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    Q. y=arccosx , -1<x<1 and 0<y<pi ( the signs should have the equal to or more than...signs on them)

    express arcsinx in terms of y


    so far i have done cos y = x but i dont know what to do next. can someone help me please.
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    It might help you to know that \cos y = \sin \left( \frac{\pi}{2} - y \right).
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    (Original post by klgyal)
    Q. y=arccosx , -1<x<1 and 0<y<pi ( the signs should have the equal to or more than...signs on them)

    express arcsinx in terms of y

    so far i have done cos y = x but i dont know what to do next. can someone help me please.
    Remember that siny = sqrt{(1 - cos^{2}y)} from cos^{2} + sin^{2} = 1

    See how this helps??
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    (Original post by nuodai)
    It might help you to know that \cos y = \sin \left( \frac{\pi}{2} - y \right).
    Where does that come from ??
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    (Original post by klgyal)
    Where does that come from ??
    It's just a graph transformation that relates y = \cos x with y = \sin x.

    That is, f(x) \mapsto f \left( \frac{\pi}{2} - x \right) denotes a positive translation of \frac{\pi}{2} parallel to the y-axis followed by a reflection in the y-axis (or a reflection in the x-axis followed by a negative translation parallel to the x-axis; both give the same result), and it just so happens that if you do this with the graph of \sin x you get the graph of \cos x
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    (Original post by nuodai)
    It's just a graph transformation that relates y = \cos x with y = \sin x.

    That is, f(x) \mapsto f \left( \frac{\pi}{2} - x \right) denotes a positive translation of \frac{\pi}{2} parallel to the y-axis followed by a reflection in the y-axis (or a reflection in the x-axis followed by a negative translation parallel to the x-axis; both give the same result), and it just so happens that if you do this with the graph of \sin x you get the graph of \cos x
    thank-you
 
 
 
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