Derivative of The Sine Function

In my last note I proved that lim⁡θ→0θsinθ=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.

lim⁡θ→0θsinθ=? \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.

lim⁡Δθ→0sin⁡(θ+Δθ)−sin⁡θΔθ\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).

=lim⁡Δθ→0sin⁡θ×cos⁡Δθ+sin⁡Δθ×cos⁡θ−sin⁡θΔθ=\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⁡θ\sin\theta

=lim⁡Δθ→0sin⁡θ×cos⁡Δθ−sin⁡θ+sin⁡Δθ×cos⁡θΔθ=\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}

=lim⁡Δθ→0sin⁡θ(cos⁡Δθ−1)+sin⁡Δθ×cos⁡θΔθ=\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

=lim⁡Δθ→0sin⁡θ(cos⁡Δθ−1)Δθ+sin⁡Δθ×cos⁡θΔθ=\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}

=lim⁡Δθ→0sin⁡θ×(cos⁡Δθ−1)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

lim⁡Δθ→0sinΔθΔθ×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 lim⁡Δθ→0sinΔθΔθ=1\displaystyle\lim_{\Delta\theta\to 0}\frac{sin\Delta\theta}{\Delta\theta}=1

=lim⁡Δθ→01×cosθ=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} 1 \times cos\theta

=lim⁡Δθ→0cosθ=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} cos\theta

That solves that limit now the other limit.

lim⁡Δθ→0sin⁡θ×(cos⁡Δθ−1)Δθ\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta}

divide out the sin⁡θ\sin\theta

=sin⁡θlim⁡Δθ→0(cos⁡Δθ−1)Δθ=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos \Delta\theta - 1)}{\Delta\theta}

multiply by the conjugate

=sin⁡θlim⁡Δθ→0(cos⁡Δθ−1)Δθ×cos⁡Δθ+1cos⁡Δθ+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}

=sin⁡θlim⁡Δθ→0(cos⁡2Δθ−1)Δθ(cos⁡Δθ+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⁡2θ+sin⁡2θ=1 \cos^{2}\theta + \sin^{2}\theta=1

=sin⁡θlim⁡Δθ→0−sin⁡2ΔθΔθ(cos⁡Δθ+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

=sin⁡θlim⁡Δθ→0(−sinΔθ)×sin⁡ΔθΔθ×1cos⁡Δθ+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

=sin⁡θ×(−sin0)×1×1cos⁡0+1=\qquad\qquad\qquad\quad\displaystyle \sin\theta \times (-sin0) \times 1 \times \frac{1}{\cos0 + 1}

=sin⁡θ×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=lim⁡Δθ→0sin⁡θ×(cos⁡Δθ−1)Δθ+lim⁡Δθ→0sinΔθΔθ×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+cos⁡θ \displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad0 + \cos\theta

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

#Calculus #Derivatives #DerivativeByFirstPrinciple

Note by Brody Acquilano
6 years, 1 month ago

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