1. alexcross's Avatar
    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 PM
  2. jdoc77's Avatar
    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 PM
  3. alexcross's Avatar
    Just kidding I figured it out... simple pythagorean equation duhh
    02-04-09 10:17 PM
  4. jdoc77's Avatar
    yar.. that's what I meant.. by square root of x/y ...
    02-04-09 10:29 PM
  5. alexcross's Avatar
    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) =
    But that is the only one I could figure out
    02-04-09 10:50 PM
  6. wnm's Avatar

    Solve for the 1 - 6 then express in terms of u.

    Just kidding I figured it out... simple pythagorean equation duhh
    Even the Scarecrow knew that one.
    02-05-09 07:02 AM
  7. Leon_Ledbetter's Avatar
    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.

    Good luck!
    02-05-09 02:42 PM