Honors Algebra II Unit 12 Review: Trigonometric Equations & Applications

These two laws let you solve oblique triangles, meaning triangles with no right angle. Label the angles $A$, $B$, $C$ and the sides across from them $a$, $b$, $c$.

The Law of Sines says $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Use it when you know an angle and its opposite side, which happens with AAS, ASA, and SSA.

The Law of Cosines says $c^2 = a^2 + b^2 - 2ab\cos C$. Use it for SAS to find the third side and for SSS to find an angle. If $C = 90^\circ$, then $\cos C = 0$ and the formula turns back into the Pythagorean Theorem.

SSA is the ambiguous case. Suppose you're given an acute angle $A$ and sides $a$ and $b$. Compute the height $h = b\sin A$. If $a < h$, no triangle exists. If $a = h$, you get one right triangle. If $h < a < b$, two triangles work. If $a \ge b$, only one does.

Worked example. A triangle has $a = 5$, $b = 7$, $c = 8$. Find $B$. Rearranging the Law of Cosines gives $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{25 + 64 - 49}{2(5)(8)} = \frac{40}{80} = 0.5$, so $B = 60^\circ$. Now you have an angle-side pair, so switch to the Law of Sines. $\sin A = \frac{5\sin 60^\circ}{7} \approx 0.619$, which gives $A \approx 38.2^\circ$ and then $C \approx 180^\circ - 60^\circ - 38.2^\circ = 81.8^\circ$.

Polar coordinates $(r, \theta)$ locate a point using a radius and an angle from the positive $x$-axis. When $r$ is negative, the point lies in the direction opposite $\theta$. To go from polar to rectangular, use $x = r\cos\theta$ and $y = r\sin\theta$. To go from rectangular to polar with a nonnegative radius, use $r = \sqrt{x^2 + y^2}$. When $x \ne 0$, use $\tan\theta = \frac{y}{x}$ and check the quadrant. If $x < 0$, add $\pi$ to the calculator's $\tan^{-1}(y/x)$ result. When $x = 0$ and $y \ne 0$, use $\theta = \frac{\pi}{2}$ for $y > 0$ or $\theta = \frac{3\pi}{2}$ for $y < 0$. At the origin, any angle works.

Some curves are simpler in polar form. The circle $x^2 + y^2 = 9$ is just $r = 3$. If $a \ne 0$ and $n$ is a positive integer, $r = a\cos(n\theta)$ is a rose with $n$ petals when $n$ is odd and $2n$ petals when $n$ is even.

A complex number $a + bi$ plots as the point $(a, b)$, so it has a polar form too, $z = r(\cos\theta + i\sin\theta)$. To multiply two complex numbers in this form, you multiply their moduli and add their arguments. Repeating that gives De Moivre's Theorem, $z^n = r^n(\cos n\theta + i\sin n\theta)$ for positive integers $n$. The formula also works for other integers when $z \ne 0$.

Worked example. Write $z = -1 + \sqrt{3}\,i$ in trig form and find $z^3$. First, $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2$. The calculator gives $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$, but $a = -1 < 0$, so $\theta = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}$. That makes $z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$. By De Moivre, $z^3 = 2^3(\cos 2\pi + i\sin 2\pi) = 8(1 + 0i) = 8$.

To solve a trig equation, treat the trig function like a variable. Isolate it, find a reference angle, then use the unit circle to find every angle in the interval with that value.

For $\sin x = k$ or $\cos x = k$ with $-1 < k < 1$, there are two solutions per $2\pi$ cycle. If $|k| > 1$, there are none, since sine and cosine never leave $[-1, 1]$. For $\tan x = k$, there's one solution per $\pi$ cycle. To write every solution, add $2\pi n$ (sine, cosine) or $\pi n$ (tangent), where $n$ is any integer.

If the equation looks like a quadratic, factor it. If it mixes functions, use an identity such as $\sin^2 x + \cos^2 x = 1$ or $\sin 2x = 2\sin x\cos x$ to rewrite it in terms of one function. If you square both sides, plug each answer back in, because squaring can create extraneous solutions.

Worked example. Solve $2\sin^2 x - \sin x - 1 = 0$ on $[0, 2\pi)$. Let $u = \sin x$, so $2u^2 - u - 1 = 0$ factors as $(2u + 1)(u - 1) = 0$. If $\sin x = -\frac{1}{2}$, the reference angle is $\frac{\pi}{6}$ and sine is negative in Quadrants III and IV, giving $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$. If $\sin x = 1$, then $x = \frac{\pi}{2}$. That makes three solutions, $\frac{\pi}{2}$, $\frac{7\pi}{6}$, and $\frac{11\pi}{6}$.