Trigonometric Function Derivatives

The derivative of the sine function is the cosine function.

\[ \lim_{x \to 0} \frac{d}{dx} sin x = cos x \]

The derivative of the cosine function is the negative of the sine funtion.

\[ \lim_{x \to 0} \frac{d}{dx} cos x = -sin x \]

Algebraic Proof

We can find the derivatives of these trigonometric functions using the \( f(x)’ = \frac{f(x + \Delta x) - f(x)}{\Delta x} \) formula and substituting in the sine and cosine functions.

Derivative of sine function

First substitute \( sin(x) \) into the derivative formula.

\[ \frac{d}{dx} sin(x) = \frac{sin(x + \Delta x) - sin(x)}{\Delta x} \]

At this point theres not much more we can do algebraically manipulate this expression, so we use the sum the difference rules from the trigonometric identities to expand the expression.

\[ \frac{sin(x)cos(\Delta x) + cos(x)sin(\Delta x) - sin(x)}{\Delta x} \]

We re-arrange to group the \( -sin(x) \) onto the left hand expression so that we can factorise easily.

\[ \frac{sin(x)cos(\Delta x) - sin(x) + cos(x)sin(\Delta x)}{\Delta x} \]

Then factorise into two distinct expressions.

\[ sin(x)\frac{cos(\Delta x) - 1)}{\Delta x} + cos(x)\frac{sin(\Delta x)}{\Delta x} \]

The left hand expression evaluates to \( sin(x) \cdot 0 \) which equals \( 0 \) and the right hand expression evaluates to \( cos(x) \cdot 1 \) which equals \( cos(x) \). Therefore the whole expression evalutes to \( 0 + cos(x) \). Therefore:

\[ \lim_{x \to 0} \frac{d}{dx} sin x = cos x \]

But why does \( \frac{cos(\Delta x) - 1)}{\Delta x} \) evaluate to \( 0 \) but \( \frac{sin(\Delta x)}{\Delta x}) \) evaluates to \( 1 \)? Its not immediately clear as it seems like both of them just approach something like \( \frac{0}{0} \) which would be \( undefined \). The important intuition here is the ratios between the denominator and numerator and the rate at which \( cos \) and \( sin \) approach the respective limits in comparison to \( \Delta x \).

In the \( sin \) case, both \( sin(\Delta x) \) and \( \Delta x \) approach zero at the same rate even at very small amounts. For example, \( sin(0.005) \approx 0.0049999 \) and \( \frac{0.0049999}{0.005} \approx 1 \) there this part of the expression approximates \( 1 \) as the \( \lim_{\Delta x \to 0} \) but never for the actual case of \( 0 \).

In the \( cos \) case, \( \frac{cos(\Delta x) - 1)}{\Delta x} \) and \( \Delta x \) diverge. The difference between \( \Delta x \) and \( \frac{cos(\Delta x) - 1)}{\Delta x} \) is large, with the subtractive numerator function being much much smaller than the denominator. So when they are divided you get an even more tiny number that gets closer and closer to \( 0 \). For example \( cos(0.001) = 0.9999995 \) but \( cos(0.001) - 1 = -4.99999958e-7 \) an extremely tiny number, while \( \Delta x \) is still just \( 0.001 \), so, \( \frac{-4.99999958e-7}{0.001} \) is another tiny number that just gets closer and closer to zero.

We can also leverage a visual interpretation of the \( sin(\Delta x) \) to see why dividing by \( \Delta x \) evaluates to \( 1 \). In the example below \( \theta \) represents the arc length of the circle in radians which, if we view its change as \( \Delta x \) has the same relationship with \( sin(\theta ) \) as \( \frac{sin(\Delta x)}{\Delta x} \). From the visualisation below, we can see that as \( \theta \) reaches very small values the line representing \( sin(\theta) \) overlaps the arc almost exactly and becomes almost equal to it. And, dividing to things that are almost equal, almost equals \( 1 \). With limits we can simply evaluate this to \( 1 \).

This visual interpretation can also be extended to the \( cos(\Delta x) - 1 \) case. In this case, the horizontal distance between the \( x \) coordinate of a point on the perimeter of the unit circle and the edge of the circle is computed by \( 1 - cos(\theta) \) because the radius of the cirlce in \( 1 \) and taking the \( cos(\theta) \) gives the horizontal distance from the center ton the perimeter point. The distace from that point to the edge of the circle approaches \( 0 \) as the arc length \( \theta \) gets closer and closer to \( 0 \) and becomes more and more vertical. So, the length of the horizontal which is computeded as \( 1 - cos(\theta) \) shrinks much more quickly than \( \theta \). And dividing this small number by a larger number approaches \( 0 \).

We can re-write \( \frac{cos(\Delta x) - 1}{\Delta x} \) as \( \frac{cos(\theta) - 1}{\theta} \) to show the equivalence between the geometric and algebraic interpretations. Then, by multiplying by \( -1 \) we change it to \( \frac{1 - cos(\theta)}{\theta} \). The two expressions are equivalent but are much easier to visualise on the unit circle and show the same ratio relationship which makes the calculations about how things evaluate in the algebraic solution make sense.

Derivative of cosine function

The derivative for cosine follows essentially the same logic as the sine function and leverages the same conceptual framework. In this case we use the sum rule for cosine to expand the the derivative formula substitution.

\[ \frac{d}{dx} cos(x) = \frac{cos(x + \Delta x) - cos(x)}{\Delta x} \]

\[ = \frac{cos(x)cos(\Delta x) - sin(x)sin(\Delta x) - cos(x)}{\Delta x} \]

\[ = cos(x)\frac{cos(\Delta x) - 1}{\Delta x} - sin(x)\frac{sin(\Delta x)}{\Delta x} \]

\[ = -sin(x) \]