Math Help - Small angle approximation for sinx

1. 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.

2. Re: Small angle approximation for sinx

Originally Posted by Googl
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

3. Re: Small angle approximation for sinx

Originally Posted by skeeter
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?

4. Re: Small angle approximation for sinx

Consider

as $\displaystyle x \to 0\implies \sin x = x \implies \frac{\sin x }{x} = 1$

5. Re: Small angle approximation for sinx

Originally Posted by Googl
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.

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

$\lim_{\theta \to 0} \frac{1 - \cos{\theta}}{\theta} = 0$

6. Re: Small angle approximation for sinx

Originally Posted by skeeter
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.