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Math Help - a couple of questions...thanks

  1. #1
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    a couple of questions...thanks

    I dont know how to find the exact value of something(trig)...e.g.

    sec(13Pi/6) I get a long decimal(1.31...) but how am I meant to get the exact value answer which in this case is 2/3 of the square root of 3. (which is 1.31 when u calculate it but when I see 1.31 how am I meant to know that its 2/3 the square root of 3?

    question 2:
    3(tan^2 )theta - sec theta = 1


    for 0 <= theta =< 2PI

    can anyone solve that for me? thanks


    oh and for the following graph sketches:

    y = sin3x
    y = |sin3x|
    y = sin|3x|

    so like I was confused to whether they amplitude was 3 times bigger or the frequency is like 3 times smaller for them. can someone tell me what the difference of those curves are from a normal sinx
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  2. #2
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    Quote Originally Posted by grammar View Post
    I dont know how to find the exact value of something(trig)...e.g.

    sec(13Pi/6) I get a long decimal(1.31...) but how am I meant to get the exact value answer which in this case is 2/3 of the square root of 3. (which is 1.31 when u calculate it but when I see 1.31 how am I meant to know that its 2/3 the square root of 3?

    [snip]
    \sec \left(\frac{13 \pi}{6} \right) = \sec \left( \frac{(12 + 1) \pi}{6} \right) = \sec \left(\frac{12 \pi}{6} + \frac{\pi}{6}\right)


     = \sec \left(2 \pi + \frac{\pi}{6}\right) = \sec \left(\frac{\pi}{6}\right) = \frac{1}{\cos \left(\frac{\pi}{6}\right)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}.
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  3. #3
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    Quote Originally Posted by grammar View Post
    [snip]
    question 2:
    3(tan^2 )theta - sec theta = 1


    for 0 <= theta =< 2PI

    can anyone solve that for me? thanks

    [snip]
    Substitute (from the Pythagorean Identity) \tan^2 \theta = \sec^2 \theta - 1:

    3 (\sec^2 \theta - 1) - \sec \theta = 1 \Rightarrow 3 \sec^2 \theta - \sec \theta - 4 = 0.

    Let x = \sec \theta to see how to factorise:

    3x^2 - x - 4 = 0 \Rightarrow (3x - 4)(x + 1) = 0.

    Therefore either  x = \frac{4}{3} or x = -1.

    Therefore either  \sec \theta = \frac{4}{3} or \sec \theta = -1.

    The latter can be exactly solved easily. The former can only be found as either a generic exact answer or a decimal approximation.
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  4. #4
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    Quote Originally Posted by grammar View Post
    [snip]
    oh and for the following graph sketches:

    y = sin3x
    y = |sin3x|
    y = sin|3x|

    so like I was confused to whether they amplitude was 3 times bigger or the frequency is like 3 times smaller for them. can someone tell me what the difference of those curves are from a normal sinx
    Draw the graph of y = |\sin (3x)| by reflecting around the x-axis the parts of the graph of y = \sin (3x) that lie below the x-axis. So there are salient points at (0, \, \pm n \pi) where n is an integer.

    Draw the graph of y = \sin |3x| by reflecting around the x-axis the part of the graph of y = \sin (3x) that lies to the left of the y-axis. So there's a salient point at (0, 0).

    Note: \sin |3x| = \sin (3x) for x \geq 0 and \sin |3x| = \sin (-3x) = -\sin (3x) for x < 0.


    Salient point: Salient Point -- from Wolfram MathWorld
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