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Math Help - Inverse Expressions and Finding the exact solution

  1. #1
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    Inverse Expressions and Finding the exact solution

    Ok, I have a few problems I just cant seem to understand.

    1. Find the Exact Value of the Expression
    Cos(tan^-1(4/3))

    I know your supposed to write tangent in terms of cosine but im completely lost.

    2. Find all exact solutions (in simplest form) to the following equations.

    a) 2cosxsinx +cosx = 0

    b)sin2x = tanx

    I tried on each, and failed miserably on each. Any help is appericiated.
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  2. #2
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    Quote Originally Posted by Byud View Post
    1. Find the Exact Value of the Expression
    Cos(tan^-1(4/3))
    By definition, tan^(-1)(4/3) = (theta) means that tan(theta) = 4/3. So draw a right triangle, label the one angle as "theta", and fill in the "opposite" and "adjacent" values. Use the Pythagorean Theorem to find the length of the remaining side. Then take the cosine.

    Quote Originally Posted by Byud View Post
    2. Find all exact solutions (in simplest form) to the following equations.

    a) 2cosxsinx +cosx = 0
    How would you start the solution to 2XY + X = 0? Do the same factorization here.

    Quote Originally Posted by Byud View Post
    b)sin2x = tanx
    Convert the left-hand side to sines and cosines, using the double-angle identity they gave you. Convert the right-hand side to sines and cosines using the definition of tangent. See where that leads....

    Quote Originally Posted by Byud View Post
    I tried on each, and failed miserably on each.
    How so? (You might have been "nearly there", but until we can see your work, we just can't know where you're needing the help.)

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  3. #3
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    Quote Originally Posted by Byud View Post
    Ok, I have a few problems I just cant seem to understand.

    1. Find the Exact Value of the Expression
    Cos(tan^-1(4/3))

    I know your supposed to write tangent in terms of cosine but im completely lost.

    2. Find all exact solutions (in simplest form) to the following equations.

    a) 2cosxsinx +cosx = 0

    b)sin2x = tanx

    I tried on each, and failed miserably on each. Any help is appericiated.
    y = \tan^{-1}\left(\frac{4}{3}\right)

    \tan{y} = \frac{4}{3} = \frac{opp}{adj} ... hypotenuse = 5

    \cos{y} = \frac{adj}{hyp} = \frac{3}{5}



    solutions 0 \leq x < 2\pi

    2\cos{x}\sin{x} + \cos{x} = 0

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

    \cos{x} = 0 ... x = \frac{\pi}{2} , x = \frac{3\pi}{2}

    \sin{x} = -\frac{1}{2} ... x = \frac{7\pi}{6} , x = \frac{11\pi}{6}



    \sin(2x) = \tan{x}

    2\sin{x}\cos{x} = \frac{\sin{x}}{\cos{x}}

    2\sin{x}\cos{x} - \frac{\sin{x}}{\cos{x}} = 0

    \frac{2\sin{x}\cos^2{x}}{\cos{x}} - \frac{\sin{x}}{\cos{x}} = 0

    2\sin{x}\cos^2{x} - \sin{x} = 0

    \sin{x}(2\cos^2{x} - 1) = 0

    \sin{x} = 0 ... x = 0 , x = \pi

    \cos{x} = \pm \frac{\sqrt{2}}{2} ... x = \frac{\pi}{4} , x = \frac{3\pi}{4} , x = \frac{5\pi}{4} , x = \frac{7\pi}{4}
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  4. #4
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    Damn, that seems much easier now, I didnt even thinking about factoring out the trig functions, I was just trying to use those damned double angle formulas.

    I did the first one the harder way, but I guess I did get it right!
    my solution was...

    Cos(tan^-1(4/3))

    tan^-1(4/3) = u

     \cos{u} = \frac{1}{\sqrt{1+\tan{u^2}}}

     \cos{u} = \frac{1}{\sqrt{1+\frac{4}{3}^2}}

     \cos{u} = \frac{1}{\sqrt{1+\frac{16}{9}}}

     \cos{u} = \frac{1}{\sqrt{\frac{25}{9}}}

     \cos{u} = \frac{1}{\frac{5}{3}}

     \cos{u} = \frac{3}{5}
    CHEER!
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