Trigonometric Arithmetic IV

Calculus Level 5

Evaluate

$\frac{1}{\pi} \sum _{ n=2 }^{ 101 }{ \int _{ 1 }^{ n }{ \left(2\tan ^{ -1 }{ (x) } +\sin ^{ -1 }{ \left(\frac { 2x }{ 1+x^{ 2 } } \right)}\right)dx } } .$

The answer is 5050.

2 solutions

Ronak Agarwal
Sep 26, 2014

While using inverse trignometric functions we need to be very careful about the domain and range of the function. We begin by writing :(I am everywhere specifying the domain and range )

${tan}^{-1}(x)=\theta$ (i) Domain and range $x \epsilon R,\theta \epsilon (-\frac{\pi}{2},\frac{\pi}{2})$

Taking tangent on both sides we get :

$tan(\theta )=x$ (ii)

We will now write :

$sin(2\theta )=\frac{2x}{1+{x}^{2}}$ (iii)

Taking ${sin}^{-1}$ on both sides we get :

${sin}^{-1}(sin(2\theta ))={sin}^{-1}(\frac{2x}{1+{x}^{2}})$

But now we have got a problem the range of ${sin}^{-1}(x)$ is

$(-\frac{\pi}{2},\frac{\pi}{2})$ . Whereas $2\theta \epsilon (-\pi,\pi)$

We realize that for $x \ge 1$ we have $\frac{\pi}{4} \le \theta < \frac{\pi}{2}$

$\Rightarrow \frac{\pi}{2} \le 2\theta < \pi$

Note that $sin(\theta)=sin(\pi-2\theta)$ also we have :

$0 < \pi - 2\theta \le \frac{\pi}{2}$ Which is in our range

In (ii) we write :

${sin}^{-1}(sin(\pi - 2\theta ))={sin}^{-1}(\frac{2x}{1+{x}^{2}})$

$\Rightarrow \pi - 2\theta = {sin}^{-1}(\frac{2x}{1+{x}^{2}})$

Using (i) we get :

$2{tan}^{-1}(x)+{sin}^{-1}(\frac{2x}{1+{x}^{2}})=\pi$

Using this our given expression in the question becomes :

$\sum _{ n=2 }^{ 101 }{ \int _{ 1 }^{ n }{ dx } } =\sum _{ n=2 }^{ 101 }{ n-1 } =\sum _{ n=1 }^{ 100 }{ n } =5050$

i didn't get the line over there-We realize that for x>or=1 we have pi/4</=thita</=pi/2...

Cestrum Nocturnam - 6 years, 8 months ago

Yeah me neither. How did you get $\frac{\pi}{4}\leq \theta \leq \frac{\pi}{2}$ ?

John M. - 6 years, 8 months ago

@John Muradeli

$\theta = {tan}^{-1}(x)$ hence it is easy to see that :

for $x \ge 1 , \frac{\pi}{4} \le {tan}^{-1}(x) \le \frac{\pi}{2}$

Since $tan(\theta) \ge 1$ for $\frac{\pi}{4} \le \theta \le \frac{\pi}{2}$

Hope this helps.

Ronak Agarwal - 6 years, 8 months ago

John M.
Sep 22, 2014

The key here is to realize that

$2\tan ^{ -1 }{ (x) } +\sin ^{ -1 }{ \left(\frac { 2x }{ 1+x^{ 2 } } \right)}=\pi$

for all $x\geq 1$ .

Does anyone know how to prove this identity? The original problem was, in fact, to prove it, but I just took it and turned it into an integration problem.

Cheers

Note that if $\alpha = 2\tan^{-1}(x)$ , then $\tan(\alpha) = \frac{2x}{1-x^2}$ . Also, if $\sin(\theta) = \frac{2x}{1+x^2}$ , then $\tan(\theta) = \frac{2x}{x^2-1}$ , for $x \ge 1$ . So we are looking at $\alpha + \theta$ , where $\tan(\alpha) = -\tan(\theta)$ . Note that $\alpha \in (\pi/2,\pi)$ and $\theta \in (0,\pi/2)$ . These facts imply that $\alpha + \theta = \pi$ .

Patrick Corn - 6 years, 8 months ago

Hm, great! But I lost you on some of the steps:

Why does $\tan{(\alpha)}=\frac{2x}{1-x^2}$ ? Could you show your steps?
Why does the tangent in your second sentence equal that "for $x\geq 1$ "?
How do the boundaries for $\alpha$ and $\theta$ imply that their sum is $\pi$ ?

That'd be great. Thanks!

John M. - 6 years, 8 months ago

This is just the double-angle formula for tangent.

The range of arcsin, which is where $\theta$ lives, is $[-\pi/2,\pi/2]$ , so $\cos(\theta)$ is positive, so in this case $\tan(\theta)$ is positive. By the usual identities, $\tan(\theta) = \pm \frac{2x}{x^2-1}$ , and so the fact that $x \ge 1$ means that it has to be the plus sign.

If one angle is in the first quadrant and one angle is in the second quadrant and their tangents are negatives of each other, this should imply that their sines are equal and their cosines are negatives of each other. So the angles sum to $\pi$ .

Patrick Corn - 6 years, 8 months ago

@Patrick Corn – Thanks for your response. I get 1 and 2, but not 3. You can probably explain it much faster than I can figure it out. Please elaborate further.

Thanks,

John M. - 6 years, 8 months ago

@John M. – If their tangents are opposites, then their tangents squared are the same. Then their secants squared are the same (because $\tan^2 x + 1 = \sec^2 x$ ). So their cosines are either equal or opposite. In this case their cosines are opposites because of the domains. And their tangents are opposites, so their sines are the same.

Patrick Corn - 6 years, 8 months ago

@Patrick Corn – Oh no no I absolutely get that part with the quadrants explanation (All Students Take Calculus - All Sine Tangent Cosine). What I don't get is how any of those things imply that the sum of alpha and theta is $\pi$ .

John M. - 6 years, 8 months ago

@John M. – Well, you can use the sum formulas to see that the sine and cosine of $\alpha + \theta$ are $0$ and $-1$ respectively.

Patrick Corn - 6 years, 8 months ago

@Patrick Corn – $\sin{(\alpha+\theta)}=0$ and $\cos{(\alpha+\theta)}=-1$ $\Rightarrow \alpha+\theta=\pi$ ?

Alright forget it. Thanks.

John M. - 6 years, 8 months ago

@John M. – Well, it implies that $\alpha + \theta = \pi + 2k\pi$ for some integer $k$ , and the boundaries for $\alpha$ and $\theta$ imply that $k = 0$ .

Patrick Corn - 6 years, 8 months ago

@Patrick Corn – Yeah... If only I could see how... I'm sorry but I can make absolutely no connection to any of these things. And I know trig pretty well, but I guess not that well to be able to understand such implications.

That's alright. Don't wanna be a bother. I'll ask around. Thanks again.

John M. - 6 years, 8 months ago

I understood the following:

$\tan \alpha = -\tan \theta$ implies either:

$\alpha = \pi - \theta$ or $\alpha = -\theta$

but using the boundaries for $\alpha$ and $\theta$ , we find that both are positive hence we take $\alpha = \pi - \theta$ .

I hope this is what you meant , did you?

Hasan Kassim - 6 years, 8 months ago

Ok PLEASE explain how $\tan{\alpha}=-\tan{\theta}$ implies $\alpha=\pi-\theta$ !

John M. - 6 years, 8 months ago

@John M. – Man its a trig identity!

All Students Take Calculus => Tangent is positive in first quadrant and negative in the second quadrant, and $\pi-\theta$ is in second quadrant.

Want another proof? take a look:

$\tan \alpha = \tan (\pi - \theta) = \frac{\tan \pi - \tan \theta}{1+\tan \pi \tan \theta} = \frac{0-\tan \theta}{1+0} = -\tan \theta$

Hasan Kassim - 6 years, 8 months ago

@Hasan Kassim – Ooh of course... $0\leq \theta \leq \pi / 2$ . Got it.

Thanks, Hassan!

John M. - 6 years, 8 months ago

@Hasan Kassim – Woah we're getting somewhere... Now really quickly tell me why $\pi - \theta$ is in the second quadrant? Yea I know Ima dumbo, but my time flows quicker than space around a black hole, so no time to dig into my brain! Not that you do, but eh I'll take my chances.

THANKS!!!

John M. - 6 years, 8 months ago

Hey, King Integral , any ideas?

@hasan kassim

John M. - 6 years, 8 months ago

Here is the proof of your identity:

Let $u=a+b$ , where $a= 2\arctan x , b= \arcsin (\frac{2x}{1+x^2})$ .

$\tan u = \frac{\tan a +\tan b}{1-\tan a \tan b }$

$* \tan a = \tan(2\arctan x) = \frac{2\tan \arctan x}{1-\tan^2 \arctan x} = \frac{2x}{1-x^2}$

$* \tan b = \frac{\sin b}{\sqrt{1-\sin^2 b}}= \frac{\frac{2x}{1+x^2}}{\sqrt{1-(\frac{2x}{1+x^2})^2}}$

After reducing:

$\tan b= \frac{2x}{|1-x^2|}$

Now substitute $\tan a$ and $\tan b$ in the expression of $\tan u$ :

$\tan u = \frac{\frac{2x}{1-x^2} +\frac{2x}{|1-x^2|} }{ 1-\frac{2x}{1-x^2} \frac{2x}{|1-x^2|} }$

Now we check values:

For $|x|>1$ : $\tan u = 0 => u=\pi$

For $|x|<1$ : $\tan u$ will depend on $x$ .

Try $x= \pm 1$ By hand (Using the original form of the identity) To get $\pm \pi$

Hasan Kassim - 6 years, 8 months ago

Meh this looks mad complicated :O so many SUBS!

Can you interpret Patrick Corn's explanation? Not that yours isn't good, but I think his explanation is simpler and uses less operational steps and more conceptual ones.

Thanks for your proof, Hasan.

John M. - 6 years, 8 months ago

Actually This is the first time I see this identity.

I used substitution $x=\tan u$ And ended up with a bothering summation:

$\displaystyle \sum_{n=2}^{101} n\arctan n - \frac{1}{2} \ln (1+n^2)$

If any one could solve it, then thanks!

Hasan Kassim - 6 years, 8 months ago

Trigonometric Arithmetic IV

The answer is 5050.

2 solutions

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