terminal side

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October 3rd 2012, 10:57 AM
RiderMind

Hello. I am having difficulty with this problem:
Find the point on the terminal side of θ = -3pi/4 that has an x coordinate of -1

I was hoping that someone here would be able to help me out? I am actually trying to figure out what I am doing so if you could explain how you found the answer, I would really appreciate it. Thank you so much guys.
October 3rd 2012, 11:01 AM
MarkFL

Re: terminal side

θ = ?
October 3rd 2012, 11:02 AM
RiderMind

Re: terminal side

oops, sorry.

θ=-3pi/4

Thanks
October 3rd 2012, 11:10 AM
MarkFL

Re: terminal side

From the given information, we know:

$r\cos\left(-\frac{3\pi}{4} \right)=-1$

$r=\sqrt{2}$

The $y$ -coordinate is then:

$r\sin\left(-\frac{3\pi}{4} \right)=?$
October 3rd 2012, 11:58 AM
RiderMind

Re: terminal side

0?
October 3rd 2012, 12:15 PM
MarkFL

Re: terminal side

No, what is $\sin\left(-\frac{3\pi}{4} \right)=-\sin\left(\frac{3\pi}{4} \right)=-\sin\left(\pi-\frac{3\pi}{4} \right)=-\sin\left(\frac{\pi}{4} \right)$ ?
October 3rd 2012, 12:30 PM
RiderMind

Re: terminal side

-45 degrees?

or sqrt2
October 3rd 2012, 12:37 PM
MarkFL

Re: terminal side

No, the sine of an angle will not return an angle. $\frac{\pi}{4}$ is a special angle for which you should know the trig. functions at that angle.

$-\sin\left(\frac{\pi}{4} \right)=-\frac{1}{\sqrt{2}}$

So, the $y$ -coordinate of the point is:

$y=r\cdot\left(-\frac{1}{\sqrt{2}} \right)$

Recall we found $r=\sqrt{2}$ hence:

$y=\sqrt{2}\cdot\left(-\frac{1}{\sqrt{2}} \right)=-1$

And so, the point in question is (-1,-1). Try drawing a diagram, and you will easily see that the $y$ -coordinate has to be equal to the $x$ -coordinate, as the given angle lies along the line $y=x$ .
October 3rd 2012, 12:52 PM
RiderMind

Re: terminal side

I see what you mean now. Thanks for telling me to draw a diagram. That made it easier for me to understand. So -pi/4 is one of those I just need to memorize then right?

Thanks so much bro.
October 3rd 2012, 01:06 PM
HallsofIvy

Re: terminal side

Well, it's a lot easier to memorize if you understand it. Imagine a right triangle having one angle of $\pi/4$ and one leg of length 1. Since $\pi/2= 2(\pi/4)$ , so that the other angle is also $\pi/4$ which means that the other leg also has length 1. By the Pythagorean theorem, the length of the hypotenuse is given by $c^2= 1^2+ 1^2= 2$ so that $c= \sqrt{2}$ . That gives $sin(\pi/4)= 1/\sqrt{2}= \frac{\sqrt{2}}{2}$ .
October 3rd 2012, 06:02 PM
Nervous

Re: terminal side

Could someone send me a link to teach me how to do this? I'm afraid I don't even understand it...