The sine and cosine functions are periodic functions, meaning that the function cycles, or repeats.
The period of both the basic sine function and the cosine functions, \sin\left(x\right) and \cos\left(x\right) is 2\pi.
From the graph, it is clear that the gradient of the sine function also varies periodically with the same period as the original function.
x | Slope at x |
---|---|
0 | ⬚ |
\dfrac{\pi}{2} | ⬚ |
\pi | ⬚ |
\dfrac{3\pi}{2} | ⬚ |
2\pi | ⬚ |
From completing the activity above we have evidence suggesting that the derivative of f\left(x\right)=\sin\left(x\right) is in fact f'\left(x\right)=\cos\left(x\right). What do you suspect might be the derivative of f\left(x\right)=\cos\left(x\right)?
Another way to obtain the derivative of \sin\left(x\right) is to use first principles as follows:
f'\left(x\right) = \lim_{h\rightarrow 0} \dfrac{\sin\left(x+h \right )-\sin\left(x\right)}{h}
To evaluate this limit algebraically we will need to manipulate it and know some key limits involving trigonometric functions. Before we look further into that, let's explore the approximate value of the limit above by using selected values of x and an appropriately small value of h. and compare it to values of \cos\left(x\right). Note: that values of x and h will be in radians.
A convenient way to investigate this will be to set up a spreadsheet.
Step 1. Create the column headings for x, h, the fraction \dfrac{\sin\left(x+h\right)-\sin\left(x\right)}{h} and \cos \left(x\right).
Step 2. Create a list of varying x values in column A - these can be positive, negative, go up in increments of \dfrac{\pi}{4} or be any varying set. Also, set the value of h, in column B, as a convenient number close to zero, such as 0.001 or 0.0001. Both the values of x and h can be altered later to further investigate the limit.
Step 3. Enter the following formula in cell C2 to evaluate the given fraction: =(sin(A2+B2)-sin(A2))/B2. Then drag the formula to fill down the column.
Step 4. Enter the following formula in cell D2 to evaluate cosine of the corresponding x value for comparison: =cos(A2). Then drag the formula to fill down the column.
We have observed that the derivative of f\left(x\right)=\sin\left(x\right) appears to be f'\left(x\right)=\cos\left(x\right). We can set about proving this using a geometric or algebraic approach. Let's look at both approaches.
Let's look at an outline of the geometric proof.
Using pen and paper or a geometry program, follow the construction steps below:
Step 1. In a unit circle label point A at an angle of \theta from the positive x-axis and construct the triangle ABC as shown below:
Step 2. Place a second point D on the unit circle at an angle of \theta+\Delta\theta from the positive x-axis. Then construct the triangle ADE as shown below.
We are interested in \dfrac{d}{d\theta}\sin \left(\theta \right)=\lim_{\Delta \theta \rightarrow 0} \dfrac{\sin\left(\theta+\Delta\theta\right)-\sin \left(\theta \right)}{\Delta \theta} Consider the following:
The applet below demonstrates the crux of the geometric proof. Select a value for \theta then use the slider for \Delta \theta to observe the given ratios as \Delta \theta approaches zero. Retry for different values of \theta.
The applet illustrates the following:
\dfrac{d}{d\theta}\left(\sin\left(\theta \right)\right)= \dfrac{dy}{d\theta}=\lim_{\Delta \theta \rightarrow 0} \dfrac{\Delta y}{\Delta \theta}\approx \dfrac{DE}{DA}\approx \dfrac{BC}{AB}=\cos\left(\theta \right)
Our algebraic proof relies on us knowing the value of the following limits \lim_{\theta\rightarrow 0} \dfrac{\sin\left(\theta \right)}{\theta} and \lim_{\theta\rightarrow 0} \dfrac{\cos\left(\theta \right)-1}{\theta}. Let's first look at establishing the values of these limits.
\lim_{\theta\rightarrow 0} \dfrac{\sin\left(\theta \right)}{\theta}=1
Follow the explanation given below and fill in the gaps to complete the proof.
To prove the value of this limit we will consider the following diagram and show that the limit as \theta\rightarrow 0 from above and below is 1.
Let's first look at the case where 0<\theta<\dfrac{\pi}{2}. That is, we will look to show that \lim_{\theta\rightarrow 0^+} \dfrac{\sin\left(\theta \right)}{\theta}=1.
Consider the following diagram:
Triangle AOC:
Area \Delta AOC | = | \dfrac{1}{2}\text{base}\times \text{height} |
= | \dfrac{1}{2} ⬚ \times ⬚ |
Triangle DOC:
Area \Delta DOC | = | \dfrac{1}{2}\text{base}\times \text{height} |
= | \dfrac{1}{2} ⬚ \times ⬚ |
Sector AOC:
Area sector DOC | = | (Fraction of the circle)\times (Area of the circle) |
|
= | ⬚ \times ⬚ |
Remember to write the fraction of a circle in terms of radians |
|
= | ⬚ \times ⬚ |
Simplify |
From the figure we can see that the following inequality holds:
Area triangle AOC | \le | Area sector AOC | \le | Area triangle |
|
⬚ | \le | \dfrac{\theta}{2} | \le | ⬚ |
Enter the formulas for the areas found above |
⬚ | \le | \theta | \le | ⬚ |
Multiply through by 2 |
⬚ | \le | \dfrac{\theta}{\sin\left(\theta \right)} | \le | ⬚ |
Divide through by \sin\left(\theta \right) and simplify. (See note below) |
⬚ | \ge | \dfrac{\sin\left(\theta \right)}{\theta} | \ge | ⬚ |
Invert each term, this will reverse the inequality signs |
Note: As 0<\theta<\dfrac{\pi}{2}, \sin \left(\theta \right)>0 and thus, dividing by \sin\left(\theta \right) will not impact the direction of the inequalities.
Hence, we should arrive at the following:
\cos\left(\theta \right)\le \dfrac{\sin\left(\theta \right)}{\theta} \le 1
So we have the value of \dfrac{\sin\left(\theta \right)}{\theta} lies between \cos\left(\theta \right) and 1. Since both \lim_{\theta\rightarrow 0^+} \cos\left(\theta \right)=1 and \lim_{\theta\rightarrow 0^+} 1=1, by the sandwich theorem (also known as the squeeze theorem) our limit of interest must also be 1. That is \lim_{\theta\rightarrow 0^+} \dfrac{\sin\left(\theta \right)}{\theta}=1.
Lastly, as \dfrac{\sin\left(\theta \right)}{\theta}=\dfrac{\sin\left(-\theta\right)}{-\theta} and using the inequality above, we also have \lim_{\theta\rightarrow 0^-} \dfrac{\sin\left(\theta \right)}{\theta}=1. Thus completing our proof.
\lim_{\theta\rightarrow 0} \dfrac{\cos\left(\theta \right)-1}{\theta}=0
Follow the explanation given below and fill in the gaps to complete the proof.
To prove this we are required to algebraically manipulate the fraction to a point we can evaluate the limit.
\lim_{\theta\rightarrow 0} \dfrac{\cos\left(\theta \right)-1}{\theta} | = | \lim_{\theta\rightarrow 0} \dfrac{\cos\left(\theta \right)-1}{\theta} \times \dfrac{\left(\cos\left(\theta \right)+1\right)}{\left(\cos\left(\theta \right)+1\right)} |
Multiply the numerator and denominator by \left(\cos\left(\theta \right)+1\right) |
= | \lim_{\theta\rightarrow 0} \dfrac{⬚}{\theta \left(\cos \left(\theta \right) +1 \right)} |
Combine and expand the numerator |
|
= | \lim_{\theta\rightarrow 0} \dfrac{⬚}{\theta \left(\cos\left(\theta\right)+1\right)} |
Simplify the numerator using the identity \sin^2\left(\theta \right)+\cos^2\left(\theta \right)=1 |
|
= | \lim_{\theta\rightarrow 0} \dfrac{⬚}{\theta}\times\lim_{\theta\rightarrow 0} \dfrac{⬚}{ \cos\left(\theta\right)+1} |
Split the fraction and apply the limit to each term |
Let's look at the two limits separately for a moment:
\lim_{\theta\rightarrow 0}\dfrac{-\sin\left(\theta \right)}{\theta} | = | ⬚ |
Does the limit look familiar? Use Theorem 1 to evaluate. |
|
|||
\lim_{\theta\rightarrow 0}\dfrac{\sin\left(\theta \right)}{\cos\left(\theta \right)+1} | = | \dfrac{⬚}{⬚} |
Evaluate the limit by letting \theta=0 |
= | 0 |
|
Hence,
\lim_{\theta\rightarrow 0} \dfrac{\cos\left(\theta \right)-1}{\theta} | = | ⬚ \times ⬚ |
From the limits given above |
= | ⬚ |
Simplify |
This concludes the proof of the two limits required for our final proof of the derivative of f\left(x\right)=\sin \left(x\right).
If f\left(x\right)=\sin\left(x\right) then f'\left(x\right)=\cos\left(x\right).
Follow the explanation given below and fill in the gaps to complete the proof. We will prove this theorem using first principles, algebraic manipulation and the limits from theorem 1 and 2.
For f\left(x\right)=\sin\left(x\right) we have:
f'\left(x\right) | = | \lim_{h\rightarrow 0}\dfrac{f\left(x+h\right)-f\left(x\right)}{h} |
|
= | \lim_{h\rightarrow 0} \dfrac{⬚ - ⬚}{h} |
|
|
= | \lim_{h\rightarrow 0} \dfrac{⬚ - \sin \left(x\right)}{h} |
Using the angle addition formula expand the term \sin\left(x+h\right). See note 1 below. |
|
= | \lim_{h\rightarrow 0} \dfrac{\sin \left(x\right) \left(⬚\right)- ⬚}{h} |
Group the terms with \sin \left(x\right) together and factorise |
|
= | \lim_{h\rightarrow 0} \sin \left(x\right) \dfrac{⬚}{h} + \lim_{h\rightarrow 0}\cos \left(x\right) \dfrac{⬚}{⬚} |
Split the fraction in two and apply the limit to each term |
|
= | \sin \left(x\right) \times \lim_{h\rightarrow 0}\dfrac{⬚}{h} + \cos \left(x\right) \times \lim_{h\rightarrow 0}\dfrac{⬚}{h} |
See note 2 below. Do the limits look familiar? |
|
= | \sin \left(x\right) \times ⬚ + \cos \left(x\right) \times ⬚ |
Use theorem 1 and 2 to evaluate the limits |
|
\therefore f'\left(x\right) | = | \cos \left(x\right) |
Simplify |
Note 1. Angle addition formula tells us that \sin\left(A+B\right)=\sin \left(A\right) \cos \left(B\right) + \cos \left(A\right) \sin \left(B \right).
Note 2. As \sin \left(x\right) and \cos \left(x\right) do not depend on h, these terms can be moved to the front of the limits.
Thus, we have formally proven that the derivative of f\left(x\right)=\sin \left(x\right) is indeed f'\left(x\right)=\cos \left(x\right).
If f\left(x\right)=\cos\left(x\right) then f'\left(x\right)=-\sin\left(x\right).
Now that we have proven the derivative of \sin \left(x\right), we can use this to prove the derivative of \cos \left(x\right).
We will use the fact that the function f\left(x\right)=\cos\left(x\right) is equivalent to f\left(x\right)=\sin\left(\dfrac{\pi}{2}-x\right) and the chain rule.
\dfrac{d}{dx}\left(\cos x\right) | = | \dfrac{d}{dx}\left(\sin\left(\dfrac{\pi}{2}-x\right)\right) |
|
= | -1\times \cos\left(\dfrac{\pi}{2}-x\right) |
By chain rule |
|
= | -\sin\left(x\right) |
Using the identity \cos\left(\dfrac{\pi}{2}-x\right)=\sin \left(x\right) |
We used radians to evaluate the limits in this investigation. Do the limits change if the argument of the trigonometric function is in degrees? What impact if any is there on the derivative functions?