Derivatives of Circular Functions (VCE SSCE Mathematical Methods): Revision Notes

Derivatives of Circular Functions

Introduction

In this section, we learn how to find the derivatives of the circular functions: sine, cosine and tangent. These are fundamental results that allow us to work with trigonometric functions in calculus.

The key derivatives we will establish are:

• The derivative of $\sin(k\theta)$ is $k\cos(k\theta)$
• The derivative of $\cos(k\theta)$ is $-k\sin(k\theta)$
• The derivative of $\tan(k\theta)$ is $k\sec^2(k\theta)$

The derivative of $\sin(k\theta)$

Basic result for sine

The fundamental result for the sine function is:

If $f : \mathbb{R} \to \mathbb{R}$, where $f(\theta) = \sin \theta$, then f'(θ) = cos θ

This can be proved using first principles and relies on an important limit result.

Important limit result

This result can be understood geometrically using the unit circle. Consider a point $K$ on the unit circle where the angle $\angle KOH = \theta$.

By comparing areas, we can show that for $0 < \theta < \frac{\pi}{2}$:

$$\sin \theta \leq \theta \leq \tan \theta$$

Dividing through by $\sin \theta$ (which is positive in this range):

$$1 \leq \frac{\theta}{\sin \theta} \leq \frac{1}{\cos \theta}$$

As $\theta$ approaches $0$, $\cos \theta$ approaches $1$. This squeezes $\frac{\theta}{\sin \theta}$ towards $1$, which means $\frac{\sin \theta}{\theta}$ approaches $1$.

Proving the derivative of sine

Using the limit result above, we can prove that the derivative of $\sin \theta$ is $\cos \theta$ using first principles.

We also need the limit:

$$\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$$

The proof uses the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$ and considers the gradient of the secant between points $P(\theta, \sin \theta)$ and $Q(\theta + h, \sin(\theta + h))$.

General case using chain rule

For the more general function $f(\theta) = \sin(k\theta)$, we use the chain rule.

Let $y = \sin(k\theta)$ and let $u = k\theta$

Then $y = \sin u$, and:

$$\frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta} = \cos u \cdot k = k\cos(k\theta)$$

The derivative of $\cos(k\theta)$

Complementary angle relationships

To find the derivative of cosine, we use these complementary angle identities:

$$\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$$ $$\sin \theta = \cos\left(\frac{\pi}{2} - \theta\right)$$

Derivation

Let $y = \cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$

Let $u = \frac{\pi}{2} - \theta$

Then $y = \sin u$

Using the chain rule:

$$\frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta} = \cos u \cdot (-1) = -\cos\left(\frac{\pi}{2} - \theta\right) = -\sin \theta$$

General formula

The derivative of $\tan(k\theta)$

The secant function

For working with tangent derivatives, we introduce the secant function:

$$\sec \theta = \frac{1}{\cos \theta}$$

Derivation from first principles

The derivative of $\tan \theta$ can be found using first principles. Consider points $P(\theta, \tan \theta)$ and $Q(\theta + h, \tan(\theta + h))$ on the graph.

The gradient of the secant $PQ$ is:

$$\frac{\tan(\theta + h) - \tan \theta}{h}$$

After algebraic manipulation using $\tan \theta = \frac{\sin \theta}{\cos \theta}$, this simplifies to:

$$\frac{\sin h}{h \cos(\theta + h) \cos(\theta)}$$

As $h \to 0$:

• $\cos(\theta + h) \to \cos \theta$
• $\frac{\sin h}{h} \to 1$

Therefore:

$$f'(\theta) = \frac{1}{\cos^2 \theta} = \sec^2 \theta$$

General formula using chain rule

For $f(\theta) = \tan(k\theta)$, using the chain rule gives:

Worked examples

Solution:

a Let $y = \sin(2\theta)$

Using the derivative formula for sine:

$$\frac{dy}{d\theta} = 2\cos(2\theta)$$

b Let $y = \tan(3\theta)$

Using the derivative formula for tangent:

$$\frac{dy}{d\theta} = 3\sec^2(3\theta)$$

c Let $y = \sin^2(2\theta)$ and $u = \sin(2\theta)$

Then $y = u^2$

Using the chain rule:

$$\frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta}$$ $$= 2u \cdot 2\cos(2\theta)$$ $$= 4\sin(2\theta)\cos(2\theta)$$

d Let $y = \sin^2(2\theta + 1)$ and $u = \sin(2\theta + 1)$

Then $y = u^2$

Using the chain rule:

$$\frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta}$$ $$= 2u \cdot 2\cos(2\theta + 1)$$ $$= 4\sin(2\theta + 1)\cos(2\theta + 1)$$

e Let $y = \cos^3(4\theta + 1)$ and $u = \cos(4\theta + 1)$

Then $y = u^3$

Using the chain rule:

$$\frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta}$$ $$= 3u^2 \cdot (-4)\sin(4\theta + 1)$$ $$= -12\cos^2(4\theta + 1)\sin(4\theta + 1)$$

f Let $y = \tan(3\theta^2 + 1)$ and $u = 3\theta^2 + 1$

Then $y = \tan u$

Using the chain rule:

$$\frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta}$$ $$= \sec^2 u \cdot 6\theta$$ $$= 6\theta \sec^2(3\theta^2 + 1)$$

Solution:

a Let $y = \cos \theta$

Then:

$$\frac{dy}{d\theta} = -\sin \theta$$

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

$$y = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$ $$\frac{dy}{d\theta} = -\sin\left(\frac{\pi}{4}\right) = \frac{-1}{\sqrt{2}}$$

So the gradient at $\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)$ is $\frac{-1}{\sqrt{2}}$

When $\theta = \frac{\pi}{2}$:

$$y = \cos\left(\frac{\pi}{2}\right) = 0$$ $$\frac{dy}{d\theta} = -\sin\left(\frac{\pi}{2}\right) = -1$$

The gradient at $\left(\frac{\pi}{2}, 0\right)$ is $-1$

b Let $y = \tan \theta$

Then:

$$\frac{dy}{d\theta} = \sec^2 \theta$$

When $\theta = 0$:

$$y = \tan(0) = 0$$ $$\frac{dy}{d\theta} = \sec^2(0) = 1$$

The gradient at $(0, 0)$ is $1$

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

$$y = \tan\left(\frac{\pi}{4}\right) = 1$$ $$\frac{dy}{d\theta} = \sec^2\left(\frac{\pi}{4}\right) = 2$$

The gradient at $\left(\frac{\pi}{4}, 1\right)$ is $2$

Summary of formulas

Remember!