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Math Help - Limits of trigonometric functions

  1. #1
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    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)
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  2. #2
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    Hello, tariq_h_tauheed!

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


    \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}}}

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  3. #3
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    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}}
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