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# Derivative of The Sine Function

In my last note I proved that $$\quad\displaystyle \lim_{\theta\to 0}\frac { \theta}{sin\theta} = 1$$

If you have not read that I would recommend reading that first as this limit is used in proving the derivative of the sine function. Here is the link.

$$\qquad \qquad\qquad\qquad\qquad\qquad \displaystyle \lim_{\theta\to 0}\frac { \theta}{sin\theta} = ?$$

To prove the derivative of the sine function we will be using the first principles of derivatives.

$$\qquad\qquad\qquad\qquad\qquad\displaystyle \lim_{\Delta\theta\to 0}\frac{\sin(\theta + \Delta\theta) - \sin\theta}{\Delta\theta}$$

Use the addition formula for $$\sin(\theta + \Delta\theta)$$.

$$=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta \times \cos \Delta\theta + \sin\Delta\theta \times \cos\theta - \sin\theta}{\Delta\theta}$$

rearrange the equation and factor out $$\sin\theta$$

$$=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta \times \cos \Delta\theta - \sin\theta + \sin\Delta\theta \times \cos\theta }{\Delta\theta}$$

$$=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta( \cos \Delta\theta - 1) + \sin\Delta\theta \times \cos\theta }{\Delta\theta}$$

break apart the fraction

$$=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\frac{\sin\theta( \cos \Delta\theta - 1)}{\Delta\theta} + \frac{\sin\Delta\theta \times \cos\theta }{\Delta\theta}$$

$$=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta} + \frac{sin\Delta\theta}{\Delta\theta} \times cos\theta$$

We wIll now evaluate the two limits seperately

$$\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} \frac{sin\Delta\theta}{\Delta\theta} \times cos\theta$$

In my last note I proved that $$\displaystyle\lim_{\Delta\theta\to 0}\frac{sin\Delta\theta}{\Delta\theta}=1$$

$$=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} 1 \times cos\theta$$

$$=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} cos\theta$$

That solves that limit now the other limit.

$$\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta}$$

divide out the $$\sin\theta$$

$$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos \Delta\theta - 1)}{\Delta\theta}$$

multiply by the conjugate

$$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos \Delta\theta - 1)}{\Delta\theta} \times \frac{\cos\Delta\theta+1}{\cos\Delta\theta+1}$$

$$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos^{2} \Delta\theta - 1)}{\Delta\theta(\cos\Delta\theta+1)}$$

Rewrite the the equation using the Pythagorean identity. $$\cos^{2}\theta + \sin^{2}\theta=1$$

$$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{-\sin^{2} \Delta\theta }{\Delta\theta(\cos\Delta\theta+1)}$$

break apart the fraction

$$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} (-sin\Delta\theta) \times \frac{\sin\Delta\theta}{\Delta\theta} \times \frac{1}{\cos\Delta\theta + 1}$$

evaluate the limits

$$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \times (-sin0) \times 1 \times \frac{1}{\cos0 + 1}$$

$$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \times 0 \times 1 \times \frac{1}{2}$$

$$=\qquad\qquad\qquad\quad\displaystyle 0$$

Now add both limts.

$$\displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta} + \lim_{\Delta\theta\to 0}\frac{sin\Delta\theta}{\Delta\theta} \times cos\theta$$

$$\displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad0 + \cos\theta$$

$$\displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad \cos\theta$$

Note by Brody Acquilano
2 years, 7 months ago

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