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Math Help - any help would be great

  1. #1
    bssm19
    Guest

    any help would be great

    ok, thanks for taking a look for starters,
    here we go,
    i have two functions y= sinx-cosx and y=2cos2x
    ive selected an area that has been "trapped" by these two functions and must find the domain and range of the area selected, how do i go about doing this
    i aslo need to find the points of intersection of the two functions in the area i chose , do i start by making both functions equal each other?, if so how do i solve that?
    thanks again
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  2. #2
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    Hello, bssm19!

    I have two functions: y\:=\:\sin x - \cos x and y\:=\:2\cos2x
    I've selected an area that has been "trapped" by these two functions
    and must find the domain and range of the area selected.
    How do i go about doing this?

    Not a clue . . .
    Since the area is some constant, I don't see how it can have a doman and range.


    I also need to find the points of intersection of the two functions in the area i chose.
    Do i start by making both functions equal each other? . . . Yes!
    If so, how do i solve that? . . . It's not easy

    I'll get you started . . .

    Equate the two functions: . 2\underbrace{\cos2x}\;=\;\sin x - \cos x

    Double-angle identity: . 2(\overbrace{\cos^2x - \sin^2x})\;=\;\sin x - \cos x

    We have: . 2(\cos^2x - \sin^2x) + \cos x - \sin x \;= \;0

    Factor: . 2(\cos x - \sin x)(\cos x + \sin x) + (\cos x - \sin x)\;=\;0

    Factor: . (\cos x - \sin x)\,\left[2(\cos x + \sin x) + 1\right]\;=\;0


    And we have two equations to solve:

    (a)\;\;\cos x - \sin x\:=\:0\quad\Rightarrow\quad \sin x\:=\:\cos x\quad\Rightarrow\quad \frac{\sin x}{\cos x} \:=\:1\quad\Rightarrow\quad\tan x\:=\:1
    . . . Hence: . \boxed{x\:=\:\frac{\pi}{4} + \pi n} *

    (b)\;\;2(\cos x + \sin x) + 1\:-\:0\quad\Rightarrow\quad \sin x + \cos x \:=\:-\frac{1}{2}

    . . . Square both sides: . \sin^2x + 2\sin x\cos x + \cos^2x\:=\:\frac{1}{4}

    . . . Since \sin^2x + \cos^2x\:=\:1 and 2\sin x\cos x\:=\:\sin2x
    . . . . we have: . \sin2x + 1 \:= \:\frac{1}{4}\quad\Rightarrow\quad\sin2x\:=\:\text  {-}\frac{3}{4}

    . . . Hence: . 2x\:=\:\sin^{-1}\!\left(\text{-}\frac{3}{4}\right) + 2\pi n\quad\Rightarrow\quad \boxed{x\:=\:\frac{1}{2}\sin^{-1}\!\left(\text{-}\frac{3}{4}\right) + \pi n} *

    [Note: Since we squared the equation, check for extraneous roots.]

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

    * Of course, the value of n depends on what area you selected.

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