Draw the triangle with sides 1, 3, and sqrt(10)

[powerless]

Evaluate sin(tan^{-1}(\frac{1}{3}))



3. The Attempt at a Solution

Let y = tan^{-1}(\frac{1}{3})

<=> tan(y)=1/3

y = ?

sin(tan^{-1}(\frac{1}{3})) = sin y

I think for solution to exist tan^{-1}(\frac{1}{3}) must be in the range of sin, (that is \left\frac{-\pi}{2}, \frac{\pi}{2}[\right]

i'm not sure how to evaluate this. :(

 

[mjsd]

if \tan(\theta) = 1/3 then \theta = \tan^{-1} (1/3) and you need
\sin (\theta) but what is \theta? well, it doesn't matter, since you want \sin (\theta) and we know

\sin (\theta) =\frac{\text{opposite}}{\text{hypothenuse}}

AND


\tan (\theta) =\frac{\text{opposite}}{\text{adjacent}} = \frac{1}{3} (given!)

ie. you have been given the value for opposite side and adjacent sides! surely you can work out the hypothenuse eh?

 

[powerless]

\tan (\theta) =\frac{\text{opposite}}{\text{adjacent}} = \frac{1}{3} (given!)

ie. you have been given the value for opposite side and adjacent sides! surely you can work out the hypothenuse eh?

Thanks for your message! So can I use √1² + 3²? (using pythagoras to find the hypothenuse)

I'm not sure what I should do next.

 

[snipez90]

Well you should have drawn the triangle with sides 1, 3, and sqrt(10). It's easy to see sin(theta) = 1/sqrt(10). But you're not done yet. On the coordinate plane, tan(theta) is positive in the first and third quadrant. In the third quadrant, the x and y values will be negative and so tan(theta) = -1/-3 = 1/3. And so sin(theta) = -1/sqrt(10).

Anyways the above explanation is very loose with definitions. It makes more sense if you've studied the unit circle.

 

[powerless]

Where can I learn about the unit circle(from the very basics)? some links would be helpful!

Because my textbooks haven't elaborated much on the basics of the unit circle, i don't know where to learn it from.

 

[Mr.Magnetar]

www.themathpage.com/aTrig/unit-circle.htm
www.analyzemath.com/unitcircle/unitcircle.html
www.sosmath.com/trig/Trig2/trig2/trig2.html
www.humboldt.edu/~dlj1/PreCalculus/Images/UnitCircle.html[/URL]
mathmistakes.info/facts/TrigFacts/learn/uc/uc.html
[url]www.clarku.edu/~djoyce/trig[/url]
[url]www.senocular.com/flash/source.php?id=0.114[/url]



Keep in mind these are all google and most likely put in the simplest terms. If you want math talk then try an actual pre-cal book.

Peace

 

[snipez90]

I like the first and third links especially. The first one may tell you facts you already know but they are worth rehashing. It eases into the unit circle so read all of it if you can.

Once you're ready you'll see that the motivation for the unit circle is allow us to find trig values of obtuse angles or worse. The fundamental relation is that the coordinate (x,y) on the unit circle is defined to be (cos(theta), sin(theta)). Then from symmetry on the coordinate plane you'll be able to find trigonometric values for many angles.

 

[powerless]

Thank you guys for the links!

Once you're ready you'll see that the motivation for the unit circle is allow us to find trig values of obtuse angles or worse. The fundamental relation is that the coordinate (x,y) on the unit circle is defined to be (cos(theta), sin(theta)). Then from symmetry on the coordinate plane you'll be able to find trigonometric values for many angles.

Hi snipez! I'm starting to understand these. And yes I like the hird link as well.


Here's the 2nd part of my question:

Simplify cos(2tan^{-1}(x))

y = 2tan^{-1}(x) => cos(y)


cos 2\theta = cos^2 \theta - sin^2 \theta
= 1-2sin^2\theta

cos (2 tan^{-1}(x))= 1-2[sin(tan^{-1}(x))]^2

sin(tan^{-1}(x))
y = tan^{-1}x
tan(y) = x

Now drawing the triangle we have the adjacent and opposite sides now I can work out the hypothenuse;
sin(y) = \frac{x}{\sqrt{x^2 +1}}

sin(y) = sin(sin^{-1} \frac{x}{\sqrt{x^2 +1}}

= \frac{x}{\sqrt{x^2 +1}}

pluging this back into the equation above;

cos (2 tan^{-1}(x))= 1-2[\frac{x}{\sqrt{x^2 +1}}]^2

I need some help now because this looks very messy!
I'm not sure if it's right though!

 

[snipez90]

Hi powerless! You're learning quickly, good job :).

Anyways I've skimmed over the work and it looks good. Let's just clear up the notation a bit. Hats off to you for noticing the cosine double angle formula. Now instead of using y over and over, let's be a bit more clear.

Let arctan(x) = \theta \Rightarrow\ tan(\theta) = x

where arctan is the same as your inverse tangent function. But I just used arctan because I couldn't find the inverse tan :P. However, the great thing about using this notation is that it's all you really need to define.

After you've defined the above, the problem reduces to finding cos(2*theta) and therefore finding sin(arctan(x)). Then you proceed with the triangle and as you can see, our definition provides that info for us on the right hand side of the implication sign (the right arrow). This makes for a more "clean" solution. Use a definition like the above, list the useful trig identity, then find sin(theta) from our knowledge that tan(theta) = x and then you're done! This way you don't have to use y at all!

But I think you got the hang of it and the right final answer. Don't worry with practice you'll get cleaner solutions :).