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

2 Answers
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.

(1-sin theta)(1+sin theta) = 1- sin^2 theta = cos^2 theta(1sinθ)(1+sinθ)=1sin2θ=cos2θ

So let's multiply by (1+sin theta)/(1+sin theta)1+sinθ1+sinθ.

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

cos(2 theta) = cos^2 theta - sin^2 thetacos(2θ)=cos2θsin2θ " " that doesn't look helpful

= 1-2sin^2 theta=12sin2θ " " might be helpful, but we're going to get cos thetacosθ on top, so let's go with

cos(2 theta) = 2cos^2 theta - 1cos(2θ)=2cos2θ1 (And the 11'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)1sinθ1+cos(2θ)=1sinθ1+(2cos2θ1)1+sinθ1+sinθ

= (1-sin^2 theta)/(2cos^2 theta (1+sin theta))=1sin2θ2cos2θ(1+sinθ)

= (cos^2 theta) / (2cos^2 theta (1+sin theta))=cos2θ2cos2θ(1+sinθ)

= 1/(2(1+sin theta)=12(1+sinθ)

Taking the limit as x rarr pi/2xπ2, we get 1/414

Alternative Solution using cos(2 theta) = 1-2sin^2 thetacos(2θ)=12sin2θ

(1-sin theta)/(1+cos(2 theta)) = (1-sin theta)/(1+(1-2sin^2 theta)1sinθ1+cos(2θ)=1sinθ1+(12sin2θ)

= (1-sin theta)/(2(1-sin^2 theta))=1sinθ2(1sin2θ)

Now 1-sin^2 theta1sin2θ is a difference of squares, so we can factor it and reduce the fraction.

= (1-sin theta)/(2(1+sin theta)(1-sin theta)=1sinθ2(1+sinθ)(1sinθ)

= 1/(2(1+sin theta)=12(1+sinθ)

Taking the limit as x rarr pi/2xπ2, we get 1/414

Aug 24, 2016

1/414.

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.!.