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Math Help - sine inverse

  1. #1
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    Question sine inverse

    Hey
    how do i solve this:
    sinInverse (sqrt(3/2))

    since its range is not between -1 and 1
    ?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taurus View Post
    Hey
    how do i solve this:
    sinInverse (sqrt(3/2))

    since its range is not between -1 and 1
    ?
    i suppose you meant \frac {\sqrt{3}}2

    Let \sin^{-1} \left( \frac {\sqrt{3}}2\right) = x

    \Rightarrow \sin x = \frac {\sqrt{3}}2 .........remember the range of arcsine

    now, what does x have to be?
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  3. #3
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    oops yea, it was this other question i had problems with:

    secInverse (sqrt(2))
    ?


    i know sec(x)=1/cos(x) but i dont know secInverse and how to do the sqrt(2)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    umm...you didn't answer my last question

    Quote Originally Posted by taurus View Post
    oops yea, it was this other question i had problems with:

    secInverse (sqrt(2))
    ?


    i know sec(x)=1/cos(x) but i dont know secInverse and how to do the sqrt(2)
    Let \sec^{-1}(\sqrt{2}) = x

    \Rightarrow \sec x = \sqrt{2}

    \Rightarrow \frac 1{\cos x} = \sqrt{2}

    \Rightarrow \cos x = \frac 1{\sqrt{2}}

    Now what does x have to be for this to happen?


    I ask you these questions because these are special angles that you SHOULD and MUST know
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  5. #5
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    45 correct?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taurus View Post
    45 correct?
    yes (this is for the cosine question, correct? i asked you two questions, you should be specific as to which you are answering). that is in degrees. you should tell these answers in radians i believe, unless otherwise instructed. that's what they use in college math. so the answer would be \frac {\pi}4
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