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Math Help - Identities and solve for x

  1. #1
    Junior Member Freaky-Person's Avatar
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    Identities and solve for x

    Man our whole class struggled at this one the whole period.

    Solve for x

    1. tanx + sec2x = 1


    Prove

    2. (\frac{\sin4x}{1-cos4x})*(\frac{1-cos 2x}{cos2x}) = tanx


    3. cot^2A * sin^2 (\frac{\pi}{2}-A) = tan(\frac{\pi}{2}- A) - cosA
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  2. #2
    Junior Member Freaky-Person's Avatar
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    OOOhh! nevermind question 1!

    I just remembered that after 2 hours of deliberation between everyone in the class + 1, I finally got it.

    I got up to \frac{(cos2x + 1)(sin2x)}{cos^2 2x} = tan x on question 2

    And \frac{\frac{cos^3 A}{sinA}}{1+sinA} = \frac{sinA + sin^2 A - 1}{cosA}

    On the last one.

    PLEASE HELP T.T
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  3. #3
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    Hello, Freaky-Person!

    I assume you know the Double- and Half-angle Identities.
    . . Here's #2 . . .


    2) Prove: . \frac{\sin 4x}{1-\cos4x}\cdot \frac{1-\cos 2x}{\cos2x} \;= \;\tan x

    We have: . \frac{2\sin2x\cos2x}{2\sin^2\!2x}\cdot\frac{2\sin^  2\!x}{\cos2x} \;=\;\frac{2\sin^2\!x}{\sin2x} \;= \;\frac{2\sin^2\!x}{2\sin x\cos x} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x

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  4. #4
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by Freaky-Person View Post
    3. cot^2A * sin^2 (\frac{\pi}{2}-A) = tan(\frac{\pi}{2}- A) - cosA
    My calculator says this is false.
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    My calculator says this is false.
    yeah, i think..
    i was able to show
    cot^2A - cos^2A \neq cot A - cos A
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  6. #6
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    Hello, Freaky-Person!

    1. Solve for x\!:\;\;\tan x + \sec2x \:= \:1

    We have: . \frac{\sin x}{\cos x} + \frac{1}{\cos2x} \;=\;1

    . . \sin x\cos2x + \cos x \;=\;\cos x\cos2x

    . . \sin x(1 - 2\sin^2\!x) + \cos x \;=\;\cos x(1 - 2\sin^2\!x)

    . . \sin x - 2\sin^3\!x + \cos x \;=\;\cos x - 2\sin^2\!x\cos x

    . . 2\sin^3\!x - 2\sin^2\!x\cos x - \sin x \;=\;0

    . . \sin x\left(2\sin^2\!x - 2\sin x\cos x - 1\right) \;=\;0


    And we have two equations to solve . . .


    [1]\;\;\sin x \:=\:0\quad\Rightarrow\quad\boxed{ x \:=\:0,\,\pm\pi,\,\pm2\pi, \cdots}


    [2]\;\;2\sin^2x - 2\sin x\cos x - 1 \;=\;0

    Divide by \cos^2\!x\!:\;\;2\tan^2\!x - 2\tan x - \sec^2\!x \;=\;0

    . . 2\tan^2\!x - 2\tan x - (\tan^2\!x + 1) \;=\;0

    . . \tan^2\!x - 2\tan x - 1 \;=\;0

    Quadratic Formula: . \tan x \;=\;\frac{2 \pm\sqrt{2^2-4(1)(-1)}}{2(1)} \;=\;1 \pm \sqrt{2}

    . . Hence: . x \;=\;\arctan\left(1 \pm\sqrt{2}\right)

    Therefore: . \boxed{x \:=\:\begin{Bmatrix}\frac{3\pi}{8} + 2\pi n \\ \text{-}\frac{\pi}{8} + 2\pi n \end{Bmatrix}}

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