Thread: Trivial Trigonometry Question, Need Help Understanding Concept

  1. #1
    Join Date
    Nov 2002

    Trivial Trigonometry Question, Need Help Understanding Concept

    I have my final tommorow and I don't understand the concept of how to do these problems. Here are sample problems and the answers, I just don't know how to go about solving these, I tried the book, my notes, and nothing works. I know it's easy, even the book says it's easy, I just can't figure it out. I have everything else down pretty much for the final except this.

    I will use sqrt(expression) to indicate the square root of whatever, keyboard needs a square root key hehe

    I understand how to do like this...

    cos[arctan(sqrt3)] becomes cos(pi/3) and then 1/2 is the answer

    But when it's a value that I don't recognize, like 1/3, I get totally blown away. Here are two I can't figure out how to do.

    cos[2arcsin(1/3)] .... The answer is 7/9 but I don't know how to get there I get stuck right at the beginning, trying to figure out the arcsin of 1/3. If I could figure that out I could do the rest probably. I was thinking of maybe incorporating an identity for this one because cos2x = 1 - 2sin^2x but I don't think it's needed?

    Also this one

    tan[arcsin(2/5)] .... The answer is 2/sqrt21, but why?

    I can't use a decimal answer for these, any help would be great.
    Last edited by SourceCode; 12-14-2003 at 05:04 PM.

  2. #2
    & the hat of GPL slaying Thantos's Avatar
    Join Date
    Sep 2001
    First step is to draw the triangle. Solve that triangle for the 3rd side: 3^2=1^2 + x^2, x = sqrt (9-1), x = 2sqrt2.
    Now cos of that would be (2sqrt2)/3. Now solve for cos(2 * (2sqrt2)/3).

  3. #3
    Toaster Zach L.'s Avatar
    Join Date
    Aug 2001
    Think in terms of right triangles. The second problem, for example:

    You know the sin of something is opposite over hypotenuse. So, we have a triangle, the opposite side is 'a', the hypotenuse is 'c', and the adjacent side is 'b'. And, label the angle 'x' for which sin(x) = a/b (that is, the angle opposite side 'a').

    Now, we know that opposite over hypotenuse is 2/5, to a/c is 2/5, and since it is our triangle, we set a=2, and c=5. Solving for the third side ('b') using the Pythagorean theorem, we get sqrt(21). And 'b' is of course the adjacent side, and since tan equals opposite over adjacent, tan(x)=2/sqrt(21).

    So, don't solve for the angle, solve for the other side of a right triangle with the known quantities.

    Using the identity you provided for the first is a good way of going about it:
    You have cos(2x)=1-2*(sin(x))^2
    But x=arcsin(1/3), so sin(x)=1/3
    Then cos(2x)=1 - 2*1/9=7/9

    Its a bit difficult to explain without the use of some images, but I hope this helps a bit.
    The word rap as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.

  4. #4
    Join Date
    Nov 2002
    That makes perfect sense, I got it. Thanks soooo much guys

Popular pages Recent additions subscribe to a feed

Similar Threads

  1. another exercise question
    By luigi40 in forum C# Programming
    Replies: 3
    Last Post: 11-28-2005, 03:52 PM
  2. Basic Trigonometry Question
    By xddxogm3 in forum A Brief History of
    Replies: 19
    Last Post: 01-28-2005, 01:45 AM
  3. MUD Concept Question
    By mrpickle in forum Game Programming
    Replies: 3
    Last Post: 12-01-2003, 12:45 PM
  4. new question: understanding parse error
    By Pig in forum C++ Programming
    Replies: 3
    Last Post: 11-19-2003, 01:14 PM
  5. what does this warningmean???
    By kreyes in forum C Programming
    Replies: 5
    Last Post: 03-04-2002, 07:53 AM