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Math Help - I'm stumped

  1. #1
    Junior Member Freaky-Person's Avatar
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    I'm stumped

    A few question on my trig assignment... It's not going all that well.. I worked hard on the LaTex...

    1. Simplify: \frac{\tan(\frac{\pi}{4} - x) - \tan(\frac{\pi}{4} + x)}{\tan x}

    2. Solve for x, 0 \leq x \leq 2\pi

    a) \sin 2x + \cos x = 0

    b) 3\tan x = \tan 2x


    THANKSIES!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Freaky-Person View Post
    A few question on my trig assignment... It's not going all that well.. I worked hard on the LaTex...

    1. Simplify: \frac{\tan(\frac{\pi}{4} - x) - \tan(\frac{\pi}{4} + x)}{\tan x}
    Hint: Use the addition formulas for tangent

    recall: \tan ( \alpha + \beta ) = \frac {\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta} and \tan ( \alpha - \beta ) = \frac {\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}

    2. Solve for x, 0 \leq x \leq 2\pi

    a) \sin 2x + \cos x = 0
    use the double angle formula for sine

    \sin 2x + \cos x = 0

    \Rightarrow 2 \sin x \cos x + \cos x = 0

    \Rightarrow \cos x ( 2 \sin x + 1 ) = 0

    Now continue

    b) 3\tan x = \tan 2x
    use the double angle formula for tangent

    3 \tan x = \tan 2x

    \Rightarrow 3 \tan x = \frac {2 \tan x }{1 - \tan^2 x}

    Now continue.

    (alternatively, you could change everything to be in terms of sine and cosine here, but I think that would be more trouble than it's worth -- maybe not)


    By the way, good job with the LaTex!
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  3. #3
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by Jhevon View Post
    Hint: Use the addition formulas for tangent

    recall: \tan ( \alpha + \beta ) = \frac {\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta} and \tan ( \alpha - \beta ) = \frac {\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}

    use the double angle formula for sine

    \sin 2x + \cos x = 0

    \Rightarrow 2 \sin x \cos x + \cos x = 0

    \Rightarrow \cos x ( 2 \sin x + 1 ) = 0

    Now continue

    use the double angle formula for tangent

    3 \tan x = \tan 2x

    \Rightarrow 3 \tan x = \frac {2 \tan x }{1 - \tan^2 x}

    Now continue.
    That's great and all, but I have a sheet filled to the brim with all that. I got to those steps already, but I don't know where to go from there
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Freaky-Person View Post
    A few question on my trig assignment... It's not going all that well.. I worked hard on the LaTex...

    1. Simplify: \frac{\tan(\frac{\pi}{4} - x) - \tan(\frac{\pi}{4} + x)}{\tan x}
    what have you tried here? you must have gotten farther than I showed you. if nothing else you could follow your nose, add the fractions and attempt to simplify. show me where you got stuck


    2. Solve for x, 0 \leq x \leq 2\pi



    a) \sin 2x + \cos x = 0
    picking up where i left off.

    \cos x ( 2 \sin x + 1) = 0

    \Rightarrow \cos x = 0 \mbox { or } \sin x = - \frac {1}{2}

    \Rightarrow x = \frac {\pi}{2}, \frac {3 \pi}{2}, \frac {7 \pi}{6}, \mbox { and } \frac {11 \pi}{6} for 0 \leq x \leq 2 \pi

    Now check all those solutions to make sure they work


    b) 3\tan x = \tan 2x
    again, picking up where i left off

    3 \tan x = \frac {2 \tan x}{1 - \tan^2 x}

    \Rightarrow 2 \tan x = 3 \tan x - 3 \tan^3 x ............i cross-multiplied

    \Rightarrow 3 \tan^3 x - \tan x = 0

    \Rightarrow \tan x \left( 3 \tan^2 x - 1 \right) = 0 .................i factored out the common tan(x)

    \Rightarrow \tan x = 0 \mbox { or } \tan x = \frac {1}{\sqrt {3}}

    \Rightarrow x = 0, \pi, 2 \pi, \frac {\pi}{6} \mbox { and } \frac {7 \pi}{6} for 0 \leq x \leq 2 \pi

    Check those solutions as well.
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