# Confusion regarding the derivation of $\cos(x)$ when differentiating $\sin(x)$

The textbook I'm reading derives it like this:

\eqalign{ & y = \sin x \Rightarrow \left( 1 \right) \cr & y + \delta y = \sin (x + \delta x) \Rightarrow (2) \cr}

Subtracting equation (1) from (2):

\eqalign{ & y + \delta y - y = \sin (x + \delta x) - \sin x \cr & \delta y = \sin (x + \delta x) - \sin x \cr & = 2\cos (x + {1 \over 2}\delta x)\sin \left({1 \over 2}\delta x\right) \cr}

Without completing the whole "proof" I can tell you where I am confused.

$\delta y = \sin (x + \delta x) - \sin x$

to this:

$2\cos (x + {1 \over 2}\delta x)\sin \left({1 \over 2}\delta x\right)$

What happened here? It looks like the double angle identity was used? Isn't the double angle formula used for when two angles are the same? I know dx is very small, but still, the value of x has been altered so shouldn't the "angle addition" identity be used instead? Further more what happened to the -sin(x)? How was it incorporated into $2\cos (x + {1 \over 2}\delta x)\sin \left({1 \over 2}\delta x\right)$ ?

This is confusing me quite a bit, any help would be greatly appreciated, thank you.

EDIT: Furthermore, 1/2(dx): where does this come from?

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You need the formula for sinus of the sum of two angles, not the twice angle formula! You certainly find that in wikipedia. –  kjetil b halvorsen Apr 16 '13 at 11:02

To prove that formula just use

Let $\alpha = A+B, \beta = A-B$. Then note that $\alpha +\beta =2A$ and $\alpha - \beta=2B$. Now you have $A=\displaystyle \frac{\alpha+\beta}{2}$ and $B=\displaystyle \frac{\alpha-\beta}{2}$

\begin{align*} \sin(\alpha)-\sin(\beta) &= \sin(A+B) - \sin(A-B) \\ &= \sin(A)\cdot\cos(B) + \cos(A)\cdot \sin(B) - \sin(A)\cdot\cos(B)+\cos(A)\cdot\sin(B) \\ &=2 \cdot \cos(A) \cdot \sin(B) \\ &= 2 \cdot \cos\left(\frac{\alpha+\beta}{2}\right) \cdot \sin\left(\frac{\alpha-\beta}{2}\right) \end{align*}

$\Large\text{Method 2}$

\begin{align*} f'(x) &= \lim_{\delta x\to 0} \frac{\sin(x+\delta{x}) -\sin(x)}{\delta{x}} \\ &=\lim_{\delta{x}\to0} \frac{\sin(x)\cdot\cos(\delta{x}) +\cos(x)\cdot\sin\delta{x} -\sin(x)}{\delta{x}} \\ &=\lim_{\delta{x}\to0} \frac{\sin(x)\cdot \bigl(\cos(\delta{x})-1\bigr)}{\delta{x}} + \lim_{\delta{x}\to 0}\frac{\cos{x}\cdot \sin(\delta{x})}{\delta{x}} \\ &= \lim_{\delta{x}\to 0} \frac{\sin(x)\cdot \bigl(\cos(\delta{x})-1\bigr)}{\delta{x}} + \cos(x) \qquad \Bigl[\because \small{\lim_{x\to 0} \frac{\sin{x}}{x}=1} \Bigr] \end{align*}

Now i leave it to you to prove $\displaystyle\lim_{\delta{x}\to 0}\frac{\sin(x)\cdot \bigl(\cos(\delta{x})-1\bigr)}{\delta{x}} =0$. This can be proved by observing that $1-\cos(2x) = 2\cdot \sin^{2}(x)$.

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There was used a well-known formula $$\sin{\alpha}-\sin{\beta}=2\cos{\dfrac{\alpha+\beta}{2}}\sin{\dfrac{\alpha-\beta}{2}}.$$

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I didn't know it... –  Ben Millwood Apr 16 '13 at 11:01
I see, thank you! –  seeker Apr 16 '13 at 11:02

Recall the identities $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and its close relative $\sin(A-B)=\sin A\cos B-\cos A\sin B$. Subtract. We get $$\sin(A+B)-\sin(A-B)=2\cos A\sin B.$$ Let $A+B=s$ and $A-B=t$. Then $A=\frac{s+t}{2}$ and $B=\frac{s-t}{2}$. So we obtain $$\sin s-\sin t=2\cos\left(\frac{s+t}{2} \right) \sin\left(\frac{s-t}{2} \right).$$

Remark: We give a somewhat different derivation of the derivative formula, that I think is a little easier for students. We also change notation slightly for ease of typing.

We are interested in the behaviour of $\frac{\sin(x+h)-\sin x}{h}$ as $h$ approaches $0$. The top is $\sin x\cos h=\cos x \sin h$. Divide by $h$ and rearrange slightly. We get $$\cos x\frac{\sin h}{h} -\sin x\frac{1-\cos h}{h}.$$ Let $h\to 0$. What happens to the first part is familiar. As for $\frac{1-\cos h}{h}$, multiply top and bottom by $1+\cos h$. We get $$\frac{1}{1+\cos h}\frac{\sin^2 h}{h}.$$ Since $\frac{\sin h}{h}$ has limit $1$, we conclude that $\frac{\sin^2 h}{h}$ has limit $0$.

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Yep, got it, thanks. –  seeker Apr 16 '13 at 11:18