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Math Help - Trig expressions

  1. #1
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    Trig expressions

    (1 pt) Find the exact value (NO DECIMALS) of each expression by sketching a triangle:
    A) sin(cos^-1(3/5))
    b) cos(sin^-1(4/5))
    I have no idea where to start
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  2. #2
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    Re: Trig expressions

    Quote Originally Posted by RD33 View Post
    (1 pt) Find the exact value (NO DECIMALS) of each expression by sketching a triangle:
    A) sin(cos^-1(3/5))
    b) cos(sin^-1(4/5))
    I have no idea where to start
    A)

    Use the fact that
    cos^2(x)+sin^2(x) = 1

    We have
    cos^-1(3/5) <=> cos(x) = 3/5 => cos^2(x) = 9/25

    Using the trigonometric one we get
    9/25+sin^2(x) = 25/25 <=> sin^2(x) = 16/25 <=> sin(x) = +/- 4/5
    But since cos^-1 is defined for 0 to pi, the only solution is sin(x) = 4/5

    By this we now have the expression
    sin(cos^-1(3/5)) = sin(sin^-1(4/5)) = 4/5

    I leave the second one as exercise for you!
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  3. #3
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    Re: Trig expressions

    Thank you. I got the second one right honestly by guessing.

    Cos(sin^-1(4/5))

    I got sin(x)=4/5

    Sin^2= 16/25

    And I'm not sure where to go from there
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  4. #4
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    Re: Trig expressions

    Quote Originally Posted by RD33 View Post
    Thank you. I got the second one right honestly by guessing.

    Cos(sin^-1(4/5))

    I got sin(x)=4/5

    Sin^2= 16/25

    And I'm not sure where to go from there
    sin^2(x)+cos^2(x) = 1 => 16/25+cos^2(x) = 25/25 <=> cos^2(x) = 9/25 <=> cos(x) = +/- 3/5

    Since we know that sin^1(x) is defined for -pi<=x<=pi we know that cos^-1(3/5)

    Then we have

    Cos(sin^-1(4/5)) = cos(cos^-1(3/5))) = 3/5
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