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Thread: 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 $\displaystyle x = \frac{\pi}{2}$. 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 $\displaystyle x\!:\quad \sin x - \cos x \:=\: 1$
    I will assume that the answers are on the interval $\displaystyle [0,\:2\pi)$


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

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

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

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

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

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


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

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

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

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

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

    So, comparing coefficients:

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

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

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

    and, dividing:

    $\displaystyle \tan\alpha = 1$

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

    So the original equation becomes:

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

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

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

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

    Grandad
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