Find difference quotient for $f(x) = \sin x$

Question

I need to use the different quotient:

$\frac{f(x+h)-f(x)}{h}$

to show that

$f(x) = \sin(x)$

simplifies to

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

How do I do that? I already have, working from the start point,

$\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}$

but I don't know where to go from there. Thank you!

EDIT:

Following a hint from the first comment, I now have

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

first step: Switch the second and third term in the numerator and factor out $sinx$. Then you need to split the fraction into two fractions. — imranfat, Mar 15 '16 at 21:39
Oops, scratch that. I misread the question. It is only asking for the difference quotient, not the derivative (limit the of difference quotient as $h$ approaches $0$). — , Mar 15 '16 at 23:39
It seems to me that you already have the answer. Your goal is $$\cos(x)\frac{\sin(h)}{h} + \sin(x)\frac{\cos(h) - 1}{h}$$ and you have $$\frac{\sin(x)(\cos(h)-1)}{h} + \frac{\cos(x)\sin(h)}{h}$$ These are the same thing, written slightly differently. — , Mar 15 '16 at 23:55
@Bungo That looks like more than slightly differently to me. What am I missing? — , Mar 16 '16 at 00:03
In the second expression, write the two terms in the opposite order. Then pull the $\sin(x)$ and $\cos(x)$ outside of the fraction. — , Mar 16 '16 at 00:05
Don't overthink this. The expression $\frac{2x}{5}$ is the same thing as $2*\frac{x}{5}$ In your case, it's the same thing — imranfat, Mar 16 '16 at 00:07
@imranfat I feel like a moron but thank you so much I got it now! :) — , Mar 16 '16 at 00:48

score 0 · Accepted Answer · answered Mar 16 '16 at 07:02

0

Things are much easier if we choose to use Prosthaphaeresis Formula,

$$\sin(x+h)-\sin x=2\sin\dfrac h2\cos\dfrac{2x+h}2$$

and use $\lim_{u\to0}\dfrac{\sin u}u=1$

answered Mar 16 '16 at 07:02

lab bhattacharjee

1 Answers1