Differentiating the Trigonometric Functions Revision Notes for HSC SSCE Mathematics Advanced

Differentiating the Trigonometric Functions

Introduction

Building on the fundamental limit from earlier work, we can now establish the derivatives of the three main trigonometric functions: $\sin x$ , $\cos x$ , and $\tan x$ . These derivatives form essential standard results that you need to know by heart. The proofs are quite advanced, but using these formulas to differentiate more complex trigonometric functions is straightforward once you master the basic patterns.

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While the mathematical proofs of these derivative formulas are quite sophisticated, you don't need to memorize the proofs. However, you absolutely must memorize the three standard derivative formulas themselves, as they form the foundation for all trigonometric differentiation.

Standard derivatives of trigonometric functions

Here are the three fundamental derivative formulas for trigonometric functions. These must be memorised:

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Standard derivatives:

$\frac{d}{dx} \sin x = \cos x$
$\frac{d}{dx} \cos x = -\sin x$
$\frac{d}{dx} \tan x = \sec^2 x$

Critical requirement: These formulas only work when $x$ is measured in radians, not degrees. This is one of the key reasons why radians are the standard unit for angles in calculus.

Notice that the derivative of sine is cosine, but the derivative of cosine introduces a negative sign. The derivative of tangent involves secant squared, which may seem unusual at first but follows from the quotient rule applied to $\tan x = \frac{\sin x}{\cos x}$ .

Graphical understanding of the derivative of sin x

Before diving into calculations, it helps to understand visually why the derivative of $\sin x$ is $\cos x$ .

Consider the graph of $y = \sin x$ . If we sketch its derivative by examining where the gradient is zero, maximum, or minimum:

At $x = 0$ , the sine curve has its steepest positive gradient
At $x = \frac{\pi}{2}$ , the curve reaches a maximum, so the gradient is zero
At $x = \pi$ , the gradient is steepest in the negative direction
At $x = \frac{3\pi}{2}$ , the curve reaches a minimum, so the gradient is zero again

When we plot these gradient values, we get a wave that looks exactly like the cosine function. Furthermore, we know from limits that the gradient of $y = \sin x$ at the origin equals exactly $1$ , which matches the maximum value of $\cos x$ at $x = 0$ .

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This graphical approach doesn't constitute a rigorous proof, but it provides strong visual evidence for the relationship between sine and its derivative. It's an excellent way to build intuition about why $\frac{d}{dx} \sin x = \cos x$ makes sense.

Working with the standard forms

Now let's see how to apply these standard derivatives in practice.

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Worked Example: Basic differentiation

Differentiate these functions and find the gradient when $x = \frac{\pi}{4}$ :

a) $y = \sin x + \cos x$

Applying the standard derivatives:

$y' = \cos x - \sin x$

When $x = \frac{\pi}{4}$ :

$y' = \cos \frac{\pi}{4} - \sin \frac{\pi}{4}$

$y' = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}$

$y' = 0$

This tells us that at $x = \frac{\pi}{4}$ , the function has a horizontal tangent.

b) $y = x - \tan x$

Differentiating:

$y' = 1 - \sec^2 x$

When $x = \frac{\pi}{4}$ :

$y' = 1 - \sec^2 \frac{\pi}{4}$

$y' = 1 - (\sqrt{2})^2$

$y' = 1 - 2 = -1$

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Worked Example: Finding the tangent line

If $f(x) = \sin x$ , find $f'(0)$ and determine the equation of the tangent to $y = \sin x$ at the origin.

Step 1: Note that $f(0) = \sin 0 = 0$ , so the curve passes through the origin.

Step 2: Differentiate: $f'(x) = \cos x$

Step 3: Evaluate at $x = 0$ : $f'(0) = \cos 0 = 1$

Step 4: This means the tangent line at the origin has gradient 1. Using the point-gradient form:

$y - 0 = 1(x - 0)$

$y = x$

This simple result confirms that radian measure is the natural choice for calculus. The tangent to the sine curve at the origin has the beautifully simple equation $y = x$ .

Using the chain rule to generate more standard forms

When the angle in a trigonometric function is not simply $x$ , but a linear expression like $3x + 4$ , we need to use the chain rule. This creates a predictable pattern.

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Worked Example: Chain rule with linear functions

a) Differentiate $y = \cos(3x + 4)$

Let $u = 3x + 4$ , so $y = \cos u$

Then $\frac{du}{dx} = 3$ and $\frac{dy}{du} = -\sin u$

By the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

$\frac{dy}{dx} = -\sin(3x + 4) \times 3$

$\frac{dy}{dx} = -3\sin(3x + 4)$

b) Differentiate $y = \tan(5x - 1)$

Let $u = 5x - 1$ , so $y = \tan u$

Then $\frac{du}{dx} = 5$ and $\frac{dy}{du} = \sec^2 u$

By the chain rule:

$\frac{dy}{dx} = \sec^2(5x - 1) \times 5$

$\frac{dy}{dx} = 5\sec^2(5x - 1)$

c) Differentiate $y = \sin(ax + b)$

Let $u = ax + b$ , so $y = \sin u$

Then $\frac{du}{dx} = a$ and $\frac{dy}{du} = \cos u$

By the chain rule:

$\frac{dy}{dx} = \cos(ax + b) \times a$

$\frac{dy}{dx} = a\cos(ax + b)$

This last result generalises to give us extended standard forms.

Standard derivatives of functions of ax + b

When the angle is a linear function $ax + b$ , we can use these extended formulas:

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Extended standard forms:

$\frac{d}{dx} \sin(ax + b) = a\cos(ax + b)$
$\frac{d}{dx} \cos(ax + b) = -a\sin(ax + b)$
$\frac{d}{dx} \tan(ax + b) = a\sec^2(ax + b)$

The pattern is clear: differentiate as normal, but multiply by the coefficient of $x$ .

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Worked Example: Using extended standard forms

a) Differentiate $y = \cos 7x$

Here $a = 7$ and $b = 0$ :

$\frac{dy}{dx} = -7\sin 7x$

b) Differentiate $y = 4\sin(3x - \frac{\pi}{3})$

Here $a = 3$ and $b = -\frac{\pi}{3}$ :

$\frac{dy}{dx} = 4 \times 3\cos(3x - \frac{\pi}{3})$

$\frac{dy}{dx} = 12\cos(3x - \frac{\pi}{3})$

c) Differentiate $y = \tan \frac{3}{2}x$

Here $a = \frac{3}{2}$ and $b = 0$ :

$\frac{dy}{dx} = \frac{3}{2}\sec^2 \frac{3}{2}x$

Using the chain rule with more complex functions

For functions that don't fit the $ax + b$ pattern, we apply the chain rule in its general form.

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Worked Example: Chain rule with powers and composite functions

a) Differentiate $y = \tan^2 x$

Let $u = \tan x$ , so $y = u^2$

Then $\frac{du}{dx} = \sec^2 x$ and $\frac{dy}{du} = 2u$

By the chain rule:

$\frac{dy}{dx} = 2u \times \sec^2 x$

$\frac{dy}{dx} = 2\tan x \sec^2 x$

b) Differentiate $y = \sin(x^2 - \frac{\pi}{4})$

Let $u = x^2 - \frac{\pi}{4}$ , so $y = \sin u$

Then $\frac{du}{dx} = 2x$ and $\frac{dy}{du} = \cos u$

By the chain rule:

$\frac{dy}{dx} = \cos u \times 2x$

$\frac{dy}{dx} = 2x\cos(x^2 - \frac{\pi}{4})$

Using the product rule with trigonometric functions

When a trigonometric function is multiplied by another function, use the product rule: $\frac{d}{dx}(uv) = v\frac{du}{dx} + u\frac{dv}{dx}$

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Worked Example: Product rule application

Differentiate $y = e^x \cos x$

Step 1: Let $u = e^x$ and $v = \cos x$

Step 2: Find the derivatives: $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = -\sin x$

Step 3: Apply the product rule:

$\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$

$\frac{dy}{dx} = e^x \cos x - e^x \sin x$

$\frac{dy}{dx} = e^x(\cos x - \sin x)$

Using the quotient rule with trigonometric functions

When one function is divided by another, use the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

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Worked Example: Quotient rule application

Differentiate $y = \frac{\sin x}{x}$

Step 1: Let $u = \sin x$ and $v = x$

Step 2: Find the derivatives: $\frac{du}{dx} = \cos x$ and $\frac{dv}{dx} = 1$

Step 3: Apply the quotient rule:

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

$\frac{dy}{dx} = \frac{x\cos x - \sin x}{x^2}$

Successive differentiation of sine and cosine

An interesting pattern emerges when we differentiate $y = \sin x$ repeatedly:

y = \sin x y' = \cos x y'' = -\sin x y''' = -\cos x y'''' = \sin x

Notice that after four differentiations, we return to the original function. This means differentiation is an order 4 operation on the sine function.

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The same pattern occurs with $y = \cos x$ . Both functions satisfy:

$y'' = -y$ (the second derivative equals the negative of the original)
$y'''' = y$ (the fourth derivative equals the original)

This cyclical behavior is unique to sine and cosine among elementary functions, and it reflects their periodic nature.

Differentiation as phase shift

Each differentiation of sine or cosine shifts the wave to the left by $\frac{\pi}{2}$ , which is one quarter of the period $2\pi$ . We say that differentiation advances the phase by $\frac{\pi}{2}$ .

Key relationships:

$\frac{d}{dx} \sin x = \cos x = \sin(x + \frac{\pi}{2})$
$\frac{d}{dx} \cos x = -\sin x = \cos(x + \frac{\pi}{2})$

The second derivative shifts the wave by $\pi$ (half a period), which reflects the graph in the $x$ -axis:

$\frac{d^2}{dx^2} \sin x = -\sin x$
$\frac{d^2}{dx^2} \cos x = -\cos x$

Four differentiations shift the wave by $2\pi$ (one full period), bringing it back to itself:

$\frac{d^4}{dx^4} \sin x = \sin x$
$\frac{d^4}{dx^4} \cos x = \cos x$

Connection between π and e

There are striking parallels between the number $\pi$ (used in radians) and the number $e$ (the natural exponential base).

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Interesting Parallels:

For exponential functions:

The derivative of $y = e^x$ is exactly $y' = e^x$
The tangent to $y = e^x$ at the $y$ -intercept has gradient exactly $1$

For trigonometric functions with radian measure:

The derivative of $y = \sin x$ is exactly $y' = \cos x$
The tangent to $y = \sin x$ at the origin has gradient exactly $1$

Both $\pi = 3.141592...$ and $e = 2.718281...$ are irrational numbers. The number $\pi$ relates to the area of circles, while $e$ relates to areas under the rectangular hyperbola. These connections hint at deeper relationships between trigonometric and exponential functions.

Interestingly, there are now four functions whose fourth derivatives equal themselves: $y = \sin x$ , $y = \cos x$ , $y = e^x$ , and $y = e^{-x}$ .

Graphs of the six trigonometric functions

It's helpful to visualise all six trigonometric functions. Here are their key properties:

Amplitude and period:

$\sin x$ and $\cos x$ each have amplitude 1
The other four functions do not have a defined amplitude
$\sin x$ , $\cos x$ , $\sec x$ , and $\text{cosec } x$ each have period $2\pi$
$\tan x$ and $\cot x$ have period $\pi$

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Key Points to Remember:

Master the three standard derivatives: $\frac{d}{dx} \sin x = \cos x$ , $\frac{d}{dx} \cos x = -\sin x$ , $\frac{d}{dx} \tan x = \sec^2 x$
For linear angles $ax + b$ : Multiply the standard derivative by the coefficient $a$
Use the chain rule when the angle is a more complex function of $x$
Successive differentiation of sine and cosine returns to the original function after four steps
Radian measure is essential for these formulas to work correctly

Differentiating the Trigonometric Functions (HSC SSCE Mathematics Advanced): Revision Notes