Learning Lab

Example 1

Given that \(\sin\theta=0.3,\) find all values of \(\theta\) in the domain \(\left\{ \theta:0^{\circ}\leq\theta\leq360^{\circ}\right\} .\)

Solution:

Draw a picture showing the possible positive values for \(\theta\) as shown below:

There are at least two solutions shown in red and blue. We have \(\theta_{r}\) (in red) which is given by \[\begin{align*} \theta_{r} & =\sin^{-1}\left(0.3\right)\\ & =17.46^{\circ}.\text{ (from calculator)} \end{align*}\] The other angle \(\theta_{b}\) (in blue) may be found by symmetry: \[\begin{align*} \theta_{b} & =180^{\circ}-\theta_{r}\\ & =180^{\circ}-17.46^{\circ}\\ & =162.54^{\circ}. \end{align*}\] Hence the solutions are \(\theta=17.46^{\circ},\,162.54^{\circ}.\)

Example 2

Given that \(\sin\theta=0.3,\) find all values of \(\theta\) such that \(0^{\circ}\leq\theta\leq500^{\circ}.\)

Solution:

Note that this is the same problem as in Example 1 except that the domain in which we are looking for solutions has been extended to \(500^{\circ}.\)

Draw a picture showing the possible positive values for \(\theta\) as shown below:

From Example 1 we know \(\theta=17.46^{\circ}\)or \(162.54^{\circ}.\) If we add \(360^{\circ}\) to each of these, they will still satisfy \(\sin\theta=0.3.\) The question is, are they in the domain of interest?

Adding \(360^{\circ}\) to the solution \(\theta=17.46^{\circ}\) found in Example 1 gives \[\begin{align*} \theta & =17.46^{\circ}+360^{\circ}\\ & =377.46^{\circ} \end{align*}\] which is less than \(500^{\circ}\) and so a solution to Example 2. Adding \(360^{\circ}\) to the other solution to Example \(1\) gives \[\begin{align*} \theta & =162.54^{\circ}+360^{\circ}\\ & =522.54^{\circ} \end{align*}\] which is bigger than \(500^{\circ}\) and so not a solution to Example 2.

Hence the solutions are \(\theta=17.46^{\circ},\,162.54^{\circ}\) and \(377.46^{\circ}.\)

Example 3

Solve \(\cos\alpha=0.5\) over the domain \(\left\{ \alpha:0\leq\alpha\leq2\pi\right\} .\)

Solution:

Draw a picture showing the possible positive values for \(\alpha\) as shown below:

Referring to the figure, \[\begin{align*} \alpha_{1} & =\cos^{-1}\left(0.5\right)\\ & =\frac{\pi}{3}\ (\text{from calculator)} \end{align*}\] and from symmetry of the unit circle, \[\begin{align*} \alpha_{2} & =2\pi-\alpha_{1}\\ & =2\pi-\frac{\pi}{3}\\ & =\frac{5\pi}{3}. \end{align*}\] Hence the solution is \(\alpha=\pi/3\) and \(5\pi/3.\)

Example 4

Solve \(\cos\alpha=0.5\) over the domain \(\left\{ \alpha:-2\pi\leq\alpha\leq2\pi\right\} .\)

Solution:

Note this is the same problem as in Example 3 but the domain has been extended.

Draw a picture showing the possible positive and negative values for \(\alpha\) as shown below:

The two positive angles (shown in red) are the same as those in Example 3, namely \[\begin{align*} \alpha_{1} & =\frac{\pi}{3}\\ \alpha_{2} & =\frac{5\pi}{3}. \end{align*}\] The negative angles (shown in blue) are \(\alpha_{3}\) and \(\alpha_{4}.\) Using the results from Example 3 and symmetry we have \[\begin{align*} \alpha_{3} & =-\alpha_{1}\\ & =-\frac{\pi}{3} \end{align*}\] and \[\begin{align*} \alpha_{4} & =-\left(2\pi-\alpha_{1}\right)\\ & =\alpha_{1}-2\pi\\ & =\frac{\pi}{3}-2\pi\\ & =-\frac{5\pi}{3}. \end{align*}\] Hence the solution is \(\alpha=-\frac{5\pi}{3},\,-\frac{\pi}{3},\,\frac{\pi}{3},\,\frac{5\pi}{3}.\)

Example 5

Solve \(\tan\left(2x\right)=-3\) over the domain \(\left\{ x:-90^{\circ}\leq x\leq180^{\circ}\right\} .\)

Solution:

Since the variable is \(2x,\) we have to redefine the domain to \(\left\{ 2x:-180^{\circ}\leq2x\leq360^{\circ}\right\} .\)

Draw a picture showing the possible positive and negative values for \(2x\) as shown below:

There are two positive angles \(\alpha_{1},\,\alpha_{2}\) (shown in red) and two negative angles \(\alpha_{3,}\,\alpha_{4}\) (shown in blue) that may lie within \(-180^{\circ}\leq2x\leq360^{\circ}.\)

We have \[\begin{align*} \tan\left(2x\right) & =-3\\ 2x & =\tan^{-1}\left(-3\right)\\ & =-71.57^{\circ}\ \text{(from calculator).} \end{align*}\] Referring to the figure, we see that \[\begin{align*} \alpha_{3} & =-71.57^{\circ} \end{align*}\] and \[\begin{align*} \alpha_{4} & =\alpha_{3}-180^{\circ}\\ & =-71.57^{\circ}-180^{\circ}\\ & =-251.57^{\circ}. \end{align*}\] Note that this is outside of the domain for \(2x\) and so is rejected. That is we discard \(\alpha_{4}.\)

Using the symmetry of the unit circle, we have \[\begin{align*} \alpha_{1} & =360^{\circ}+\alpha_{4}\\ & =360^{\circ}-251.57^{\circ}\\ & =108.43^{\circ} \end{align*}\] and using the figure, \[\begin{align*} \alpha_{2} & =\alpha_{1}+180^{\circ}\\ & =108.43^{\circ}+180^{\circ}\\ & =288.43^{\circ}. \end{align*}\] Hence \[\begin{align*} 2x & =\alpha_{3},\ \alpha_{1},\ \alpha_{2}\\ & =-71.57^{\circ},\ 108.43^{\circ},\ 288.43^{\circ} \end{align*}\] and the solution is \[\begin{align*} x & =-35.79^{\circ},\ 54.22^{\circ},\ 144.22^{\circ}. \end{align*}\]

Exercises

Solve the following equations:

\(\text{a) If $\sin\phi=0.25$ find $\phi$ for $0^{\circ}\leq\phi\leq180^{\circ}$ .}\)

\(\text{b) If $\tan\phi=0.8$ find }\phi\) for \(0^{\circ}\leq\phi\leq360^{\circ}\).

\(\text{c) If $\cos\phi=0.4$ find $\phi\ \text{for $0^{\circ}\leq\phi\leq360^{\circ}.$ }$ }\)

\(\text{d) If $\cos\phi=-0.4\ \text{find $\phi\text{ for $-180^{\circ}\leq\phi\leq360^{\circ}.$ }$ }$ }\)

\(\text{e) If $\tan\phi=-1.5\ \text{find $\phi\ \text{for $-180^{\circ}\leq\phi\leq360^{\circ}.$ }$ }$ }\)

\(\text{f) If $\cos\phi=-0.3\ \text{find $\phi$ for $0^{\circ}\leq$ $\phi\leq360^{\circ}.$ }$ }\)