Derivative of The Sine Function

In my last note I proved that limθ0θsinθ=1 \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.

limθ0θsinθ=? \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.

limΔθ0sin(θ+Δθ)sinθΔθ\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(θ+Δθ)\sin(\theta + \Delta\theta).


=limΔθ0sinθ×cosΔθ+sinΔθ×cosθsinθΔθ=\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 sinθ\sin\theta


=limΔθ0sinθ×cosΔθsinθ+sinΔθ×cosθΔθ=\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}

=limΔθ0sinθ(cosΔθ1)+sinΔθ×cosθΔθ=\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


=limΔθ0sinθ(cosΔθ1)Δθ+sinΔθ×cosθΔθ=\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}

=limΔθ0sinθ×(cosΔθ1)Δθ+sinΔθΔθ×cosθ=\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


limΔθ0sinΔθΔθ×cosθ\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 limΔθ0sinΔθΔθ=1\displaystyle\lim_{\Delta\theta\to 0}\frac{sin\Delta\theta}{\Delta\theta}=1


=limΔθ01×cosθ=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} 1 \times cos\theta

=limΔθ0cosθ=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} cos\theta


That solves that limit now the other limit.


limΔθ0sinθ×(cosΔθ1)Δθ\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta}


divide out the sinθ\sin\theta


=sinθlimΔθ0(cosΔθ1)Δθ=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos \Delta\theta - 1)}{\Delta\theta}


multiply by the conjugate


=sinθlimΔθ0(cosΔθ1)Δθ×cosΔθ+1cosΔθ+1=\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}

=sinθlimΔθ0(cos2Δθ1)Δθ(cosΔθ+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. cos2θ+sin2θ=1 \cos^{2}\theta + \sin^{2}\theta=1


=sinθlimΔθ0sin2ΔθΔθ(cosΔθ+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


=sinθlimΔθ0(sinΔθ)×sinΔθΔθ×1cosΔθ+1=\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


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

=sinθ×0×1×12=\qquad\qquad\qquad\quad\displaystyle \sin\theta \times 0 \times 1 \times \frac{1}{2}

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


Now add both limts.


ddθ=limΔθ0sinθ×(cosΔθ1)Δθ+limΔθ0sinΔθΔθ×cosθ \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

ddθ=0+cosθ \displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad0 + \cos\theta

ddθ=cosθ \displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad \cos\theta

Note by Brody Acquilano
4 years, 5 months ago

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