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Math Help - Trig Help

  1. #1
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    Trig Help

    Hey Everyone,
    First time poster. I just have a couple questions that I was wondering if anyone could help me with (big test tomorrow). Thanks in advance.

    1.
    Given tan^-1 (x) = u :

    a) State the domain of tan^-1 (x)

    b) Find the exact value of sin (2tan^-1(5)).

    c) Find an algebraic expression in x for f(x)=sin(2tan^-1(x))

    2
    Solve for x exactly:

    a) 2cos^2(x) - sin(x) = 1

    b) 2cos(x) + 1 < 0


    3. A quadrilateral ABCD with AB=10, BC=10, CD=12 and DA=15, and angle A=50 degrees. Find the area of the quadrilateral.

    Thanks again. I'll try to return the favor by answering some others questions at some point.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bluejfr09 View Post
    1.
    Given tan^-1 (x) = u :

    a) State the domain of tan^-1 (x)

    b) Find the exact value of sin (2tan^-1(5)).

    c) Find an algebraic expression in x for f(x)=sin(2tan^-1(x))
    a) For the domain we usually look for where x is not defined. So do you know of any values of x where tan^{-1}(x) does not exist?

    b)
    sin(2~tan^{-1}(5)) = 2sin(tan^{-1}(5))~cos(tan^{-1}(5))

    Now, sketch a right triangle with hypotenuse of length 1. (We can actually choose any length we like for this. I choose 1 as a standard) and let one of the non-right angles be \theta. My typical orientation is that the hypotenuse slopes up and to the right, and my \theta is in the lower left corner. Define \theta = tan^{-1}(5). This means that tan(\theta) = 5. So we know that the legs of the right triangle (x and y in an obvious orientation) are related by
    tan(\theta) = \frac{y}{x} = 5 \implies y = 5x

    So the Pythagorean theorem says that
    x^2 + y^2 = 1

    x^2 + (5x)^2 = 1

    26x^2 = 1

    x = \frac{1}{\sqrt{26}} \implies y = \frac{5}{\sqrt{26}}

    So
    sin(tan^{-1}(5)) = sin(\theta) = \frac{5}{\sqrt{26}}
    and
    cos(tan^{-1}(5)) = cos(\theta) = \frac{1}{\sqrt{26}}

    So finally:
    sin(2~tan^{-1}(5)) = 2sin(tan^{-1}(5))~cos(tan^{-1}(5)) = 2 \cdot \frac{5}{\sqrt{26}} \cdot \frac{1}{\sqrt{26}}

    = \frac{10}{26} = \frac{5}{13}

    c) This is virtually identical to b). In fact, the right triangle is exactly the same, it's just that instead of 5 you have an x. (You'll want to change the variables x and y that I used in the above derivation, of course.)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bluejfr09 View Post
    2
    Solve for x exactly:

    a) 2cos^2(x) - sin(x) = 1
    Probably the best way to tackle a) is to recall that
    cos^2(x) = 1 - sin^2(x)

    Sub this into your equation and you'll have a quadratic in sin(x) that you can solve.

    Quote Originally Posted by bluejfr09 View Post
    b) 2cos(x) + 1 < 0
    cos(x) < -\frac{1}{2}

    The easiest thing to do here would be to graph the function y = cos(x) + \frac{1}{2} and see where it is negative.

    -Dan
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  4. #4
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    Thank you!
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