MHB What is the Simplified Form of the Trigonometric Expression?

AI Thread Summary
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.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Evaluate $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{1}{\sin^2 \dfrac{3\pi}{10}}$.
 
Mathematics news on Phys.org
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.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...

Similar threads

Replies
1
Views
958
Replies
1
Views
1K
Replies
1
Views
1K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
2
Views
956
Back
Top