Trigonometry

1.5 Trigonometry

The three graphs

The transformations of Functions apply here:

• $y = a\sin x$ has amplitude $a$.
• $y = \sin bx$ has period $\dfrac{360^\circ}{b}$, so there are $b$ complete waves in 360°.
• $y = \sin x + c$ moves the wave up by $c$.
• $y = \sin(x + d)$ moves the wave left by $d$.

$\cos x$ runs from $-1$ to $1$, so the expression runs from $3 - 2 = 1$ to $3 + 2 = 5$. The range is $1 \leq y \leq 5$.

Exact values

Related angles have the same size of value with a sign that depends on the quadrant. This diagram shows which ratios are positive in each quadrant:

So $\sin 120^\circ = \sin 60^\circ = \dfrac{\sqrt{3}}{2}$, and $\cos 120^\circ = -\cos 60^\circ = -\dfrac{1}{2}$.

Inverse notation and principal values

$\sin^{-1}x$, $\cos^{-1}x$ and $\tan^{-1}x$ give one angle: the principal value, which is what the calculator shows.

The two identities

The second can be rearranged as $\sin^2\theta \equiv 1 - \cos^2\theta$ and $\cos^2\theta \equiv 1 - \sin^2\theta$.

$\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$. Cosine is negative for an obtuse angle, so $\cos\theta = -\dfrac{4}{5}$ and $\tan\theta = -\dfrac{3}{4}$.

$\text{LHS} = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} = \text{RHS}$

Solving trigonometric equations

There are usually several solutions in an interval. Find the principal value, then use the symmetry of the graph:

$x = 30^\circ$ and $x = 180^\circ - 30^\circ = 150^\circ$.

When the angle is changed

Change the interval first, solve for the whole bracket, and only then find $x$.

The interval becomes $0^\circ \leq 2x \leq 360^\circ$. The solutions are $2x = 53.13^\circ$ and $2x = 306.87^\circ$, so $x = 26.6^\circ$ or $x = 153.4^\circ$.

The interval becomes $-30^\circ \leq x - 30^\circ \leq 330^\circ$. So $x - 30^\circ = 45^\circ$ or $225^\circ$, giving $x = 75^\circ$ or $x = 255^\circ$.

Using the identities in equations

Use $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ when an equation has both sine and cosine to the same power.

Divide by $\cos x$: $\tan x = \dfrac{4}{3}$, so $x = 53.1^\circ$ or $x = 233.1^\circ$.

Use $\sin^2\theta + \cos^2\theta = 1$ when squares appear, to get a quadratic in one ratio.

$2(1 - \cos^2 x) + 3\cos x = 0 \quad\Rightarrow\quad 2\cos^2 x - 3\cos x - 2 = 0 \quad\Rightarrow\quad (2\cos x + 1)(\cos x - 2) = 0$

$\cos x = 2$ is impossible, so $\cos x = -\dfrac{1}{2}$ and $x = 120^\circ$ or $x = 240^\circ$.