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

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#

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.!.