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Math Help - Trig Help

  1. #1
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    Trig Help

    I have a collection of Trig problems that are giving me some difficulty...

    1. Finding the Period, Phase Shift, and Asymptotes for the following expressions
    3+4cot(5x-2)
    y=-3tan2(pi X-pi/4)
    2. Find the exact value for
    tan105
    sin195
    cos51cos9-sin51sin9
    3.Simplify the expression
    1+cosx + sin x
    sin x 1+cos x

    1+secx
    sinx + tanx

    1 +siny
    1 +cscy

    thanks in advance for all help...these are some HW problems that i am confused on and i have an exam in the coming week, so if explanations could be given for all problems that would be nice as well
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  2. #2
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    The first simplify expression was posted weirdly

    it is supposed to read

    1 + cosx divided by sinx + sin x divided by 1 + cosx
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  3. #3
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    Hello, mmgolf!

    There are some "tricks" you should know for #3 . . .


    3. Simplify:

    . . (a)\;\;\frac{1+\cos x}{\sin x} + \frac{\sin x}{1+\cos x}
    Multiply the second fraction by: \frac{1-\cos x}{1-\cos x}

    . . We have: . \frac{\sin x}{1 + \cos x}\cdot\frac{1-\cos x}{1-\cos x} \;=\;\frac{\sin x(1 - \cos x)}{1 - \cos^2\!x} \;=\;\frac{\sin x(1 - \cos x)}{\sin^2\!x} \;=\;\frac{1-\cos x}{\sin x}


    The problem becomes: . \frac{1+\cos x}{\sin x} + \frac{1-\cos x}{\sin x}\;=\; \frac{2}{\sin x} \;=\;2\csc x



    (b)\;\;\frac{1+\sec x }{\sin x + \tan x}
    Multiply by: \frac{\cos x}{\cos x}

    . . \frac{\cos x}{\cos x}\,\cdot\,\frac{1 + \sec x}{\sin x + \tan x} \;=\;\frac{\cos x + 1}{\sin x\cos x + \sin x} \;=\;\frac{\cos x + 1}{\sin x(\cos x + 1)} \;=\;\frac{1}{\sin x} \;=\;\csc x



    (c)\;\frac{1 + \sin y}{1 + \csc y}
    Multiply by: \frac{\sin y}{\sin y}

    . . \frac{\sin y}{\sin y}\cdot\frac{1 + \sin y}{1 + \csc y} \;=\;\frac{\sin y(1 + \sin y)}{\sin y + 1} \;=\;\sin y

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  4. #4
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    thanks...anyone else....??
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  5. #5
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    Hello again, mmgolf!

    For #2, you need some Sum/Difference formulas.


    2. Find the exact value for:
    (a)\;\;\tan105
    (b)\;\;\sin195
    (c)\;\;\cos51\cos9 - \sin51\sin9
    (a)\;\;\tan105^o

    We note that: . 105 \:=\:60 + 45
    . . and we use: . \tan(A + B)\:=\:\frac{\tan A + \tan B}{1 - \tan A\tan B}

    Hence we have: . \tan105 \:=\:\tan(60 + 45) \;=\;\frac{\tan60 + \tan45}{1 - \tan60\tan45}

    Can you finish it now?



    (b)\;\;\sin195^o

    Note that: . 195 \:=\:150 + 45
    . ,and we use: . \sin(A + B) \:=\:\sin A\cos B + \sin B\cos A

    So we have: . \sin(150 + 45) \:=\:\sin150\cos45 + \sin45\cos150 . . . etc.



    (c)\;\;\cos51\cos9 - \sin51\sin9
    You're expected to recognize that this is the right side of the formula:
    . . \cos(A + B) \:=\:\cos A\cos B - \sin A\sin B

    Got it?

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  6. #6
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    thanks soroban

    any hints on the tangent problems on how to find the asymptotes, period, and phase shift??
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