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

I cant figure out this part .

2. You just need to study the function $\displaystyle f(t) = \frac{(1+t)^2}{9+t^2}$ defined over IR

3. Originally Posted by running-gag
You just need to study the function $\displaystyle f(t) = \frac{(1+t)^2}{9+t^2}$ 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 .

4. I am just saying that you need to study the function (variations, limits, ...)
Start by calculating the derivative

5. Hello everyone
Originally Posted by running-gag
You just need to study the function $\displaystyle f(t) = \frac{(1+t)^2}{9+t^2}$ defined over IR
Here's a useful technique for dealing with this type of expression.

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

Then $\displaystyle 9y+t^2y = 1 + 2t+t^2$

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

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

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

Which simplifies to:

$\displaystyle -9y^2 +10y \ge 0$

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

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

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