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Math Help - Trigo 7

  1. #1
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    Trigo 7

    By using the formula which expresses sin a and cos a in terms of t , where t= tan (a/2) , show that

    (1+sin a)/(5+4 cos a )=(1+t)^2/(9+t^2)

    I can do this part .

    Deduce that 0\leq\frac{1+\sin a}{5+4 \cos a}\leq\frac{10}{9} for all values of a .

    I cant figure out this part .
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  2. #2
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    You just need to study the function f(t) = \frac{(1+t)^2}{9+t^2} defined over IR
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  3. #3
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    Quote Originally Posted by running-gag View Post
    You just need to study the function f(t) = \frac{(1+t)^2}{9+t^2} defined over IR

    Erm , i can see that t can be any real numbers for the function f(t) to be defined over IR . I don get what u are trying to say , sorry .
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  4. #4
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    I am just saying that you need to study the function (variations, limits, ...)
    Start by calculating the derivative
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  5. #5
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    Hello everyone
    Quote Originally Posted by running-gag View Post
    You just need to study the function f(t) = \frac{(1+t)^2}{9+t^2} defined over IR
    Here's a useful technique for dealing with this type of expression.

    Let y =\frac{(1+t)^2}{9+t^2}

    Then 9y+t^2y = 1 + 2t+t^2

    \Rightarrow t^2(y-1) - 2t + (9y-1) =0

    This is a quadratic in t with real roots. So using the discriminant:

    (-2)^2 - 4(y-1)(9y-1) \ge 0

    Which simplifies to:

    -9y^2 +10y \ge 0

    \Rightarrow y(10-9y) \ge 0

    \Rightarrow 0 \le y \le \frac{10}{9}

    \Rightarrow 0 \le \frac{(1+t)^2}{9+t^2} \le\frac{10}{9}

    Grandad
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