Question

Sin(arctan(-0.2))

Explanation?

Answer 1

Do you know what the functions sine and arctangent (inverse tangent) mean, and where to find them on your calculator?

This is not one of the "common" unit circle angle values, so you would need to use a calculator to evaluate. You could either enter it all in one step just as it is above; or first evaluate \(\large \tan^{-1}(-0.2)\) and then evaluate the sine of the result.

Answer 2

Is there a way you could do it without a calculator? Using a unit or circle or something?

Answer 3

Do you think you're supposed to it without a calculator? hmmm.........

Answer 4

Unfortunately my teacher thinks I am supposed to

Answer 5

Well, lessee.... maybe I'm missing something.

We need, first, \(\large \tan^{-1}(-0.2)=\tan^{-1}(-\dfrac{2}{10})=\tan^{-1}(-\dfrac{1}{5})\)

So that is the Q4 angle x such that:
\(\Large \tan(x)=-\dfrac{1}{5}\)

So the point (1,-5) is on the terminal side.

a HAH!!

Now I see it......

Answer 6

do you see how you can find the sine of this angle? :)

Answer 7

Haha. Uhhhhh...

Answer 8

It's actually a clever little problem. I didn't see it until I started to "think out loud" about what we know, from the given information.

The beauty of it is, we don't NEED to KNOW what the ANGLE is to find it's sine value. We can use the definition of sine as it relates to a point on the terminal side of the angle. And we have that, since we know the tangent of the angle, that we want the sine of! :)

Answer 9

I'm really trying to understand...

Answer 10

Do you know the equations for the trig functions, using a point on the terminal side of the angle?

\(\large \sin\alpha=y/r\) where \(\large r=\sqrt{x^2+y^2}\)

Does that look familiar?

Answer 11

Yes

Answer 12

Actually, maybe I should back up. Do you understand the picture I drew above? how I got the point on the terminal side of the angle, using the fact that:

We want: \(\large \sin x\)
where: \(\large x=\tan^{-1}(-0.2)\)

Do you follow how I got the diagram of the angle that I drew, from that?

Answer 13

Yes because Tangent is y/x so you just plugged in 1 and 5, right?

Answer 14

-5*

Answer 15

Exactly. Once the decimal was converted to a reduced fraction, you can use that ratio to give you a point on the terminal side.

Answer 16

And how does that lead us to Sine?

Answer 17

(actually, even unreduced fraction would have worked - I would just have used the point (2, -10) in that case.)

OK, here is how that gets us to sine: we want the sine of THAT angle, the angle that I drew above.

We know that
\(\large \sin\alpha=y/r\) where \(\large r=\sqrt{x^2+y^2}\)

So now, just compute r using that point (1, -5), and then use y and r to find the sine of the angle! :)

Answer 18

Once you have a point on the terminal side of an angle, that's all you need - you can get all 6 of the trig functions from that point.

Answer 19

So I find r using the pythagorean theorem, plug that into "-5/r" and find the sine of that fraction?

Answer 20

When you plug r into -5/r, that IS the SINE of that ANGLE. There is nothing else to do.

Answer 21

Can you solve it and see if we get the same answer?

Answer 22

\(\large \sin\alpha=y/r\) where \(\large r=\sqrt{x^2+y^2}\) and (x,y) is a point on the terminal side of \(\alpha\).

You have the point, (x,y). All that's left is to compute r and then plug it into \(\large \sin\alpha=y/r\).

Yes, I've already solved it. Let me know what you get, I'll tell you if you are right. :)

Answer 23

I got \[(-5\sqrt{26})/26\]
Is that right?

Answer 24

Yes, that's it! And you even rationalized the denominator - good for you. :)

Answer 25

Thank you so much!

Answer 26

You're welcome!

and welcome to Open Study! :)

Answer 27

Thanks!

Answer 28

Hopefully you still see this, why could the point on the unit circle not be (-1,5) instead of (1,-5)?

Answer 29

Ah, great question! Here is why:

You might recall, when you learned about the inverse functions, you (hopefully, lol) learned that each one has a SPECIFIC and defined range to go with it.

E.g., the function y=tan(x) is, in general NOT an invertible function because it isn't one-to-one (you can get the same tangent value for multiple values of the input, x).

How do we get around that, and make it so that we can have inverse functions? Why, we restrict the domain of the function y=tan(x) over which we will invert it! That way, the inverse function IS, well, a function (because to be a function, it must be true that every value in the domain can map to ONLY ONE value in the range. So by restricting the range of the original function that we are inverting, we restrict the range of the inverse.

So for y=arctan(x), the domain of y=tan(x) is restricted to \(-\pi/2<x<\pi/2\), in other words, quadrants 4 and 1.

Thus, the range of y=arctan(x) is \(-\pi/2<x<\pi/2\). So, y=arctan(x) MEANS "that angle y in Q4 or Q1 that has tan(y)=x". Since the value here was negative, we know that we are in Q4.

Answer 30

Ahhhhhh, right. So we use the domain of the inside trig function? So in this problem it would be tangent instead of sine?

Answer 31

So if I had:\[\tan (\cos^{-1} (-12/13)\] it would be in the second quadrant with a point at (-12,13)?

Answer 32

Yes, that's exactly right, because the range of the inverse cosine function is 0 to pi (Q1 and Q2), so arccos(-12/13) is an angle in Q2. :)

Answer 33

Sorry I meant a point at (-12,5) with a radius of 13

Answer 34

Or was I right the first time?