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Math Help - arcos(sinθ)

  1. #1
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    arcos(sinθ)

    If 3 \pi < \theta < \frac{7 \pi}{2}, what is arccos(sin\theta) equal to?

    Attempt:

    Since 3 \pi < \theta < \frac{7 \pi}{2}, \theta = \frac{\pi}{2} so:

    arccos(1)

    = 0

    but think I made a mistake without considering the interval it in. What should I do
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  2. #2
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    Quote Originally Posted by SyNtHeSiS View Post
    If 3 \pi < \theta < \frac{7 \pi}{2}, what is arccos(sin\theta) equal to?

    Attempt:

    Since 3\pi < \theta < \frac{7 \pi}{2}, \theta = \frac{\pi}{2} so:

    arccos(1)

    = 0

    but think I made a mistake without considering the interval it in. What should I do
    3{\pi}\rightarrow\ 3.5{\pi} is the third quadrant, 2{\pi}+{\pi}\rightarrow\ 2{\pi}+{\pi}+\frac{\pi}{2}

    360^o+180^o\rightarrow\ 360^o+180^o+90^o

    If you draw a unit-radius circle, then in that 3rd quadrant, pick a point on the circumference.
    draw a right-angled triangle by drawing lines from the point directly up to the x-axis
    and directly from the point to the circle centre.

    For this triangle, \alpha=angle at the centre inside the triangle.

    adjacent =|x|

    opposite =|y|

    hypotenuse =1

    sin\theta=sin\alpha=-y

    Draw a second right-angled triangle, whose base is |y| and whose opposite is |x| in that same 3rd quadrant

    arccos(-y)=3{\pi}+\frac{\pi}{2}-\alpha

    which is the angle after one revolution in the 3rd quadrant.
    There is another angle in the 2nd quadrant, as cosine is negative there also.
    From there, the range of angles can be deduced from 0\rightarrow\ 2{\pi}
    Last edited by Archie Meade; November 9th 2010 at 05:12 PM.
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  3. #3
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    Quote Originally Posted by SyNtHeSiS View Post
    If 3 \pi < \theta < \frac{7 \pi}{2}, what is arccos(sin\theta) equal to?
    you can only determine the range of values for \arccos(\sin{\theta}) .

    if \displaystyle 3 \pi < \theta < \frac{7 \pi}{2} , then -1 < \sin{\theta} < 0 . Subsequently ...

    \displaystyle \frac{\pi}{2} < \arccos(\sin{\theta}) < \pi .
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    Quote Originally Posted by Archie Meade View Post
    Draw a second right-angled triangle, whose base is |y| and whose opposite is |x| in that same 3rd quadrant

    arccos(-y)=3{\pi}+\frac{\pi}{2}-\alpha
    Why did you swap the x and y around on the triangle when working out the arccos? Also can you say arccos(-y) = \pi - \beta, since its in the 3rd quadrant?
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  5. #5
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    Quote Originally Posted by SyNtHeSiS View Post
    If 3 \pi < \theta < \frac{7 \pi}{2}, what is arccos(sin\theta) equal to?

    Attempt:

    Since 3 \pi < \theta < \frac{7 \pi}{2},\;\;range\;\;of\;\;\theta = \frac{\pi}{2} so:

    arccos(1)

    = 0

    but think I made a mistake without considering the interval it in. What should I do
    Better to start from scratch,

    we can ascertain the graph of acos(sinx)

    u=sinx

    Notice that \frac{d}{dx}acos(sinx)=\frac{d}{dx}acos(u)=\frac{d  }{du}acos(u)\frac{d}{dx}sinx

    \displaystyle\ =-\frac{1}{\sqrt{1-u^2}}cosx=-\frac{1}{\sqrt{1-sin^2x}}cosx=-\frac{cosx}{|cosx|}

    This means the slope of the graph is \pm1, depending on whether cosx is positive or negative.

    Hence, it is a periodic triangle wave.

    For 3{\pi}<\theta<3.5{\pi}

    \displaystyle\ sin\theta goes from 0\rightarrow-1

    Therefore acos(sin\theta) goes from acos(0)\rightarrow\ acos(-1) which is from \frac{\pi}{2}\rightarrow\ {\pi}

    If you have an exact value for \theta, then since the graph is rising linearly from 3{\pi}\rightarrow\ 3.5{\pi}, then

    acos\left(sin\theta}\right)=\frac{\pi}{2}+\theta-3{\pi}
    Attached Thumbnails Attached Thumbnails arcos(sin&#952;)-acos-sinx-.jpg  
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