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Math Help - Small angle approximation for sinx

  1. #1
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    Small angle approximation for sinx

    Hi all.

    I am working on a problem but I am stuck.

    Prove that dcosx/dx = -sinx using Small angle approximation and the addition formula for sin(x)

    My try:

    (sin(x+h) - sin(x))/h

    I know I am supposed to expand here using the addition formula and the small angle approxima... but I am not too confident.
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  2. #2
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    Re: Small angle approximation for sinx

    Quote Originally Posted by Googl View Post
    Hi all.

    I am working on a problem but I am stuck.

    Prove that dcosx/dx = -sinx using Small angle approximation and the addition formula for sin(x)

    My try:

    (sin(x+h) - sin(x))/h

    I know I am supposed to expand here using the addition formula and the small angle approxima... but I am not too confident.
    why the difference quotient w/ sine? should be cosine.

    \frac{\cos(x+h)-\cos{x}}{h} =

    \frac{\cos{x}\cos{h} - \sin{x}\sin{h}-\cos{x}}{h} =

    \frac{\cos{x}(\cos{h}-1) - \sin{x}\sin{h}}{h} =

    \frac{\cos{x}(\cos{h}-1)}{h} - \frac{\sin{x}\sin{h}}{h}

    you should see two basic trig limits in each term ... now take the limit as h approaches 0
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  3. #3
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    Re: Small angle approximation for sinx

    Quote Originally Posted by skeeter View Post
    why the difference quotient w/ sine? should be cosine.

    \frac{\cos(x+h)-\cos{x}}{h} =

    \frac{\cos{x}\cos{h} - \sin{x}\sin{h}-\cos{x}}{h} =

    \frac{\cos{x}(\cos{h}-1) - \sin{x}\sin{h}}{h} =

    \frac{\cos{x}(\cos{h}-1)}{h} - \frac{\sin{x}\sin{h}}{h}

    you should see two basic trig limits in each term ... now take the limit as h approaches 0
    Is that really correct? how does that equal to -sinx finally?
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  4. #4
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    Re: Small angle approximation for sinx

    Consider

    as \displaystyle x \to 0\implies \sin x = x \implies \frac{\sin x }{x} = 1
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  5. #5
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    Re: Small angle approximation for sinx

    Quote Originally Posted by Googl View Post
    Is that really correct? how does that equal to -sinx finally?
    as I said previously ... there are two basic trig limits with which you should be familiar.

    \lim_{\theta \to 0} \frac{\sin{\theta}}{\theta} = 1

    \lim_{\theta \to 0} \frac{1 - \cos{\theta}}{\theta} = 0
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  6. #6
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    Re: Small angle approximation for sinx

    Quote Originally Posted by skeeter View Post
    as I said previously ... there are two basic trig limits with which you should be familiar.

    \lim_{\theta \to 0} \frac{\sin{\theta}}{\theta} = 1

    \lim_{\theta \to 0} \frac{1 - \cos{\theta}}{\theta} = 0
    Thanks a lot. I managed to understand it.
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