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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,

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  • $\begingroup$ Factor a $\sin(x)$ out of the first and third terms in the numerator. $\endgroup$ Oct 31, 2017 at 17:02

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\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}

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