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Wave Function0%

Trigonometry · Topic 15 of 19

Wave Function

Video lesson · from 52:433 worked examples

One lesson video covers all of Trigonometry, so it opens at 52:43 for this topic — not from the beginning.

Theory

We can combine everything learned so far to rewrite trigonometric sums as a single wave function.

The point is this: an expression like $a\sin x^\circ + b\cos x^\circ$ is the sum of two waves, which is awkward to work with. It can always be rewritten as a single wave — and once it is, the maximum, the minimum and the solutions of equations can all be read off almost immediately.

There are four possible target forms, and the question will always tell you which one to use:

k\sin(x + \alpha)^\circ \qquad k\sin(x - \alpha)^\circ \qquad k\cos(x + \alpha)^\circ \qquad k\cos(x - \alpha)^\circ

The method is the same every time: expand the target form using the appropriate addition formula, then equate coefficients with the expression you were given. Taking $k\cos(x - \alpha)^\circ$ as the example:

k\cos(x - \alpha)^\circ = k\cos x^\circ\cos\alpha^\circ + k\sin x^\circ\sin\alpha^\circ

Comparing that with $a\cos x^\circ + b\sin x^\circ$ gives two equations:

k\cos\alpha^\circ = a \qquad\text{and}\qquad k\sin\alpha^\circ = b

Finding $k$ : square both equations and add them. Because $\cos^2\alpha^\circ + \sin^2\alpha^\circ = 1$ , this leaves

k^2 = a^2 + b^2 \quad\Longrightarrow\quad k = \sqrt{a^2 + b^2}

Always take the positive root — $k$ is an amplitude.

Finding $\alpha$ : divide one equation by the other, so that $k$ cancels:

\frac{k\sin\alpha^\circ}{k\cos\alpha^\circ} = \tan\alpha^\circ = \frac{b}{a}

On its own $\tan^{-1}$ cannot tell you which angle you want — it only ever returns one of two possibilities. The quadrant is decided by the signs of $k\sin\alpha^\circ$ and $k\cos\alpha^\circ$ ; since $k$ is positive, those are simply the signs of $b$ and $a$ . Both positive puts $\alpha$ in the first quadrant, and so on round the CAST diagram.

The Golden Rule: find $k$ by squaring and adding, find $\alpha$ by dividing — then let the signs of the two equations, not your calculator, decide the quadrant.

⚠️ Common Examiner Traps

The quadrant is the marks: $\tan^{-1}$ returns only one of two possible angles. Read the signs of $k\cos\alpha$ and $k\sin\alpha$ and use CAST — taking the calculator value on trust is the most common error in the topic.
Match the form you were asked for: $k\cos(x-\alpha)$ and $k\sin(x+\alpha)$ give different values of $\alpha$ . Expand the form the question specifies, not the one you find easiest.
Equate coefficients, and show it: the marks are for lining up the $\cos x$ and $\sin x$ terms. Write that comparison down as its own line.
Take the positive root for $k$ : it is an amplitude, so it is never negative.
Watch the units: if the question is set in radians, $\alpha$ must be in radians too — mixing them within a line of working is a frequent and expensive error.

Worked examples

Example 1

Write $5\cos x^\circ + 12\sin x^\circ$ in the form $k\cos(x - \alpha)^\circ$ where $k > 0$ and $0 \leq \alpha^\circ \leq 360$ .

Expand $k\cos(x - \alpha)$ and equate coefficients:

k\cos(x - \alpha) = k\cos x^\circ\cos\alpha^\circ + k\sin x^\circ\sin\alpha^\circ

\begin{aligned} k\cos\alpha &= 5 \\ k\sin\alpha &= 12 \end{aligned}

Find $k$ :

k = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13

Find $\alpha$ :

\tan\alpha = \frac{k\sin\alpha}{k\cos\alpha} = \frac{12}{5}

Base angle: $\tan^{-1}\left(\frac{12}{5}\right) \approx 67.4^\circ$ .

Since $\sin\alpha > 0$ and $\cos\alpha > 0$ , it's in the first quadrant, so $\alpha = 67.4^\circ$ .

Final wave function:

13\cos(x - 67.4)^\circ

Example 2

Write $5\cos x - 3\sin x$ in the form $k\cos(x - \alpha)$ where $k > 0$ and $0 \leq \alpha \leq 2\pi$ .

Expand $k\cos(x - \alpha)$ and equate coefficients:

k\cos(x - \alpha) = k\cos x\cos\alpha + k\sin x\sin\alpha

\begin{aligned} k\cos\alpha &= 5 \\ k\sin\alpha &= -3 \end{aligned}

Find $k$ :

k = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}

Find $\alpha$ :

\tan\alpha = \frac{-3}{5}

Base angle: $\tan^{-1}\left(\frac{3}{5}\right) \approx 0.540$ rad.

$\cos\alpha > 0$ (Q1, Q4), $\sin\alpha < 0$ (Q3, Q4). Both are true in Q4.

\alpha = 2\pi - 0.540 = 5.743 \text{ rad}

Final wave function:

\sqrt{34}\cos(x - 5.743)

Example 3

Write $4\cos x^\circ + 3\sin x^\circ$ in the form $k\sin(x + \alpha)^\circ$ where $k > 0$ and $0 \leq \alpha^\circ \leq 360$ .

Expand $k\sin(x + \alpha)$ and equate coefficients:

k\sin(x + \alpha)^\circ = k\sin x^\circ\cos\alpha^\circ + k\cos x^\circ\sin\alpha^\circ

Rearrange the given expression to match $\sin$ first: $3\sin x^\circ + 4\cos x^\circ$

\begin{aligned} k\cos\alpha &= 3 \\ k\sin\alpha &= 4 \end{aligned}

Find $k$ :

k = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Find $\alpha$ :

\tan\alpha = \frac{4}{3}

Base angle: $\tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ$ . Q1.

Final wave function:

5\sin(x + 53.1)^\circ