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- Alright so I'm really frustrated with this trig problem and I can't get help from any math forums because they suck, and, since this one is active I was hoping maybe somebody could answer a quick and simple question!

Okay I understand the majority of this problem, but there's just one part that I don't get!

Problem: Given that Θ is in Quadrant II, and cscΘ=3, find the values of all six trigonometric functions of Θ.

So when i draw the triangle in Quadrant II, I got y=1, r=3, and on the key Θ=square root of 8. I understand how y and r were found, but how was Θ found?02-04-09 09:06 PMLike*0* - Jeez.. Trig and calc were SOOO long ago, someone else is going to have to help you. Isn't it just a function/formula? Theta = the arc sin or something... man that was too long ago.

Something like cos(theta) = sqrt(x)/y ?

*McCoy Voice*

what am I a doctor or a mathematician!!02-04-09 09:22 PMLike*0* - Okay just in case somebody good with math catches this, a problem I really can't figure out is....

Suppose Cos11°=u. Write each of the following in terms of u.

1) Sin79°=?

2) Sin11°=?

3) Tan11°=?

4) Sec11°=?

5) cos169°=?

6) sec709°=?

How do you know when to use which identity? I understand for Sin79° you use the Cofunction Identity, making it

Sin79° =

cos(90°-79°) =

cos(11°) =

u

But that is the only one I could figure out02-04-09 10:50 PMLike*0* - Since you are in quad. II sine is positive (as is your "y" value), but your "x" value - root 8, is
. Punch (on your calculator) inverse cosine of NEGATIVE root 8 divided by 3 into your calculator and it will spit out 160.5 degrees (a second quadrant angle). If you punch up inverse sine 1 divided by 3 you will get 19.5 degrees. But that is a FIRST quadrant angle. Your triangle is in QUADRANT II. Convert 19.5 degrees into a quadrant two angle (subtract it from 180 degrees) and you will again get 160.5 degrees.**negative**

Good luck!02-05-09 02:42 PMLike*0*

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