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 May 9th, 2013, 04:01 PM #1 Member   Joined: Mar 2013 Posts: 36 Thanks: 0 Find sine and cosine of given value Hi, If $sin t= -\frac{2sqrt{2}}{3}$ and $cos t= \frac13$, find the sine and cosine of the given value. OK, the point lies in quadrant IV. I know this because sine is negative and cosine is positive. $\text{a. } t + \pi$ $\left(-\frac13, \frac{2sqrt{2}}{3}\right)$ Moved ? units anticlockwise from quadrant IV. $\text{b. } -t$ $\left(\frac13, \frac{2sqrt{2}}{3}\right)$ The x-coordinate stays the same, but the sign of the y-coordinate changes. $\text{c. } t + \frac\pi2$ $\left(-\frac13, \frac{2sqrt{2}}{3}\right)$ Moved 90 degrees from quadrant IV anticlockwise. $\text{d. } -t + \frac\pi2$ $\left(-\frac13, \frac{2sqrt{2}}{3}\right)$ Moved 90 degrees anticlockwise from quadrant I.
 May 9th, 2013, 04:43 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,739 Thanks: 1810 Have you checked your answers by using a calculator?
May 9th, 2013, 05:44 PM   #3
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 Originally Posted by skipjack Have you checked your answers by using a calculator?
I am not sure how to.
For d. the calculator gave two positive answers. Is that correct?

The concepts aren't really firm in my mind. If I sort that out, the calculator steps should follow.

 May 10th, 2013, 12:27 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,739 Thanks: 1810 Are you saying you don't know how to use your calculator to find a value for t in the fourth quadrant that satisfies the equation $sin t= -\frac{\small2sqrt{2}}{\small3}$? It would suffice to evaluate $\sin^{\small-1}$$-\frac{\small2sqrt{2}}{\small3}$$.$ If you get that right, you should find that cos(t) = 1/3. In your answers, are you listing the sine first or the cosine first?
May 10th, 2013, 01:05 AM   #5
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Quote:
 Originally Posted by skipjack Are you saying you don't know how to use your calculator to find a value for t in the fourth quadrant that satisfies the equation $sin t= -\frac{\small2sqrt{2}}{\small3}$? It would suffice to evaluate $\sin^{\small-1}$$-\frac{\small2sqrt{2}}{\small3}$$.$ If you get that right, you should find that cos(t) = 1/3. In your answers, are you listing the sine first or the cosine first?
Thank you.

Isn't there a particular concept behind this sort of problem?

I was focused on solving it another way. Can't it be done without a calculator?
That would be ideal for my learning.

Cosine first.

May 10th, 2013, 01:35 AM   #7
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 Originally Posted by skipjack I assumed that you had already obtained your answers without use of a calculator (probably by using standard trigonometric identities). That's why I suggested using a calculator to check your answers. If you post all your reasoning (for example, explaining how you used "move 90° anticlockwise), the specific errors you have made can be identified.
$\text{a. } t + \pi$

$\left(-\frac13, \frac{2sqrt{2}}{3}\right)$

I think of the point t as being somewhere in quadrant 4 because of the signs of the coordinates.
Starting at initial angle of 2?, then moving in an anticlockwise direction because ? is positive.
This leads me to quadrant II where cosine is negative and sine positive.

$\text{b. } -t$

$\left(\frac13, \frac{2sqrt{2}}{3}\right)$ For this I used the cosine and sine function only because it's the only way I know to solve this type of problem.

$\text{c. } t + \frac\pi2$

$\left(-\frac13, \frac{2sqrt{2}}{3}\right)$ Move a quarter of the way around the circle from (1,0) in an anticlockwise direction because ?/2 is positive.

$\text{d. } -t + \frac\pi2$

$\left(-\frac13, \frac{2sqrt{2}}{3}\right)$ This is point is in quadrant II, so moving from there to in an anticlockwise direction by ?/2 units leaves in quadrant III.

May 10th, 2013, 03:03 AM   #9
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b. cos (-t) = cos t; sin(-t) = -sin t.
$\left(\frac13, \frac{2sqrt{2}}{3}\right)$

c. ?/2 lands in the first quadrant.

$\left(\frac13, \frac{2sqrt{2}}{3}\right)$

d. Adding ?/2 to -t in the first quadrant brings me to quadrant II.

$\left(-\frac13, \frac{2sqrt{2}}{3}\right)$

 May 10th, 2013, 03:52 AM #10 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond Re: Find sine and cosine of given value For c, see here. See if you can figure out why these identities are true.

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