Limits of trigonometric functions

• Sep 12th 2008, 01:41 PM
tariq_h_tauheed
Limits of trigonometric functions
I have to find the limit of the following function. Can i hav the procedure plzz? Plzz use identities that are not too complex. I am a +2 student.

Limit 1 - sin(x/2)
x--> pie cos(x/2)[cos(x/4-sin(x/4)
• Sep 12th 2008, 03:07 PM
Soroban
Hello, tariq_h_tauheed!

We'll need this identity: . $\cos^2\!A - \sin^2\!A \:=\:\cos2A$

Quote:

$\lim_{x\to\pi}\:\frac{1-\sin\frac{x}{2}} {\cos\frac{x}{2}\left(\cos\frac{x}{4} - \sin\frac{x}{4}\right)}$
Multiply top and bottom by $\left(\cos\frac{x}{4} + \sin\frac{x}{4}\right)\!:$

$\frac{1-\sin\frac{x}{2}}{\cos\frac{x}{2}\left(\cos\frac{x} {4} - \sin\frac{x}{4}\right)} \cdot {\color{blue}\frac{\cos\frac{x}{4}+ \sin\frac{x}{4}}{\cos\frac{x}{4} + \sin\frac{x}{4}}} \;\;=$ . $\frac{(1-\sin\frac{x}{2})(\cos\frac{x}{4} + \sin\frac{x}{4})}{\cos\frac{x}{2}\left(\cos^2\!\fr ac{x}{4} - \sin^2\!\frac{x}{4}\right)}$

. . $= \;\;\frac{(1-\sin\frac{x}{2})(\cos\frac{x}{4} + \sin\frac{x}{4})}{\cos\frac{x}{2}\cdot\cos\frac{x} {2}} \;\;=\;\;\frac{(1-\sin\frac{x}{2})(\cos\frac{x}{4} + \sin\frac{x}{4})}{\cos^2\!\frac{x}{2}}$

. . $=\;\;\frac{(1-\sin\frac{x}{2})(\cos\frac{x}{4} + \sin\frac{x}{4})} {1-\sin^2\!\frac{x}{2}} \;\;=$ . $\frac{(1-\sin\frac{x}{2})(\cos\frac{x}{4} + \sin\frac{x}{4})}{(1 - \sin\frac{x}{2})(1 + \sin\frac{x}{2})} \;= \;\frac{\cos\frac{x}{4} + \sin\frac{x}{4}}{1 + \sin\frac{x}{2}}$

Therefore: . $\lim_{x\to\pi}\:\frac{\cos\frac{x}{4} + \sin\frac{x}{4}}{1 + \sin\frac{x}{2}} \;\;=\;\;\frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} {1 + 1} \;\;=\;\;\frac{\frac{2}{\sqrt{2}}}{2} \;\;=\;\;\boxed{\frac{1}{\sqrt{2}}}$

• Sep 13th 2008, 02:38 AM
Moo
Hello,

Another way of doing it...

$t=\frac x4 \implies t \to \frac \pi 4$

So we're looking for $\lim_{t \to \frac \pi 4} \frac{1-\sin(2t)}{\cos(2t) (\cos(t)-\sin(t))}$

Note that $1-\sin(2t)=\underbrace{\cos^2(t)+\sin^2(t)}_{=1}-\underbrace{2 \sin(t) \cos(t)}_{=\sin(2t)}=(\cos(t)-\sin(t))^2$ and $\cos(2t)=\cos^2(t)-\sin^2(t)=(\cos(t)-\sin(t))(\cos(t)+\sin(t))$

The limit is now :

$\lim_{t \to \frac \pi 4} \frac{{\color{red}(\cos(t)-\sin(t))^2}}{{\color{red}(\cos(t)-\sin(t))}(\cos(t)+\sin(t)){\color{red}(\cos(t)-\sin(t))}}$

Simplify :

$\lim_{t \to \frac \pi 4} \frac{1}{\cos(t)+\sin(t)}=\frac{1}{\frac{\sqrt{2}} {2}+\frac{\sqrt{2}}{2}}=\frac{1}{\sqrt{2}}$