Question

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

Answer 1

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.

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

So let's multiply by `(1+sin theta)/(1+sin theta)`.

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

`cos(2 theta) = cos^2 theta - sin^2 theta` `" "` that doesn't look helpful

`= 1-2sin^2 theta` `" "` might be helpful, but we're going to get `cos theta` on top, so let's go with

`cos(2 theta) = 2cos^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.

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

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

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

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

Taking the limit as `x rarr pi/2`, we get `1/4`

Alternative Solution using `cos(2 theta) = 1-2sin^2 theta`

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

`= (1-sin theta)/(2(1-sin^2 theta))`

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 rarr pi/2`, we get `1/4`

Answer 2

`1/4`.

Explanation:

We know that, `lim_(trarr0)sint/t=1`.

Reqd. Limit`=lim_(thetararrpi/2)[(1-sintheta)/(1+cos2theta)]`

Let `theta=pi/2+h, "so that, as" theta rarrpi/2, hrarr0`.Hence,

`"The Reqd. Lim."=lim_(hrarr0)[(1-sin(pi/2+h))/(1+cos2(pi/2+h))]`

`=lim_(hrarr0)[(1-cosh)/(1+cos(pi+2h))]`

`=lim_(hrarr0)[(1-cosh)/(1-cos2h)]`

`=lim_(hrarr0)[(1-cosh)/(2sin^2h)]`

`=1/2[lim_(hrarr0){cancel((1-cosh))/(cancel((1-cosh))(1+cosh))}]`

`=1/2(1/(1+1))`

`=1/4`, as Respected Jim H. , and, Debarun M. , have

obtained!

Enjoy maths.!.