Find the value of $4\cos 6\cos 42 \cos 60 \cos 66 \cos 78$ All angles in degrees

Question

I have absolutely no idea on how to go about this, except that cos 60 =1/2 :)

The usual way of multiplying by a sine doesn’t work here, and I feel very dumb for not being able to solve it. However, help would be appreciated.

Answer 1

$\begin{align}P &= 4\cos6°\cos42°\cos60°\cos66°\cos78° \cr &= {1\over2}(2\cos6° \cos66°)(2\cos42° \cos78°) \cr &= {1\over2}(\cos72° + \cos60°)(\cos120° + \cos36°)\cr &= {1\over8}(2\cos72° + 1)(-1 + 2\cos36°) \end{align}$

From Golden triangle, $\quad\quad2\cos36° = \phi = \large{1+\sqrt5 \over 2}$

From Golden ratio identity, $\phi^2 = \phi+1$

$2\cos72°+1 = 2(2\cos^2 36°-1) + 1 = \phi^2-1 = \phi$

$$P = {\phi(\phi-1) \over 8} = {\phi^2-\phi \over 8} = {1\over8}$$

Answer 2

Hint :-

Use- $\cos(a)\cos(60+a)\cos(60-a)=\frac14\cos3a$

Answer 3

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1. A trigonometric equation

2. Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$

3. Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$

4. Prove that $\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}$

Observe that $$\cos5x=\cos30^\circ$$ for $x=6^\circ,78^\circ,150^\circ,222^\circ=(180+42)^\circ,294^\circ=(360-66)^\circ$

Now $\cos5x+\cos x=2\cos3x\cos2x=2(4\cos^3x-3\cos x)(2\cos^2x-1)$

$\cos5x=16\cos^5x-(12+8)\cos^3x+(6-1)\cos x$

So, if $\cos5x=\cos5A,5x=360^\circ n\pm5A$ where $n$ is any integer

$x=72^\circ n+ A$ where $n=0,1,2,3,4$

So, the roots of $$16c^5-20c^3+5c-\cos5A=0$$ are $\cos(72^\circ n+ A)$ where $n=0,1,2,3,4$

$$\implies\prod_{n=0}^4\cos(72^\circ n+ A)=\dfrac{\cos5A}{16}$$

Can you complete from here?

See also :

Prove that $\tan55^\circ\cdot\tan65^\circ\cdot\tan75^\circ=\tan85^\circ$

Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$