deduce that $\cos 6° \cos42° \cos66° \cos78°= \frac{1}{16}$

Question

Prove that $$4 \cos\theta \cos(\frac{\pi}{3}-\theta) \cos(\frac{\pi}{3}+\theta)= \cos 3\theta$$ and deduce that $$\cos 6° \cos42° \cos66° \cos78°= \frac{1}{16}$$

I have proved by using $2 \cos A \cos B= \cos (A+B)+ \cos (A-B) $

But I cant deduce $1/16$

P.S. we have to use prove that to deduce $1/16$

You can apply the identity with $\theta = 18^\circ$ and with $\theta=6^\circ$. — Jonas Meyer, Jan 02 '15 at 07:05
Interesting product there. Looks like it can be rewritten as $\sin96^\circ\sin48^\circ\sin24^\circ\sin12^\circ$. — Mike, Jan 02 '15 at 08:31
@Freddy, Related: http://math.stackexchange.com/questions/455070/proving-a-fact-tan6-circ-tan42-circ-tan12-circ-tan24-cir — lab bhattacharjee, Jan 04 '15 at 15:22

score 2 · Accepted Answer · answered Jan 02 '15 at 07:20

2

$$\begin{align} \cos6\cos42\cos66\cos78&=\frac{1}{\cos54\cos18}\left(\cos6\cos54\cos66\right)\left(\cos18\cos42\cos78\right)\\ &=\frac{1}{\cos54\cos18}\frac{\cos18}{4}\frac{\cos54}{4}\\ &=\frac{1}{16} \end{align}$$

answered Jan 02 '15 at 07:20

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deduce that $\cos 6° \cos42° \cos66° \cos78°= \frac{1}{16}$

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