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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?

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\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}

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

Like prove that : cosx.cos(x-60).cos(x+60)= (1/4)cos3x

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

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

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