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Math Help - inverse trig?

  1. #1
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    inverse trig?

    can someone help me simplify these? this is soo frusrating, im about to cry.

    cos(arctan(-1))

    arccos(sin(-pie/2))

    cot(arcsin 3/5)

    sin (arccos (t))


    thanks soo much!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by akilele View Post
    arccos(sin(-pie/2))
    Mmmmm pie...

    Anyway,  \sin(-\frac{\pi}{2}) = -1 .

     \arccos(-1) = \pi .

    Is this enough explanation, or would you like more?
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  3. #3
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    I get how you got -1 from sin(-pi/2) but i'm confused how you got pi. arccos(-1)=pi. But I thought cos(-1) got you pi. So what does Arc do to the equation?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by akilele View Post
    I get how you got -1 from sin(-pi/2) but i'm confused how you got pi. arccos(-1)=pi. But I thought cos(-1) got you pi. So what does Arc do to the equation?
    arccos is the inverse of cos.

    That means cos(x)=y implies arccos(y)=x

    so cos(pi)=-1 implies arccos(-1)=pi
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  5. #5
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    from what you gave me, i was able to do the first problem but im having trouble with the last 2 because i can't find 3/5 on the unit circle. and i don't know what t is supposed to be.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by akilele View Post
    from what you gave me, i was able to do the first problem but im having trouble with the last 2 because i can't find 3/5 on the unit circle. and i don't know what t is supposed to be.
    Remember that  \sin^2(t)+\cos^2(t)=1 .

    Let  y=\arccos(t) \implies t=\cos(y) .

    Therefore  \sin^2(y)+\cos^2(y)=\sin^2(y)+t^2=1 .

    Thus  \sin^2(y)=1-t^2 \implies \sin(y)=\sqrt{1-t^2} .

    Now substitute  y=\arccos(t) back in to get  \sin(\arccos(t))=\sqrt{1-t^2} .
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