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Trigonometry challenge

Pranav

Well-known member
Nov 4, 2013
428
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
Feb 7, 2012
2,725
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
Nov 4, 2013
428
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 $$\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}} = \frac{40}3 - \frac43 - 2 = 10.$$
Thanks Opalg for your participation, your answer is correct. :)

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
Feb 14, 2012
3,755
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
Nov 4, 2013
428
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! :cool:

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
Feb 14, 2012
3,755
Excellent! :cool:

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.(Thinking)
 

Pranav

Well-known member
Nov 4, 2013
428
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
Feb 14, 2012
3,755
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?(Devil)
 
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Pranav

Well-known member
Nov 4, 2013
428
yes, there are still a few more in my notebook, but, what would I get in return, hehehe...perhaps a cup of coffee?(Devil)
I am not sure what you meant there. (Thinking)
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
I am not sure what you meant there. (Thinking)
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...:eek:

Night night MHB!(Sleepy)
 

Pranav

Well-known member
Nov 4, 2013
428
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...:eek:

Night night MHB!(Sleepy)
Please take your time and good night. :)
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
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
Nov 4, 2013
428
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! :)