# Trigo 7

• Aug 16th 2009, 02:46 AM
thereddevils
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 .
• Aug 16th 2009, 04:08 AM
running-gag
You just need to study the function $f(t) = \frac{(1+t)^2}{9+t^2}$ defined over IR
• Aug 16th 2009, 06:50 AM
thereddevils
Quote:

Originally Posted by running-gag
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 .
• Aug 16th 2009, 10:44 AM
running-gag
I am just saying that you need to study the function (variations, limits, ...)
Start by calculating the derivative (Wink)
• Aug 16th 2009, 11:04 AM
Hello everyone
Quote:

Originally Posted by running-gag
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:

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

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

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

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