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Trig Help

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm wondering how you would find values of sin, cos, tan, sec, csc, and cot without a calculator. I don't have a specific problem but how would you solve things like tan 45 degrees, sec 30 degrees, and cos -30 degrees?

    2. Relevant info

    I have the values of sin and cos memorized for 0, 30, 45, 60, 90, 180, and 270 degrees. There is a technique to solving the problems above knowing whether the trig functions are positive or negative in specific quadrants. How is that relevant to solving for the values?


    3. The attempt at a solution

    I know how to solve some of the above using identities. For example, tan 45 is just sin 45/cos 45 which is 1. For sec 30, I believe it's just 1/cos 45 so it'd be 1/([tex]\sqrt{2}[/tex]/2).

    I don't know how to solve cos -30. How would you solve the problems without identities? There is a way knowing when the functions are positive/negative in the quadrants. It also has something to do with adding and subtracting the angle measurement. Can someone explain the technique please?
     
  2. jcsd
  3. Feb 9, 2010 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Ryuk1990! :smile:

    (have a square-root: √ and a degree: º and a theta: θ :wink:)

    Use sin = opp/hyp, cos = adj/hyp, tan = opp/adj, plus the fact that a 45º triangle is half a square, and a 30º or 60º triangle is half an equilateral triangle. :smile:

    (but sec30º = 1/cos30º, of course)
    Personally, I always use the formula for cos(180º ± θ), also cos(-θ) = cosθ, sin(-θ) = -sinθ.

    But you can also do it by drawing the angle on a graph, and using x = rcosθ, y = rsinθ (so eg in the second quadrant, x is negative but y is positive, so cos is negative but sin is positive).
     
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