Verifying that the tangent is zero (sine function)

Mar 2017
356
3
Massachusetts
Hi,

I hope someone can help. I'm currently stuck on a question which requires me to verify that a given function's tangent is zero, by using the difference quotient. As a reminder, the difference quotient follows this form: f(x+h) + f(x)/h.

The thing is that I know how to use the difference quotient for quadratic functions, but not for trig functions. I'm really looking for guidance on how to do this (i.e. how trig functions look like in difference quotient form). I attached the questions that I need guidance on in this post (5c and 5b). If someone could show the first few steps of 5c and 5b, that would be wonderful.

Cheers,
- Olivia
Screen Shot 2017-03-25 at 6.27.21 PM.png
 

romsek

MHF Helper
Nov 2013
6,734
3,032
California
for $\sin(x)$ we have a difference quotient of

$\begin {align*}
\displaystyle \lim_{h \to 0}~&\dfrac{\sin(x+h) - \sin(x)}{h} = \\

\lim_{h \to 0}~&\dfrac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h} = \\

\lim_{h \to 0} ~&\dfrac{\sin(x)(1-\cos(h))+ \cos(x)\sin(h)}{h} = \\

\lim_{h \to 0}~&\sin(x)\dfrac{1-\cos(h)}{h}+\cos(x) \dfrac{\sin(h)}{h}

\end{align*}$

$\displaystyle \lim_{h \to 0} \dfrac{1-\cos(h)}{h}=0$

$\displaystyle \lim_{h \to 0} \dfrac{\sin(h)}{h}=1$

$\displaystyle \lim_{h \to 0}~\dfrac{\sin(x+h) - \sin(x)}{h} = \cos(x)$

$\cos(90^\circ)=0$
 
Mar 2017
356
3
Massachusetts
Thank you for responding! So 1) you're saying that the difference quotient for sine function is equal to cos(x)? and 2) if that statement is true then wouldn't 5c be 5cos(90)?
 

romsek

MHF Helper
Nov 2013
6,734
3,032
California
Thank you for responding! So 1) you're saying that the difference quotient for sine function is equal to cos(x)? and 2) if that statement is true then wouldn't 5c be 5cos(90)?
I worked out the difference coefficient for $\sin(x)$

to obtain it for $5\sin(x)$ just multiply by 5 to obtain $5\cos(x)$

5 times 0 is still 0.
 
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romsek

MHF Helper
Nov 2013
6,734
3,032
California
Thank you for responding! So 1) you're saying that the difference quotient for sine function is equal to cos(x)? and 2) if that statement is true then wouldn't 5c be 5cos(90)?
my bad, the difference quotient is just

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

the limit of this as $h \to 0$ evaluated at the the specified point gets you the tangent.