What is the Simplified Form of the Trigonometric Expression?

[anemone]

Evaluate $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{1}{\sin^2 \dfrac{3\pi}{10}}$.

 

[kaliprasad]

I would use degrees to solve

we shall use the identity below to solve it

We have

Spoiler

$sin\, 72^0$
$= 2 \cos\, 36^0 \sin\, 36^0$ 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

Spoiler

$\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

Spoiler

$\frac{1}{\sin ^2 18^0} + \frac{1}{\sin ^2 54^0}$
=$\frac{\sin ^2 54^0 + \sin ^2 18^0}{\sin ^2 18^0\sin ^2 54^0}$
=$\frac{\cos ^2 36^0 + \sin ^2 18^0}{\sin ^2 18^0\sin ^2 54^0}$
=$\frac{\cos ^2 36^0 + 1- \cos ^2 18^0}{\sin ^2 18^0\sin ^2 54^0}$
$= 16( \cos^2 36^0 +1 - \cos ^2 18^0)$
$= 16 ( 1- \frac{1}{4})= 12$

 

[RLBrown]

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

 

[kaliprasad]

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.

 

[CellCoree]

I would approach this trigonometric challenge by first recognizing that the expression can be simplified using the trigonometric identity $\sin^2x + \cos^2x = 1$. By substituting $\cos^2x = 1 - \sin^2x$, we can rewrite the expression as $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{1}{(1-\sin^2 \dfrac{\pi}{10})}$. Simplifying further, we get $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{1}{\cos^2 \dfrac{\pi}{10}}$.

Next, I would use the double angle formula for cosine, $\cos2x = 1-2\sin^2x$, to rewrite the expression as $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{1}{1-\dfrac{1}{2}(1-\cos\dfrac{\pi}{5})}$. Simplifying again, we get $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{1}{\dfrac{1}{2}\cos\dfrac{\pi}{5}}$.

Using the half angle formula for sine, $\sin\dfrac{x}{2} = \pm\sqrt{\dfrac{1-\cos x}{2}}$, we can rewrite the expression as $\dfrac{1}{\sin^2 \dfrac{\pi}{10}}+\dfrac{2}{\cos\dfrac{\pi}{5}\sqrt{1-\cos\dfrac{\pi}{5}}}$.

Finally, I would use the Pythagorean identity, $\sin^2x + \cos^2x = 1$, to simplify the expression to $\dfrac{1+\sin\dfrac{\pi}{5}}{\sin^2\dfrac{\pi}{10}\cos\dfrac{\pi}{5}}$. This is the final form of the expression and cannot be simplified any further without specific values for the trigonometric functions.