Can You Verify This Trigonometric Identity Involving Cosecant Functions?

[anemone]

Here is this week's POTW:

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Prove that $\csc 6^{\circ}+\csc 78^{\circ}- \csc 42^{\circ} - \csc 66^{\circ}=8$.

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[bob012345]

I hoped the popular math book I have Trigonometric Delights by Eli Maor would come in handy someday..

Here is my proof.

Spoiler

Since csc is just inverse sine we have to prove

$$ \frac{1}{sin{(6)}} + \frac{1}{sin{(78)}} - \frac{1}{sin{(42)}} - \frac{1}{sin{(66)}} = 8$$

grouping as

$$ \left(\frac{1}{sin{(6)}} - \frac{1}{sin{(66)}}\right) + \left(\frac{1}{sin{(78)}} - \frac{1}{sin{(42)}}\right)$$

$$\left(\frac{sin(66) - sin(6)}{sin(6)sin(66)}\right) - \left(\frac{sin(78) - sin(42)}{sin(42)sin(78)}\right)$$

using the identities
$$sin(A) - sin(B) = 2 cos(\frac{A + B)}{2})sin(\frac{(A - B)}{2}$$
$$sin(A)sin(B) = \frac{sin(A + B) + sin(A - B)}{2}$$
we have
$$ sin(66) - sin(6) = 2cos(\frac{72}{2})sin(\frac{60}{2}) = 2cos(36)sin(30)$$
$$sin(6)sin(66) = \frac{cos(60) - cos(72)}{2}$$
$$ sin(78) - sin(42) = 2cos(60)sin(18) = 2cos(60)cos(72)$$
since $sin(18)=cos(90-18) = cos(72)$ and
$$sin(78)sin(42) = \frac{cos(36) - cos(120)}{2}$$

putting it together the terms are


$$ \frac{4cos(36)sin(30)}{cos(60) - cos(72) } = \frac{4cos(36)}{1 - 2cos(72) }$$

$$ \frac{4cos(60)cos(72)}{cos(36) - cos(120) } = \frac{4cos(72)}{1 + 2cos(36) }$$

so we have

$$4\left(\frac{cos(36)}{1 - 2cos(72) } - \frac{cos(72)}{1 + 2cos(36) }\right)$$At this point we need an actual value since this relationship does not work for all angles otherwise we could just use more trig identities. It is well known that $$sin(54)=\frac{1 + \sqrt{5}}{4} = cos(36)$$ then we can get $$cos(72) = \frac{-1 + \sqrt{5}}{4}$$
then we have
$$4 \left( \frac{ \large\frac{ \sqrt{5} +1}{4}}{1 - 2(\frac{ \sqrt{5} -1)}{4})} - \frac{ \large\frac{ \sqrt{5} -1}{4}}{1 + 2(\frac{ \sqrt{5} +1)}{4})} \right)$$

which simplifies to

$$ 4\left( \frac{ \sqrt{5} +1}{6 - 2\sqrt{5}} - \frac{ \sqrt{5} -1}{6 + 2\sqrt{5}}\right) $$

getting a common denominator

$$4\left(\frac{(1+\sqrt{5})(6+2\sqrt{5}) + ((1-\sqrt{5})(6-2\sqrt{5})}{(6+2\sqrt{5})(6-2\sqrt{5})}\right)$$
$$4\left( \frac{6 + 8\sqrt{5} + 10 + 6 - 8\sqrt{5} +10}{36 - 12\sqrt{5} + 12\sqrt{5} - 20}\right)$$
$$ 4\left( \frac{32}{16}\right) = 8$$

[\SPOILER]