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Math Help - Trig question help, please?

  1. #1
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    Trig question help, please?

    There's a question that I just can't figure out, and I'm sure it's really simple but it's just not working for me. Some help would be greatly appreciated. I've tried looking for a similar question, but to no avail. Anyhow, here it is:

    sinx - cosx = 1

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by darkryder341 View Post
    There's a question that I just can't figure out, and I'm sure it's really simple but it's just not working for me. Some help would be greatly appreciated. I've tried looking for a similar question, but to no avail. Anyhow, here it is:

    sinx - cosx = 1

    Thanks in advance!
    Just look at the graphs of the two functions and see where they are exactly one unit apart. This occurs in many places. One such is when x = \frac{\pi}{2}. Do you see how to get all the rest?
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  3. #3
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    Hello, darkryder341!

    There are number of approaches to this problem (none are obvious).


    Solve for x\!:\quad \sin x - \cos x \:=\: 1
    I will assume that the answers are on the interval [0,\:2\pi)


    Square both sides: . (\sin x - \cos x)^2 \:=\:1^2

    We have: . \sin^2x - 2\sin x\cos x + \cos^2x \;=\;1

    . . . . . . . \underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} - \underbrace{2\sin x\cos x}_{\text{This is }\sin2x} \;=\;1

    And we have: . 1 - \sin2x \:=\:1 \quad\Rightarrow\quad \sin2x \:=\:0

    Then: . 2x \:=\:0,\:\pi,\:2\pi,\:3\pi

    Hence: . x \:=\:0,\:\tfrac{\pi}{2},\:\pi,\:\tfrac{3\pi}{2}


    But we find that x \:=\:0,\:\tfrac{3\pi}{2} are extraneous roots.

    Therefore, the answers are: . x \;=\;\frac{\pi}{2},\:\pi

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  4. #4
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    Hmm...makes sense now! Thanks very much for both of your help I really do appreciate it.
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  5. #5
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    Trig equation

    Hello darkryder341
    Quote Originally Posted by darkryder341 View Post
    There's a question that I just can't figure out, and I'm sure it's really simple but it's just not working for me. Some help would be greatly appreciated. I've tried looking for a similar question, but to no avail. Anyhow, here it is:

    sinx - cosx = 1

    Thanks in advance!
    An alternative approach.

    Expressions like \sin x - \cos x can always be expressed in terms of a single trig function, like this:

    Suppose \sin x - \cos x = r\sin(x-\alpha)

    Then \sin x - \cos x = r\sin x \cos \alpha - r\cos x \sin\alpha

    So, comparing coefficients:

    r\cos\alpha = 1 and r\sin\alpha = 1

    If we square and add, r^2(\cos^2\alpha + \sin^2\alpha) = 2

    \Rightarrow r = \sqrt 2 (taking the positive value)

    and, dividing:

    \tan\alpha = 1

    \alpha = \tfrac{\pi}{4}

    So the original equation becomes:

    \sqrt2\sin(x- \tfrac{\pi}{4}) = 1

    \Rightarrow \sin(x- \tfrac{\pi}{4}) = \frac{1}{\sqrt2}

    \Rightarrow x -\frac{\pi}{4} = \frac{\pi}{4}, \frac{3\pi}{4}, ...

    \Rightarrow x = \frac{\pi}{2}, \pi, ...

    Grandad
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