## kittiwitti1 one year ago @mathstudent55 could you help me with this one as well? Thanks in advance- cos θ = 1/2 and θ terminates in QIV. So far I have this: http://prntscr.com/8dhn18 $\text{not sure how the answer is not }\sqrt{3}\text{ as I ran through the calculations several times}$$\text{with the same result}$

1. mathstudent55

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2. mathstudent55

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3. mathstudent55

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4. kittiwitti1

$y=\sqrt{2^{2}-1^{2}}=\sqrt{4-1}=\sqrt{3}???$

5. mathstudent55

What is y?

6. kittiwitti1

0.5 ?! Wait what?! o_o

7. mathstudent55

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8. kittiwitti1

I thought this was a 30 degree angle so I went off that, and I double checked in case.... but apparently I had the wrong idea ...Isn't cos x/r?

9. mathstudent55

Your problem does state that cos(theta) = 1/2, right?

10. mathstudent55

You have the right idea. It's a 30-60-90 triangle. There is a 30 deg angle and a 60 deg angle.

11. kittiwitti1

Yes... and I got lost afterwards since the answer is not the 'obvious' root 3. e_e

12. mathstudent55

In this case, we are dealing witht he 60-deg angle.

13. kittiwitti1

I thought it was the 30 degree... (feels stupid) x_x

14. mathstudent55

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15. kittiwitti1

Oh. Yes.

16. kittiwitti1

I neglected the quadrant. Sorry e_e

17. mathstudent55

Now notice that the y-coordinate is negative, so sine and cosecant will be negative. Since the x-coordinate is positive, the cosine and secant are positive. The tangent and cotangent are negative.

18. kittiwitti1

I think my issues were due to neglecting to check which quadrant I was in... I will make sure, brb

19. mathstudent55

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20. kittiwitti1

Yup, got them right this time. Thanks for correcting my stupid mistake ^^;

21. mathstudent55

You're welcome.