin45^\circ\\ \cos20^\circ\cdot\cos50^\circ-\cos50^\circ\cdot\cos70^\circ+\frac{\cos20^\circ}{2}-\frac{\cos70^\circ}{2}&=A\cdot(-4)\cdot \sin45^\circ\\ \frac{1}{2}(\cos30^\circ+\cos70^\circ)-\frac{1}{2}(\cos20^\circ+\cos120^\circ)&=A\cdot(-4)\cdot \sin45^\circ\\ \frac{\sqrt{3}}{4}+\frac{\cos70^\circ}{2}-\frac{\cos20^\circ}{2}+\frac{1}{4}+\frac{\cos20^\circ}{2}-\frac{\cos70^\circ}{2}&=-2\sqrt{2}A\\ A&=-\frac{\sqrt{6}+\sqrt{2}}{16} \end{align} I believe that the above answer is true. But that didn't match a variant below:

A) $$-\frac{1}{8}$$

B)$$-\frac{\sqrt{3}}{8}$$

C) $$\frac{\sqrt{3}}{8}$$

D) $$-\frac{1}{8}\sqrt{2-\sqrt{3}}$$

E) $$-\frac{1}{8}\sqrt{2+\sqrt{3}}$$

I did the problem over again. After getting the same result, I thought that the apperance of my answer could be changed to match one above, so I tried to implement one of formulae involving radical numbers: all to no avail. How to change that?

## 2 Answers

\begin{align} A &= - \sqrt{\left(\frac{\sqrt{6}+\sqrt{2}}{16}\right)^2}\\ &=- \sqrt{\left(\frac{\sqrt{3}+1}{8\sqrt{2}}\right)^2}\\ &=- \frac18\sqrt{\left(\frac{3+1+2\sqrt3}{2}\right)}\\ &=- \frac18\sqrt{2+\sqrt{3}}\\ \end{align}

As $\cos175^\circ=\cos(180^\circ-5^\circ)=-\cos5^\circ,$

$$4\cos(60^\circ-5^\circ)\cos5^\circ\cos(60^\circ+5^\circ)=\cos(3\cdot5^\circ)$$

Now use $15=60-45$ or $=45-30$