# How do you find the limit of (1-sin(theta))/(1+cos(2theta)) as theta approaches pi/2?

Aug 24, 2016

If you are trying to do this without l'Hospital, see below.

#### Explanation:

We'll use some trigonometry to rewrite the expression to get a determinate form.

$\left(1 - \sin \theta\right) \left(1 + \sin \theta\right) = 1 - {\sin}^{2} \theta = {\cos}^{2} \theta$

So let's multiply by $\frac{1 + \sin \theta}{1 + \sin \theta}$.

At the same time we'll use a double angle formula for $\cos \left(2 \theta\right)$. The question is, which one?

$\cos \left(2 \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta$ $\text{ }$ that doesn't look helpful

$= 1 - 2 {\sin}^{2} \theta$ $\text{ }$ might be helpful, but we're going to get $\cos \theta$ on top, so let's go with

$\cos \left(2 \theta\right) = 2 {\cos}^{2} \theta - 1$ (And the $1$'s will cancel too.)

Try it! If it doesn't work, try something else. Do not just give up if your first attempt doesn't work.

$\frac{1 - \sin \theta}{1 + \cos \left(2 \theta\right)} = \frac{1 - \sin \theta}{1 + \left(2 {\cos}^{2} \theta - 1\right)} \cdot \frac{1 + \sin \theta}{1 + \sin \theta}$

$= \frac{1 - {\sin}^{2} \theta}{2 {\cos}^{2} \theta \left(1 + \sin \theta\right)}$

$= \frac{{\cos}^{2} \theta}{2 {\cos}^{2} \theta \left(1 + \sin \theta\right)}$

 = 1/(2(1+sin theta)

Taking the limit as $x \rightarrow \frac{\pi}{2}$, we get $\frac{1}{4}$

Alternative Solution using $\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \theta$

(1-sin theta)/(1+cos(2 theta)) = (1-sin theta)/(1+(1-2sin^2 theta)

$= \frac{1 - \sin \theta}{2 \left(1 - {\sin}^{2} \theta\right)}$

Now $1 - {\sin}^{2} \theta$ is a difference of squares, so we can factor it and reduce the fraction.

 = (1-sin theta)/(2(1+sin theta)(1-sin theta)

 = 1/(2(1+sin theta)

Taking the limit as $x \rightarrow \frac{\pi}{2}$, we get $\frac{1}{4}$

Aug 24, 2016

$\frac{1}{4}$.

#### Explanation:

We know that, ${\lim}_{t \rightarrow 0} \sin \frac{t}{t} = 1$.

Reqd. Limit$= {\lim}_{\theta \rightarrow \frac{\pi}{2}} \left[\frac{1 - \sin \theta}{1 + \cos 2 \theta}\right]$

Let $\theta = \frac{\pi}{2} + h , \text{so that, as} \theta \rightarrow \frac{\pi}{2} , h \rightarrow 0$.Hence,

$\text{The Reqd. Lim.} = {\lim}_{h \rightarrow 0} \left[\frac{1 - \sin \left(\frac{\pi}{2} + h\right)}{1 + \cos 2 \left(\frac{\pi}{2} + h\right)}\right]$

$= {\lim}_{h \rightarrow 0} \left[\frac{1 - \cosh}{1 + \cos \left(\pi + 2 h\right)}\right]$

$= {\lim}_{h \rightarrow 0} \left[\frac{1 - \cosh}{1 - \cos 2 h}\right]$

$= {\lim}_{h \rightarrow 0} \left[\frac{1 - \cosh}{2 {\sin}^{2} h}\right]$

$= \frac{1}{2} \left[{\lim}_{h \rightarrow 0} \left\{\frac{\cancel{\left(1 - \cosh\right)}}{\cancel{\left(1 - \cosh\right)} \left(1 + \cosh\right)}\right\}\right]$

$= \frac{1}{2} \left(\frac{1}{1 + 1}\right)$

$= \frac{1}{4}$, as Respected Jim H. , and, Debarun M. , have

obtained!

Enjoy maths.!.