## dontothabonbon Group Title Let θ be an angle in quadrant II such that sin one year ago one year ago

1. cwrw238 Group Title

|dw:1372535299105:dw|

2. cwrw238 Group Title

can you continue from here?

3. dontothabonbon Group Title

Thanks. I'm not sure what to do from here.

4. cwrw238 Group Title

tan x = opposite / adjacent = -1 / sqrt15 sec x = hypotenuse / adjacent = ?

5. jdoe0001 Group Title
6. jdoe0001 Group Title

|dw:1372535888225:dw|

7. dontothabonbon Group Title

Alright now I see. Thanks to both of you.

8. cwrw238 Group Title

yw

9. dontothabonbon Group Title

Would the hypotenuse become a negative when calculating sec? Or would it just be: sec x = 4/sqrt15?

10. dontothabonbon Group Title

Or better yet, why did the 1 become negative when calculating the tangent?

11. Jhannybean Group Title

Because your function is in the second quadrant where sine is positive and cosine is negative.

12. cwrw238 Group Title

the whole ratio became negative tangent : opposite = 1 , adjacent = - sqrt15 tan = 1 / -sqrt15 = -1/sqrt15

13. Jhannybean Group Title

$\large \left(-\frac{1}{\sqrt{15}}\right)$

14. dontothabonbon Group Title

So, sec = -4/sqrt15?

15. cwrw238 Group Title

yes

16. dontothabonbon Group Title

ahh okay I see now thanks.

17. cwrw238 Group Title

x values to the left of y-axis are negative y values below x axis are also negative