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Math Help - Help please: How to graph Sin x + Cos x, and Cotangent

  1. #1
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    Help please: How to graph Sin x + Cos x, and Cotangent

    Please help explain how to graph Cotangent and Sin x + cos x
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  2. #2
    MHF Contributor red_dog's Avatar
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    For cotangent use the values for \displaystyle\frac{\pi}{6}, \ \frac{\pi}{4}, \ \frac{\pi}{3}, \ \frac{\pi}{2}, \ \frac{2\pi}{3}, \ \frac{3\pi}{4}, \ \frac{5\pi}{6} and the vertical asympthotes.


    For the second one we have
    \sin x+\cos x=\sin x+\sin\left(\frac{\pi}{2}-x\right)=2\sin\frac{\pi}{4}\cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)
    Attached Thumbnails Attached Thumbnails Help please: How to graph Sin x + Cos x, and Cotangent-cotangent.jpg   Help please: How to graph Sin x + Cos x, and Cotangent-sinpluscos.jpg  
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  3. #3
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    1) to graph cot(x).

    You should know how to graph tan(x).

    cot(x) = tan(90 -x) ----in degrees.
    cot(x)
    = tan(90 -x)
    = tan[-(x -90)]
    = -tan(x -90) ------the tangent of a negative angle is the negative value of the tangent of that angle. View the angle to be in the 4th quadrant, where tangent is negative.

    So, you graph -tan(x -90) instead of cot(x).
    That is a basic tangent curve that is shifted 90 degrees to the right, and it is "inverted" because of the negative sign.

    ---------------------------------------------
    2) to graph sin(x) +cos(x)

    cos(x) = sin(90 -x) ----in degrees.
    cos(x)
    = sin(90 -x)
    = sin[-(x -90)]
    = -sin(x -90) -----same reasoning as in tan[-(x -90)].

    So,
    sin(x) +cos(x)
    = sin(x) -sin(x -90)

    Using the trig identity
    sinA -sinB = 2cos[(A+B)/2]sin[(A-B)/2],

    = 2cos[(x +x -90)/2]sin[(x -x +90)/2]
    = 2cos(x -45)sin(45)
    = 2cos(x -45)*[1 / sqrt(2)]
    = sqrt(2)cos(x -45) --------------***

    So graph sqrt(2)cos(x -45) instead of sin(x) +cos(x).

    That is a cosine curve that has an amplitude of sqrt(2), and a horizontal shift of 45 degrees to the right.
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