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How do I establish this and similar values of trigonometric functions?

$$ 4 \sin 72^\circ \sin 36^\circ = \sqrt 5 $$

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A key observation is $36 + 72 + 72 = 180$. This means $$\sin 72 = \sin (36 + 72) = \sin 36 \cos 72 + \cos 36 \sin 72$$

Using the identity $\sin 72 = 2\sin 36 \cos 36$, we get $$2\sin 36 \cos 36 = \sin 36 \cos 72 + 2 \sin 36 \cos^2 36$$

Now notice $\sin 36$ appears everywhere, so we can divide by it... $$2\cos 36 = \cos 72 + 2\cos^2 36$$

Use the identity $\cos 72 = 2 \cos^2 36 - 1$, we get $$2\cos 36 = 4\cos^2 36 - 1$$

So $\cos36$ is one of the solutions to $2x = 4x^2 - 1$. Using the quadratic formula, you see that the solutions are $$\frac {2 \pm \sqrt{20}}4 = \frac {1 \pm \sqrt 5}2$$

But you know that $\cos 36$ must be positive, so you know the correct value. Now you can get $\sin 72$ from any identity you wish, and multiply.

How to get such values "in general": This is hard. The reason $36$ and $72$ are easy values is that a small integer multiple of them gives you a multiple of $180$. If you wanted to compute such a value for let's say $\sin 1$, it would be a very nasty expression, as you would have to iterate many times until you reach a "basic" angle like $180$ (or even $30$).

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Hint: Notice that $72 = 2\cdot 36$ and $180 = 36\cdot 5$; use formulaes for halfed or doubled argument.

$$ \sin^2\frac{x}{2} = \frac{1-\cos x}{2} $$

or

$$ \sin2x = 2\sin x\cos x $$

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  • $\begingroup$ I tried using these but I don't think they're enough $\endgroup$ – John Fernley May 31 '14 at 19:01
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Using Werner Formula, $$2\sin36^\circ\sin72^\circ=\cos36^\circ-\cos108^\circ$$

$$\cos108^\circ=\cos(180^\circ-72^\circ)=-\cos72^\circ$$

Now, $\displaystyle\cos36^\circ\cos72^\circ=\frac{2\sin36^\circ\cos36^\circ}{2\sin36^\circ}\cos72^\circ=\frac{2\sin72^\circ\cos72^\circ}{4\sin36^\circ}=\frac{\sin144^\circ}{4\sin(180^\circ-144^\circ)}=\frac14$

From Help needed with trigonometric identity, $\displaystyle\cos36^\circ-\cos72^\circ=\sin54^\circ-\sin18^\circ=\frac12$

$\displaystyle\implies2\sin36^\circ\sin72^\circ=\cos36^\circ+\cos72^\circ=+\sqrt{(\cos36^\circ-\cos72^\circ)^2+4\cos36^\circ\cos72^\circ}$

$\displaystyle=\sqrt{\left(\frac12\right)^2+1}=\frac{\sqrt5}2$

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A different approach

$ a = n\pi/5$

$5a = n\pi$

$3a = n\pi-2a$

$\sin 3a = \sin{(n\pi-2a)}$

$\sin 3a = \sin2a$

Expand using identities

Remove a sin Square both sides

You get a quadratic in $\sin^2$

Let $\sin^2 a = x$

So

$16x^2 -20 a +5 = 0$

So

$\sin^2 72 \cdot\sin^2 36 = 5/16$

Hence

$4\sin72\sin36 = \sqrt5$

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