# Trigonometry challenge

#### Pranav

##### Well-known member
Evaluate:
$$\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}}$$

#### Opalg

##### MHB Oldtimer
Staff member
First, \begin{aligned}\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}} &= \Bigl(\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}} +\frac{1}{\sin^260^{\circ}} +\frac{1}{\sin^280^{\circ}}\Bigr) - \frac{1}{\sin^260^{\circ}} - \frac{1}{\cos^245^{\circ}} \\ &= \Bigl(\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}} +\frac{1}{\sin^260^{\circ}} +\frac{1}{\sin^280^{\circ}}\Bigr) - \frac43 -2.\end{aligned} Next, $\theta=20^\circ$, $\theta= 40^\circ$, $\theta= 60^\circ$ and $\theta= 80^\circ$ all satisfy $\sin(9\theta) = 0$. But $\sin(9\theta) = T_9(\sin\theta)$, where $T_9$ is the Chebyshev polynomial $T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x.$ Thus $x = \sin(20k^\circ)\ (k=1,2,3,4)$ are all roots of that polynomial. Dividing by $x$ (because we want to ignore the root $x=0$) and then replacing $x$ by $x^2$, we see that $x = \sin^2(20k^\circ)\ (k=1,2,3,4)$ are the roots of $256x^4 - 576x^3 + 432x^2 - 120x + 9.$ Then replacing $x$ by $1/x$ (and multiplying through by $x^4$), it follows that $x = \dfrac1{\sin^2(20k^\circ)}\ (k=1,2,3,4)$ are the roots of $9x^4 - 120 x^3 + 432 x^2 - 576x^3 + 256 = 0.$ The sum of the roots is $\dfrac{120}9 = \dfrac{40}3.$ Therefore $$\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}} = \frac{40}3 - \frac43 - 2 = 10.$$

#### Pranav

##### Well-known member
First, \begin{aligned}\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}} &= \Bigl(\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}} +\frac{1}{\sin^260^{\circ}} +\frac{1}{\sin^280^{\circ}}\Bigr) - \frac{1}{\sin^260^{\circ}} - \frac{1}{\cos^245^{\circ}} \\ &= \Bigl(\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}} +\frac{1}{\sin^260^{\circ}} +\frac{1}{\sin^280^{\circ}}\Bigr) - \frac43 -2.\end{aligned} Next, $\theta=20^\circ$, $\theta= 40^\circ$, $\theta= 60^\circ$ and $\theta= 80^\circ$ all satisfy $\sin(9\theta) = 0$. But $\sin(9\theta) = T_9(\sin\theta)$, where $T_9$ is the Chebyshev polynomial $T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x.$ Thus $x = \sin(20k^\circ)\ (k=1,2,3,4)$ are all roots of that polynomial. Dividing by $x$ (because we want to ignore the root $x=0$) and then replacing $x$ by $x^2$, we see that $x = \sin^2(20k^\circ)\ (k=1,2,3,4)$ are the roots of $256x^4 - 576x^3 + 432x^2 - 120x + 9.$ Then replacing $x$ by $1/x$ (and multiplying through by $x^4$), it follows that $x = \dfrac1{\sin^2(20k^\circ)}\ (k=1,2,3,4)$ are the roots of $9x^4 - 120 x^3 + 432 x^2 - 576x^3 + 256 = 0.$ The sum of the roots is $\dfrac{120}9 = \dfrac{40}3.$ Therefore $$\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}} = \frac{40}3 - \frac43 - 2 = 10.$$

Btw, is their any elementary solution to this? I wonder if the problem really involves the use of such complicated approach as it is a problem from one of my past test papers.

#### anemone

##### MHB POTW Director
Staff member
My solution:

\begin{align*}\dfrac{1}{\cos^210^{\circ}}+\dfrac{1}{\sin^220^{ \circ }}+\dfrac{1}{\sin^240^{\circ}}-\dfrac{1}{\cos^245^{\circ}}&=\sec^210^{\circ}+\csc^220^{ \circ }+\csc^240^{\circ}-2\\&=3+\tan^210^{\circ}+\cot^220^{ \circ }+\cot^240^{\circ}-2\\&=1+\tan^210^{\circ}+\tan^270^{ \circ }+\tan^250^{\circ}\\&=1+9\tan^230^{\circ}+6\\&=10 \end{align*}

#### Pranav

##### Well-known member
My solution:

\begin{align*}\dfrac{1}{\cos^210^{\circ}}+\dfrac{1}{\sin^220^{ \circ }}+\dfrac{1}{\sin^240^{\circ}}-\dfrac{1}{\cos^245^{\circ}}&=\sec^210^{\circ}+\csc^220^{ \circ }+\csc^240^{\circ}-2\\&=3+\tan^210^{\circ}+\cot^220^{ \circ }+\cot^240^{\circ}-2\\&=1+\tan^210^{\circ}+\tan^270^{ \circ }+\tan^250^{\circ}\\&=1+9\tan^230^{\circ}+6\\&=10 \end{align*}
Excellent!

But can you please explain how do you get the following:
$$1+\tan^210^{\circ}+\tan^270^{ \circ }+\tan^250^{\circ}=1+9\tan^230^{\circ}+6$$

Thanks!

#### anemone

##### MHB POTW Director
Staff member
Excellent!

But can you please explain how do you get the following:
$$1+\tan^210^{\circ}+\tan^270^{ \circ }+\tan^250^{\circ}=1+9\tan^230^{\circ}+6$$

Thanks!
Sure!

Do you know there is such an identity that tells us $\tan^2x^{\circ}+\tan^2(60-x)^{ \circ }+\tan^2(60+x)^{\circ}=9\tan^2(3x)^{\circ}+6$ ?

That is the exact identity that I used to translate the messy sum of three square terms of tangent functions into number!

But seriously though, I don't know for sure if we could just use that identity right away or we needed to prove it first before applying.

#### Pranav

##### Well-known member
Thanks anemone!

Do you know there is such an identity that tells us $\tan^2x^{\circ}+\tan^2(60-x)^{ \circ }+\tan^2(60+x)^{\circ}=9\tan^2(3x)^{\circ}+6$ ?
Nope, never heard of it, are there any more of identities of this kind? That looks very useful.

#### anemone

##### MHB POTW Director
Staff member
Thanks anemone!

Nope, never heard of it, are there any more of identities of this kind? That looks very useful.
You're welcome, Pranav...yes, there are still a few more in my notebook, but, what would I get in return, hehehe...perhaps a cup of coffee?

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#### Pranav

##### Well-known member
yes, there are still a few more in my notebook, but, what would I get in return, hehehe...perhaps a cup of coffee?
I am not sure what you meant there.

#### anemone

##### MHB POTW Director
Staff member
I am not sure what you meant there.
I was actually trying to make some fun here. I also meant I would like very much to share with you the other useful trigonometric identities that I know of but could I do that by tomorrow? Now it is pretty late here and I can barely keep my eyes open...

Night night MHB!

#### Pranav

##### Well-known member
I was actually trying to make some fun here. I also meant I would like very much to share with you the other useful trigonometric identities that I know of but could I do that by tomorrow? Now it is pretty late here and I can barely keep my eyes open...

Night night MHB!

#### anemone

##### MHB POTW Director
Staff member
Hi Pranav, here are the other trigonometric identities that I want to share with you:

$\sin x^{\circ} \cdot \sin (60-x)^{\circ} \cdot \sin (60+x)^{\circ}=\dfrac{\sin 3x^{\circ}}{4}$

$\cos x^{\circ} \cdot \cos (60-x)^{\circ} \cdot \cos (60+x)^{\circ}=\dfrac{\cos 3x^{\circ}}{4}$

hence

$\tan x^{\circ} \cdot \tan (60-x)^{\circ} \cdot \tan (60+x)^{\circ}=\tan 3x^{\circ}$

#### Pranav

##### Well-known member
Hi Pranav, here are the other trigonometric identities that I want to share with you:

$\sin x^{\circ} \cdot \sin (60-x)^{\circ} \cdot \sin (60+x)^{\circ}=\dfrac{\sin 3x^{\circ}}{4}$

$\cos x^{\circ} \cdot \cos (60-x)^{\circ} \cdot \cos (60+x)^{\circ}=\dfrac{\cos 3x^{\circ}}{4}$

hence

$\tan x^{\circ} \cdot \tan (60-x)^{\circ} \cdot \tan (60+x)^{\circ}=\tan 3x^{\circ}$
Thanks a lot anemone!