Where is the gradient of the sine function zero?
Where is the gradient of the sine function at its steepest positive slope? What is the gradient at these points? (Imagine a tangent at one of these points on the graph above, what is its slope? Hint: it is an integer. You could also check using the applet below.)
Where is the gradient of the sine function at its steepest negative slope? What is the gradient at these points?
On the interval 0\le x\le 2\pi where is the gradient positive?
On the interval 0\le x\le 2\pi where is the gradient negative?
Create and fill in the following table:

x	Slope at x
0	\editable{}
\frac{\pi}{2}	\editable{}
\pi	\editable{}
\frac{3\pi}{2}	\editable{}
2\pi	\editable{}

Do the properties and values of the slope of sin\left(x\right) match another graph you are familiar with?
Try the applet below. Use the slider the move the point along the sine curve and observe the value of the gradient. The click "Plot slopes" to plot the value of the gradient along the curve and see the gradient function take shape.

Created with Geogebra

The evidence looked at so far suggests the derivative of f\left(x\right)=sin\left(x\right) is which function?

Graphical approximation of the derivative of cosine

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)?

Example 2

Sketch the graph of y=cos\left(x\right) and observe the behaviour of the slope.
As we did in our first activity note the key features in the gradient and where they occur.
Do these features match the function you initially suspected might be the derivative? If not, what function do you now suspect is the gradient function?
The function f\left(x\right)=cos\left(x\right) is equivalent to f\left(x\right)=sin\left(x+\frac{\pi}{2}\right), that is the graph of cosine is the same as the graph of sine translated to the left by \frac{\pi}{2}. Thus it would make sense that the derivative of cosine should match the derivative of f\left(x\right)=sin\left(x+\frac{\pi}{2}\right). Our first activity suggests this should be f'\left(x\right)=cos\left(x+\frac{\pi}{2}\right). What basic trigonometric function is this equivalent to? Does this match your observations?

Approximating the derivatives using limits

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} \frac{sin\left(x+h \right )-sin\left(x\right)}{h}

Example 3

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 \frac{sin\left(x+h\right)-sin\left(x\right)}{h} and cosx.

Step 2. Create a list of varying x values in column A - these can be positive, negative, go up in increments of \frac{\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.

Discussion questions

Did the value of the limit appear to correspond closely with the value of cos\left(x\right)?
If you adjust the value of h to be closer to zero, does the value of the limit appear to approach cos\left(x\right)? How about if the value of h is close to zero but negative?
Try altering the set of x values, does the limit always approach cos\left(x\right)?
In our previous activity using the graphs to explore the derivatives we found that the derivative of f\left(x\right)=cos\left(x\right) was f'\left(x\right)=-sin\left(x\right), how can you alter your spreadsheet to explore this conjecture?

Proving the derivative of sine

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.

Geometric proof

Let's look at an outline of the geometric proof.

Example 4

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 \frac{d}{d\theta}sin\theta=\lim_{\Delta \theta \rightarrow 0} \frac{sin\left(\theta+\Delta\theta\right)-sin\theta}{\Delta \theta} Consider the following:

From the diagram explain why the length of DE=sin\left(\theta+\Delta\theta\right)-sin\theta=\Delta y
What is the length of the arc AD?
Consider as \Delta \theta approaches zero, what will the triangle ADE look like?

Will the hypotenuse AD approach the length of arc AD?
Given angle DAB approaches a right angle, show that angle ADE approaches \theta.
Hence, the two triangles approach similarity. Use similar ratios to find what value \frac{\Delta y}{\Delta \theta} approaches.

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.

Created with Geogebra

The applet illustrates the following:

\frac{d}{d\theta}\left(sin\theta\right)= \frac{dy}{d\theta}=\lim_{\Delta \theta \rightarrow 0} \frac{\Delta y}{\Delta \theta}\approx \frac{DE}{DA}\approx \frac{BC}{AB}=cos\theta

Algebraic proof

Our algebraic proof relies on us knowing the value of the following limits \lim_{\theta\rightarrow 0} \frac{sin\theta}{\theta} and \lim_{\theta\rightarrow 0} \frac{cos\theta-1}{\theta}. Let's first look at establishing the values of these limits.

Theorem 1

\lim_{\theta\rightarrow 0} \frac{sin\theta}{\theta}=1

Example 5

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<\frac{\pi}{2}. That is, we will look to show that \lim_{\theta\rightarrow 0^+} \frac{sin\theta}{\theta}=1.

Consider the following diagram:

Triangle AOC:

Area \Delta AOC	=	\frac{1}{2}\text{base}\times \text{height}
	=	\frac{1}{2} \editable{} \times \editable{}

Triangle DOC:

Area \Delta DOC	=	\frac{1}{2}\text{base}\times \text{height}
	=	\frac{1}{2} \editable{} \times \editable{}

Sector AOC:

Area sector DOC	=	(Fraction of the circle)\times (Area of the circle)
	=	\editable{} \times \editable{}	Remember to write the fraction of a circle in terms of radians
	=	\frac{\editable{}}{\editable{}}	Simplify

From the figure we can see that the following inequality holds:

Area triangle AOC	\le	Area sector AOC	\le	Area triangle
\editable{}	\le	\frac{\theta}{2}	\le	\editable{}	Enter the formulas for the areas found above
\editable{}	\le	\theta	\le	\editable{}	Multiply through by 2
\editable{}	\le	\frac{\theta}{sin\theta}	\le	\editable{}	Divide through by sin\theta and simplify. (See note below)
\editable{}	\ge	\frac{sin\theta}{\theta}	\ge	\editable{}	Invert each term, this will reverse the inequality signs

Note: As 0<\theta<\frac{\pi}{2}, sin\theta>0 and thus, dividing by sin\theta will not impact the direction of the inequalities.

Hence, we should arrive at the following:

cos\theta\le \frac{sin\theta}{\theta} \le 1

So we have the value of \frac{sin\theta}{\theta} lies between cos\theta and 1. Since both \lim_{\theta\rightarrow 0^+} cos\theta=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^+} \frac{sin\theta}{\theta}=1.

Lastly, as \frac{sin\theta}{\theta}=\frac{sin\left(-\theta\right)}{-\theta} and using the inequality above, we also have \lim_{\theta\rightarrow 0^-} \frac{sin\theta}{\theta}=1. Thus completing our proof.

Theorem 2

\lim_{\theta\rightarrow 0} \frac{cos\theta-1}{\theta}=0

Example 6

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} \frac{cos\theta-1}{\theta}	=	\lim_{\theta\rightarrow 0} \frac{cos\theta-1}{\theta} \times \frac{\left(cos\theta+1\right)}{\left(cos\theta+1\right)}	Multiply the numerator and denominator by \left(cos\theta+1\right)
	=	\lim_{\theta\rightarrow 0}\frac{\editable{}}{\theta \left(\cos \theta + 1\right)}	Combine and expand the numerator
	=	\lim_{\theta\rightarrow 0}\frac{\editable{}}{\theta \left(\cos \theta + 1\right)}	Simplify the numerator using the identity sin^2\theta+cos^2\theta=1
	=	\lim_{\theta\rightarrow 0}\frac{\editable{}}{\theta}\times \lim_{\theta\rightarrow 0} \frac{\editable{}}{\cos \theta + 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}\frac{-sin\theta}{\theta}	=	\editable{}	Does the limit look familiar? Use Theorem 1 to evaluate.

\lim_{\theta\rightarrow 0}\frac{sin\theta}{cos\theta+1}	=	\frac{\editable{}}{\editable{}}	Evaluate the limit by letting \theta=0
	=	0

Hence,

\lim_{\theta\rightarrow 0} \frac{cos\theta-1}{\theta}	=	\editable{} \times \editable{}	From the limits given above
	=	\editable{}	Simplify

This concludes the proof of the two limits required for our final proof of the derivative of f\left(x\right)=sinx.

Theorem 3

If f\left(x\right)=sin\left(x\right) then f'\left(x\right)=cos\left(x\right).

Example 7

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}\frac{f\left(x+h\right)-f\left(x\right)}{h}
	=	\lim_{h\rightarrow 0}\frac{\editable{} - \editable{}}{h}
	=	\lim_{h\rightarrow 0}\frac{\editable{} - sinx}{h}	Using the angle addition formula expand the term sin\left(x+h\right). See note 1 below.
	=	\lim_{h\rightarrow 0}\frac{sinx \left(\editable{}\right) - \editable{}}{h}	Group the terms with sinx together and factorise
	=	\lim_{h\rightarrow 0} sinx \frac{\editable{}}{h}+\lim_{h\rightarrow 0} cosx \frac{\editable{}}{\editable{}}	Split the fraction in two and apply the limit to each term
	=	sinx \times \lim_{h\rightarrow 0}\frac{\editable{}}{h}+cosx \times \lim_{h\rightarrow 0}\frac{\editable{}}{h}	See note 2 below. Do the limits look familiar?
	=	sinx \times\editable{}+cosx \times\editable{}	Use theorem 1 and 2 to evaluate the limits
\therefore f'\left(x\right)	=	cosx	Simplify

Note 1. Angle addition formula tells us that sin\left(A+B\right)=sinAcosB+cosAsinB.

Note 2. As sinx and cosx 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)=sinx is indeed f'\left(x\right)=cosx.

Theorem 4

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 sinx, we can use this to prove the derivative of cosx.

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(\frac{\pi}{2}-x\right) and the chain rule.

\frac{d}{dx}\left(cosx\right)	=	\frac{d}{dx}\left(sin\left(\frac{\pi}{2}-x\right)\right)
	=	-1\times cos\left(\frac{\pi}{2}-x\right)	By chain rule
	=	-sin\left(x\right)	Using the identity cos\left(\frac{\pi}{2}-x\right)=sinx

Discussion question

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?

INVESTIGATION: Limits and trigonometric functions

The derivatives of sin\left(x\right) and cos\left(x\right)

Graphical approximation of the derivative of sine

Example 1