MHB What is the Simplified Form of the Trigonometric Expression?

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The discussion focuses on evaluating the expression $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{1}{\sin^2 \dfrac{3\pi}{10}}$, with the solution arriving at the value of 12. The approach involves using trigonometric identities and the concept of spread polynomials, particularly for angles that are multiples of each other. A key insight is that $\sin \left(\frac{\pi}{10}\right)$ relates to the Golden Ratio, specifically $\sin \left(\frac{\pi}{10}\right)=\frac{\phi}{2}$. The final derivation shows that $s=\frac{1}{8} \left(3+\sqrt{5}\right)$, confirming the solution. The discussion highlights the utility of spread polynomials in solving such trigonometric problems.
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Evaluate $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{1}{\sin^2 \dfrac{3\pi}{10}}$.
 
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I would use degrees to solve

we shall use the identity below to solve it

We have

$sin\, 72^0$
$= 2 \cos\, 36^0 \sin\, 36^0$ using sin 2A formula
$= 2 \cos\, 36^0 ( 2\, sin \,18^0cos\ 18^0) $using sin 2A formula again
$= 4\ cos\, 36^0 sin \,18 ^0 sin \,72^0$ as cos18 = sin(90-18) = sin 72
or
$4 \cos \,36^0 sin \,18^0= 1$
or
$\sin\, 54^0 sin\ 18^0 = \frac{1}{4}$
This is in my math blog at Fun with maths: Q13/092) Prove 4 cos 36 sin 18 = 1

another identity
$\cos^2 18^0 - cos^2 36^0 = (\cos \,18^0+ \cos \,36^0)*(\cos\, 18^0 - \cos \,36^0) $
$= \cos\, 9^0 * cos\, 27^0 * 2 *\ sin\, 9^0 * sin \,27 ^0$
$= \sin\, 18^0\ sin \,54^0 = \frac{1}{4}$

hence

$\frac{1}{\sin ^2 18^0} + \frac{1}{\sin ^2 54^0}$
=$\frac{\sin ^2 54^0 + \sin ^2 18^0}{\sin ^2 18^0\sin ^2 54^0}$
=$\frac{\cos ^2 36^0 + \sin ^2 18^0}{\sin ^2 18^0\sin ^2 54^0}$
=$\frac{\cos ^2 36^0 + 1- \cos ^2 18^0}{\sin ^2 18^0\sin ^2 54^0}$
$= 16( \cos^2 36^0 +1 - \cos ^2 18^0)$
$= 16 ( 1- \frac{1}{4})= 12$
 
Here is a "working backward" approach.
kaliprasad's solution is 12 which can be checked (approximately) with a calculator.

To verify that 12 is the solution, I'll try to derive a fact about one of my favorite angles $\theta =\frac{\pi }{10}$ , and the Golden Ratio. That is: $\sin \left(\frac{\pi }{10}\right)=\frac{\phi }{2}$

When solving problems where one angle is a (natural number) multiple another, it is useful to use a "Spread Polynomial". Click Here, for a table of spread polynomials. Let: $\theta =\frac{\pi }{10}$ and $s=\sin ^2(\theta )$

Then $\sin ^2(3 \theta )$= S3(s)
Where: S3(s) = $s(3-4s)^2$

Then the problem can be re-written as...

Given: $\frac{1}{(s (3-4 s))^2}+\frac{1}{s}$ = 12
Prove: $\sqrt{s}=\frac{\phi }{2}$

Solving for s: Click Here
$s=\frac{1}{8} \left(3+\sqrt{5}\right)$

Solving for $\sqrt{s}$:
$\sqrt{s}=\frac{\phi }{2}$ QED
 
RLBrown said:
Here is a "working backward" approach.
kaliprasad's solution is 12 which can be checked (approximately) with a calculator.

To verify that 12 is the solution, I'll try to derive a fact about one of my favorite angles $\theta =\frac{\pi }{10}$ , and the Golden Ratio. That is: $\sin \left(\frac{\pi }{10}\right)=\frac{\phi }{2}$

When solving problems where one angle is a (natural number) multiple another, it is useful to use a "Spread Polynomial". Click Here, for a table of spread polynomials. Let: $\theta =\frac{\pi }{10}$ and $s=\sin ^2(\theta )$

Then $\sin ^2(3 \theta )$= S3(s)
Where: S3(s) = $s(3-4s)^2$

Then the problem can be re-written as...

Given: $\frac{1}{(s (3-4 s))^2}+\frac{1}{s}$ = 12
Prove: $\sqrt{s}=\frac{\phi }{2}$

Solving for s: Click Here
$s=\frac{1}{8} \left(3+\sqrt{5}\right)$

Solving for $\sqrt{s}$:
$\sqrt{s}=\frac{\phi }{2}$ QED

hello Brown. Thanks for introducing me to spread polynomial

This is completely new to me.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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