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Math Help - Finding x with arcsin and arccos in expression

  1. #1
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    Finding x with arcsin and arccos in expression

    If x is greater than or equal to 0 and arcsinx = arccos x (2x), then x=?

    The answer is .447.

    The book says to take the sine of both sides, then apply pythagorean identity. I know the pythagorean identity, but I don't understand how it applies here. Thank you.
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    Re: Finding x with arcsin and arccos in expression

    You draw a Right Triangle. Designate one acure angle as the angle of interest, call is S.

    When contemplating sin(arccos(x)), notice that arccos(x) is an ANGLE. Let's call it S

    If the cos(S) = x, then you can lable the leg adjacent to S, just "x" and the hypotenuse "1".

    The the Pythagorean identity sugests the opposite side is sqrt(1^2 - x^2)

    Simply write down sin(S).
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  3. #3
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    Re: Finding x with arcsin and arccos in expression

    Hello, benny92000!

    x \ge 0\:\text{ and }\,\arcsin x \:=\: \arccos 2x.\;\;\text{ Solve for }x.

    Take the sine of both sides: . \sin\big(\arcsin x\big) \:=\:\sin\big(\arccos 2x\big)

    . . and we have: . x \;=\;\sin\big(\arccos 2x\big) .[1]


    Let \theta = \arccos2x \quad\Rightarrow\quad \cos\theta = 2x

    We have: . \cos\theta \:=\:\frac{2x}{1} \:=\:\frac{adj}{hyp}

    \theta is in a right triangle with: . adj = 2x,\;hyp = 1
    Pythagorus says: . opp \,=\,\sqrt{1-4x^2}

    Then: . \sin\theta \:=\:\frac{opp}{hyp} \:=\:\sqrt{1-4x^2} \quad\Rightarrow\quad \theta \:=\:\arcsin\sqrt{1-4x^2}


    Hence, [1] becomes: . x \;=\;\sin\big(\arcsin\sqrt{1-4x^2}\big)

    n . . . . . . . . . . . . . . . x \;=\;\sqrt{1-4x^2}

    Square both sides: . . x^2 \:=\:1-4x^2

    . . . . . . . . . . . . . . . 5x^2 \:=\:1

    n . . . . . . . . . . . . . . x^2 \:=\:\frac{1}{5}

    . . . . . . . . . . . . . . . . x \:=\:\frac{1}{\sqrt{5}}

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