0
$\begingroup$

I was watching lecture 3 from MIT 1801 on single variable calculus, and I am having trouble understanding how the professor simplified the expression for the difference quotient. The original expression for the difference quotient was: $$ \frac{\sin(x + \Delta x) - \sin(x)}{\Delta x}$$

and I understand how he simplified it to: $$ \frac{\sin(x)\cos(\Delta x) + \cos(x)\sin(\Delta x) - \sin(x)}{\Delta x}$$ I just can't understand the simplification from the above expression to this: $$ \sin(x)\biggl (\frac{\cos(\Delta x) - 1}{\Delta x}\biggr) + \cos(x)\biggl(\frac{\sin(\Delta x)}{\Delta x}\biggr)$$

Could someone explain the steps involved to get to this solution? Thanks,

$\endgroup$
  • $\begingroup$ Factor a $\sin(x)$ out of the first and third terms in the numerator. $\endgroup$ – N. F. Taussig Oct 31 '17 at 17:02
0
$\begingroup$

\begin{align} \frac{\color{blue}{\sin(x)}\cos(\Delta x) + \cos(x)\sin(\Delta x) - \color{blue}{\sin(x)}}{\color{green}{\Delta x}} &= \frac{\color{blue}{\sin(x)}\cos(\Delta x) - \color{blue}{\sin(x)}+ \cos(x)\sin(\Delta x) }{\color{green}{\Delta x}}\\ &= \frac{\color{blue}{\sin(x)}(\cos(\Delta x) - 1)+ \cos(x)\sin(\Delta x) }{\color{green}{\Delta x}} \\ &=\frac{\color{blue}{\sin(x)}(\cos(\Delta x) - 1)}{\color{green}{\Delta x}}+ \frac{\cos(x)\sin(\Delta x) }{\color{green}{\Delta x}} \\ &=\color{blue}{\sin(x)}\biggl (\frac{\cos(\Delta x) - 1}{\color{green}{\Delta x}}\biggr) + \cos(x)\biggl(\frac{\sin(\Delta x)}{\color{green}{\Delta x}}\biggr) \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.