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Math Help - solved algebracially

  1. #1
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    solved algebracially

    arcsinx-arccosx = 0

    i know x is 45

    but how to do algebraccly
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  2. #2
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    Check the problem.
    If sinx - cosx = 0 then x = 45 degrees.
    Now
    arcsinx+arccosx = π/2

    arcsinx-arccosx = arcsinx+arccosx - 2arcosx = 0

    π/2 -2arccosx = 0

    Solve for x.
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  3. #3
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    Hello, aeroflix!

    Solve: . \arcsin x-\arccos x \:=\: 0

    i know x is 45 . .
    This is wrong!


    \text}Let: }\,\alpha = \arcsin x \quad\Rightarrow\quad \begin{Bmatrix}\sin\alpha \:=\: x \\ \\[-3mm] \cos\alpha \:=\: \sqrt{1-x^2} \end{Bmatrix} .[1]

    \text{Let: }\,\beta = \arccos x \quad\Rightarrow\quad \begin{Bmatrix} \cos\beta \:=\:x \\ \\[-3mm] \sin\beta \:=\:\sqrt{1-x^2} \end{Bmatrix} .[2]


    The equation becomes: . \alpha - \beta \;=\;0

    Take the sine of both sides: . \sin(\alpha - \beta) \;=\;\sin 0

    and we have: . \sin\alpha\cos\beta - \cos\alpha\sin\beta \;=\;0


    Substitute [1] and [2]: . (x)(x) - (\sqrt{1-x^2})(\sqrt{1-x^2}) \;=\;0

    . . x^2 - (1-x^2) \:=\:0 \quad\Rightarrow\quad 2x^2-1 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:\frac{1}{2}


    Therefore: . x \;=\;\pm\dfrac{1}{\sqrt{2}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I tried to check my answers and ran into a new puzzle.
    . . (New to me, anyway.)


    First, I assumed that the angles are in [0^o,\:360^o]

    Then I tested x = \frac{1}{\sqrt{2}}

    We have: . \arcsin\left(\frac{1}{\sqrt{2}}\right) - \arccos\left(\frac{1}{\sqrt{2}}\right)\quad\hdots .Does it equal 0 ?


    We know that: . \begin{Bmatrix}\arcsin(\frac{1}{\sqrt{2}}) &=& 45^o,\:135^o \\ \\[-3mm] \arccos(\frac{1}{\sqrt{2}}) &=& 45^o,\:315^o \end{Bmatrix}


    So we have: . \begin{Bmatrix}45^o \\ 135^o\end{Bmatrix} - \begin{Bmatrix}45^o \\ 315^o\end{Bmatrix} \;\begin{array}{c}_{_?} \\=\\ ^{^?}\end{array} \; 0


    The equation is true if we select 45^o from both sets.

    Query: .Are we allowed to do that?

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