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?
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.
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 out
Since you are in quad. II sine is positive (as is your "y" value), but your "x" value - root 8, is negative. 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.