Solving Simultaneous Equations

Video lesson · from 52:431 worked example

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

You are already familiar with some wave functions, namely $f(x)=\sin x$ and $f(x)=\cos x$ .

We can add these functions together, if they have the same period, to obtain $f(x)=\sin x+\cos x$ .

This is also a wave function. You can see the amplitude has changed here and there is a phase shift.

We could also express this graph in the following forms:

We are required to solve simultaneous equations of the form:

\begin{aligned} k \sin\alpha &= a \\ k \cos\alpha &= b \end{aligned}

Solving for $k$ :

\begin{aligned} (k \sin\alpha)^2 &= a^2 \\ k^2 \sin^2\alpha &= a^2 \end{aligned}

k^2 \sin^2\alpha + k^2 \cos^2\alpha = a^2 + b^2

k^2 (\sin^2\alpha + \cos^2\alpha) = a^2 + b^2

k^2 \times 1 = a^2 + b^2

k^2 = a^2 + b^2

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

Divide to eliminate $k$ : dividing one equation by the other cancels $k$ and leaves a tangent, which is what gives you the angle.
Square and add for $k$ : using $\sin^2 + \cos^2 = 1$ gives $k^2$ directly. Always take the positive root.
The signs fix the quadrant: the angle is decided by the signs of the two right-hand sides, not by the calculator value of the inverse tangent.
Keep the working consistent: set the two equations out clearly and label them, so each later line plainly follows from them.

Example 1

Solve the following for $k > 0$ and $0 < \alpha < 360^\circ$ :
$k \sin\alpha = \sqrt{3}$
$k \cos\alpha = 1$

1. Find $k$ :

\begin{aligned} k &= \sqrt{(\sqrt{3})^2 + 1^2} \\ &= \sqrt{3 + 1} \\ &= \sqrt{4} \\ &= 2 \end{aligned}

2. Find $\alpha$ by dividing $k \sin\alpha$ by $k \cos\alpha$ :

\begin{aligned} \frac{k\sin\alpha}{k\cos\alpha} &= \frac{\sqrt{3}}{1} \\ \tan\alpha &= \sqrt{3} \end{aligned}

Base angle $\alpha = \tan^{-1}(\sqrt{3}) = 60^\circ$ .

Check quadrants: $\sin\alpha$ is positive (+), $\cos\alpha$ is positive (+). They are both positive in Quadrant 1, so $\alpha = 60^\circ$ .

Solution: $k = 2, \alpha = 60^\circ$