Deduce to inequality

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July 14th 2009, 11:25 PM
arze

Deduce to inequality

By using the formulae expressing $sin\theta$ and $cos\theta$ in terms of t, where $(t\equiv tan\frac{\theta}{2})$ or otherwise, show that $\frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^ 2}{9+t^2}$ .
Deduce that $0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10} {9}$
I have done the first part using the identities $sin\theta\equiv\frac{2t}{1+t^2}$ and $cos\theta\equiv\frac{1-t^2}{1+t^2}$ where $t\equiv tan\frac{\theta}{2}$ , but I don't know how to do the second part.
Thanks
July 14th 2009, 11:38 PM
NonCommAlg

Quote:

Originally Posted by arze

By using the formulae expressing $sin\theta$ and $cos\theta$ in terms of t, where $(t\equiv tan\frac{\theta}{2})$ or otherwise, show that $\frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^ 2}{9+t^2}$ .
Deduce that $0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10} {9}$
I have done the first part using the identities $sin\theta\equiv\frac{2t}{1+t^2}$ and $cos\theta\equiv\frac{1-t^2}{1+t^2}$ where $t\equiv tan\frac{\theta}{2}$ , but I don't know how to do the second part.
Thanks

the left side of the inequality is clear. to prove the right side we have: $\frac{(1+t)^2}{9+t^2} -\frac{10}{9}=\frac{9(1+t)^2 - 10(9+t^2)}{9(9+t^2)}=\frac{-t^2+18t-81}{9(9+t^2)}=\frac{-(t-9)^2}{9(9+t^2)} \leq 0.$

but what if the question was this: find the maximum value of $f(t)=\frac{(1+t)^2}{9+t^2}, \ t \in \mathbb{R}$ ? (no calculus allowed!)
July 14th 2009, 11:39 PM
red_dog

We have to prove that $0\leq\frac{(1+t)^2}{9+t^2}\leq\frac{10}{9}$

The first inequality is obviously true.

For the second:

$\frac{(1+t)^2}{9+t^2}\leq\frac{10}{9}\Leftrightarr ow 9+18t+9t^2\leq 90+10t^2\Leftrightarrow t^2-18t+81\geq 0\Leftrightarrow(t-9)^2\geq 0$
July 15th 2009, 12:47 AM
simplependulum

$<br /> \frac{1 + \sin\theta}{5 + 4\cos{\theta}}<br />$

$= \frac{ (\sin{\theta/2} + \cos{\theta/2})^2}{1 + 4(2)(\cos^2{\theta/2})}$

$= \cos^2{\theta/2} \frac{ (1 + \tan{\theta/2})^2}{ \sin^2{\theta/2} + 9\cos^2{\theta/2}}$

$= \frac{ (1+\tan{\theta/2})^2}{ 9 + \tan^2{\theta/2}}$

for part 2 , besides using $b^2 - 4ac \geq 0$
here is another method :

$\frac{(1+t)^2 }{t^2 + 9}$

$= \frac{ t^2 + 9 + 2t - 8}{(t-4)^2 + 8(t-4) + 25}$

$= 1 + \frac{2}{ (t-4) + 8 + \frac{25}{t-4}}$

$\leq 1 + \frac{2}{2\sqrt{25} + 8}$

$= \frac{ 10}{9}$