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Thread: Trigonometry question

  1. #1
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    Trigonometry question

    [SIZE=4]If A=sin(-1)x, where x>0 sin(-1)x means the shift sin in the calulator of angle x
    Two part question
    a)show that cosA=squareroot of 1-x^2


    b)find expressions in terms of x for cosecA and cos2A[/SIZE
    ]
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kingkaisai2
    [SIZE=4]If A=sin(-1)x, where x>0 sin(-1)x means the shift sin in the calulator of angle x
    [SIZE=5]Two part question
    a)show that cosA=squareroot of 1-x^2

    If:

    $\displaystyle A=\arcsin(x)$

    then:

    $\displaystyle
    x=\sin (A)
    $

    but:

    $\displaystyle (\sin(A))^2+(\cos(A))^2=1$,

    so:

    $\displaystyle x^2+(\cos(A))^2=1$.

    Rearranging gives:

    $\displaystyle
    \cos(A)=\pm \sqrt{1-x^2}
    $

    Now as $\displaystyle x>0$, $\displaystyle A\in (0,\pi)$, and so $\displaystyle \cos(A)>0$, so we
    want the positive square root in the last equation so:

    $\displaystyle
    \cos(A)= \sqrt{1-x^2}
    $

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by kingkaisai2
    [SIZE=4]If A=sin(-1)x, where x>0 sin(-1)x means the shift sin in the calulator of angle x


    b)find expressions in terms of x for cosecA and cos2A[/SIZE
    ]
    $\displaystyle \csc(A)=1/ \sin(A)=1/x$.

    $\displaystyle \cos(2A)=2 (cos(A))^2-1=2(1-x^2)-1=1-x^2$

    RonL
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