Honors Pre-Calculus Unit 8 Review: Further Applications of Trigonometry

The Law of Sines says that in any triangle, each side divided by the sine of its opposite angle gives the same value, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. To use it, you need one complete pair, meaning an angle and the side directly across from it.

That pair shows up in AAS and ASA (two angles and a side) and in SSA (two sides and an angle opposite one of them). Three angles alone (AAA) don't determine the side lengths, because similar triangles share angles but come in every size.

SSA is called the ambiguous case. Since $\sin B = \sin(180° - B)$, an acute angle and an obtuse angle share the same sine, so you may get zero, one or two triangles.

Worked example. Suppose $A = 30°$, $a = 6$ and $b = 10$. Then $\sin B = \frac{10\sin 30°}{6} = \frac{5}{6} \approx 0.833$. The calculator gives $B_1 \approx 56.4°$, and the other option is $B_2 = 180° - 56.4° = 123.6°$. Both work, because $30° + 56.4° < 180°$ and $30° + 123.6° = 153.6° < 180°$. So two different triangles fit this information. If $\sin B$ had come out greater than 1, no triangle would exist.

For area, use two sides and the angle between them, $\text{Area} = \frac{1}{2}ab\sin C$. With $a = 8$, $b = 5$ and $C = 30°$, the area is $\frac{1}{2}(8)(5)(0.5) = 10$ square units.

The Law of Cosines connects all three sides to one angle, $c^2 = a^2 + b^2 - 2ab\cos C$. The side on the left is always across from the angle inside the cosine. If $C = 90°$, then $\cos C = 0$ and you get the Pythagorean theorem back.

Use it for SAS (two sides and the included angle) to find the third side, or for SSS (three sides) to find an angle. To find an angle, rearrange to $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and take the inverse cosine.

Worked example. In triangle $ABC$, $b = 7$, $c = 10$ and $A = 48°$. Then $a^2 = 7^2 + 10^2 - 2(7)(10)\cos 48° = 149 - 140(0.6691) \approx 55.32$, so $a \approx 7.44$.

After one Law of Cosines step, you can switch to the Law of Sines for the next angle. Solve for the smallest angle (across from the shortest side) first, since it must be acute and the inverse sine will give the correct value.

Heron's formula gives area from three sides. Let $s = \frac{a+b+c}{2}$ be the semi-perimeter, then $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$. For sides 5, 7 and 10, $s = 11$ and the area is $\sqrt{11 \cdot 6 \cdot 4 \cdot 1} = \sqrt{264} \approx 16.25$.

Polar coordinates locate a point using a signed radius $r$ and an angle $\theta$ from the positive x-axis. The origin is called the pole, and the positive x-axis is called the polar axis. The point's distance from the pole is $|r|$. Angles are usually in radians.

To go from polar to rectangular, use $x = r\cos\theta$ and $y = r\sin\theta$. To go back, use $r = \sqrt{x^2 + y^2}$ and $\tan\theta = \frac{y}{x}$. The inverse tangent returns a principal angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. If $x < 0$, add $\pi$ to place the angle in the correct quadrant. If $x = 0$ and $y \ne 0$, use $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ according to the sign of $y$. At the pole, $r = 0$ and any angle names the same point.

One point has many polar names. $(3, \frac{\pi}{4})$, $(3, \frac{9\pi}{4})$ and $(-3, \frac{5\pi}{4})$ all describe the same location, because a negative $r$ means you go in the opposite direction.

Worked example. Convert $(-3, 3)$ to polar. First, $r = \sqrt{9 + 9} = 3\sqrt{2}$. Next, $\tan^{-1}(\frac{3}{-3}) = -\frac{\pi}{4}$, which points into Quadrant IV. The point is in Quadrant II, so add $\pi$ to get $\theta = \frac{3\pi}{4}$. The answer is $(3\sqrt{2}, \frac{3\pi}{4})$.

You can also convert whole equations. For $r = 4\cos\theta$, multiply both sides by $r$ to get $r^2 = 4r\cos\theta$, which becomes $x^2 + y^2 = 4x$. Completing the square gives $(x - 2)^2 + y^2 = 4$, a circle centered at $(2, 0)$ with radius 2.

To graph $r = f(\theta)$ by hand, make a table of $\theta$ values (multiples of $\frac{\pi}{6}$ or $\frac{\pi}{4}$ work well), compute $r$ and plot each point. When $r$ is negative, plot $|r|$ in the opposite direction, which is the same as adding $\pi$ to the angle.

Symmetry tests cut the work in half. Replacing $\theta$ with $-\theta$ tests symmetry about the polar axis, so cosine graphs like $r = 2\cos\theta$ pass. Replacing $\theta$ with $\pi - \theta$ tests symmetry about the vertical line $\theta = \frac{\pi}{2}$, so sine graphs like $r = 2\sin\theta$ pass. Replacing $r$ with $-r$ tests symmetry about the pole. Passing a test proves the symmetry, but failing one doesn't rule it out, because each point has more than one polar name.

Most questions use a small set of curve families, listed in the table. For limaçons $r = a + b\cos\theta$ with $a > 0$ and $b > 0$, the ratio $\frac{a}{b}$ decides the shape. For example, $r = 1 + 2\cos\theta$ has $\frac{a}{b} = \frac{1}{2}$, so it has an inner loop. That loop appears because $r$ goes negative for some angles.

For a positive integer $n$ and nonzero $a$, rose curves $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$ have $n$ petals when $n$ is odd and $2n$ petals when $n$ is even. Each petal has length $|a|$. Odd roses finish tracing by $\theta = \pi$, so plotting past that just retraces them.

On the complex plane, $a + bi$ sits at the point $(a, b)$, with real numbers on the horizontal axis and imaginary numbers on the vertical axis. The modulus $|z| = \sqrt{a^2 + b^2}$ is its distance from the origin. For a nonzero complex number, an argument $\theta$ is its angle from the positive real axis. Zero has no defined argument.

Polar form writes the number as $z = r(\cos\theta + i\sin\theta)$, often shortened to $r\,\text{cis}\,\theta$. You convert exactly as you do with polar points, including the quadrant check on $\theta$.

Polar form makes multiplication simple. You multiply the moduli and add the arguments, so $r_1\,\text{cis}\,\theta_1 \cdot r_2\,\text{cis}\,\theta_2 = r_1 r_2\,\text{cis}(\theta_1 + \theta_2)$. To divide by a nonzero complex number, divide the moduli and subtract the arguments. Geometrically, multiplying by a complex number scales and rotates.

Repeating that rule gives De Moivre's Theorem, $z^n = r^n\,\text{cis}(n\theta)$.

Worked example. Find $(1 + i)^6$. The modulus is $\sqrt{2}$ and the argument is $\frac{\pi}{4}$. Then $(1+i)^6 = (\sqrt{2})^6\,\text{cis}(\frac{6\pi}{4}) = 8\,\text{cis}(\frac{3\pi}{2}) = 8(0 + i(-1)) = -8i$.

Roots run the theorem backward. For a positive integer $n$ and $r > 0$, the $n$th roots of $r\,\text{cis}\,\theta$ are $\sqrt[n]{r}\,\text{cis}(\frac{\theta + 2k\pi}{n})$ for $k = 0, 1, \ldots, n-1$. They sit evenly spaced on a circle. If $r = 0$, the only root is 0. The cube roots of 8 are $2\,\text{cis}\,0$, $2\,\text{cis}\frac{2\pi}{3}$ and $2\,\text{cis}\frac{4\pi}{3}$, which are 120° apart.

Parametric equations give $x$ and $y$ separately as functions of a third variable, the parameter $t$, so $x = x(t)$ and $y = y(t)$. This lets you describe curves that fail the vertical line test and model where something is at each moment.

To eliminate the parameter, solve the simpler equation for $t$ and substitute. If $x = t + 1$ and $y = t^2$, then $t = x - 1$ and $y = (x - 1)^2$. For trig equations, use an identity instead. With $x = \cos t$ and $y = \sin t$, squaring and adding gives $x^2 + y^2 = 1$.

Going the other way has no single answer. The parabola $y = x^2$ can be written as $x = t$, $y = t^2$ or as $x = 2t$, $y = 4t^2$.

Projectile motion is the classic model. If an object is launched from the origin at speed $v_0$ and angle $\theta$, and air resistance is ignored, its position is $x(t) = v_0\cos\theta \cdot t$ and $y(t) = v_0\sin\theta \cdot t - \frac{1}{2}gt^2$, with $g \approx 9.8$ m/s².

Worked example. A ball is launched from the ground at 20 m/s and 30°. Then $x = 20\cos 30° \cdot t \approx 17.32t$ and $y = 20\sin 30° \cdot t - 4.9t^2 = 10t - 4.9t^2$. It lands when $y = 0$, so $t(10 - 4.9t) = 0$ and $t \approx 2.04$ seconds. The horizontal distance is $x \approx 17.32(2.04) \approx 35.3$ meters.

To graph a parametric curve, pick values of $t$, compute $x$ and $y$ for each, and plot the points in order. Draw arrows to show the orientation, which is the direction the curve moves as $t$ increases.

For $x = t + 1$ and $y = t^2$ with $-2 \leq t \leq 2$, the points are $(-1, 4)$, $(0, 1)$, $(1, 0)$, $(2, 1)$ and $(3, 4)$. That gives an upward parabola traced left to right.

Circles and ellipses come from cosine and sine. For $r > 0$, the equations $x = h + r\cos t$ and $y = k + r\sin t$ trace a circle centered at $(h, k)$ counterclockwise as $t$ increases. At $t = 0$, the point is at the right edge. For $a > 0$ and $b > 0$, the equations $x = a\cos t$ and $y = b\sin t$ give an ellipse with horizontal semi-axis $a$ and vertical semi-axis $b$.

The interval for $t$ controls how much of the curve appears. For $x = \cos t$ and $y = \sin t$, the interval $0 \leq t \leq 2\pi$ traces the full circle once, $0 \leq t \leq \pi$ traces only the upper half, and $0 \leq t \leq 4\pi$ goes around twice.

This is why the rectangular equation can mislead you. The equations $x = \sin t$ and $y = \sin^2 t$ eliminate to $y = x^2$, but since $\sin t$ stays between $-1$ and $1$, the graph is only the piece of the parabola from $x = -1$ to $x = 1$, traced back and forth.

A vector has both magnitude (size) and direction. A scalar is a number that can be positive, negative or zero and has no direction. You can write it in component form $\langle a, b \rangle$ or as $a\mathbf{i} + b\mathbf{j}$. Vectors with the same length and direction are equal no matter where they start, so the vector from $(1, 1)$ to $(4, 5)$ equals $\langle 3, 4 \rangle$.

The magnitude is $|\vec{v}| = \sqrt{a^2 + b^2}$. For a nonzero vector with $a \ne 0$, the direction angle satisfies $\tan\theta = \frac{b}{a}$, with the same quadrant check as polar coordinates. If $a = 0$, the direction is $\frac{\pi}{2}$ when $b > 0$ or $\frac{3\pi}{2}$ when $b < 0$. A vector of length $r$ at angle $\theta$ has components $\langle r\cos\theta, r\sin\theta \rangle$. Dividing a nonzero vector by its magnitude gives a unit vector, which has length 1. The zero vector has no direction and cannot be divided by its magnitude.

To add or subtract vectors, combine matching components. Geometrically, addition places the vectors tip to tail. Multiplying by a scalar $c$ scales the length, and a negative $c$ also flips the direction.

The dot product $\vec{u} \cdot \vec{v} = a_1a_2 + b_1b_2$ gives a number, not a vector. For nonzero vectors, it also equals $|\vec{u}||\vec{v}|\cos\theta$, so a dot product of 0 means their directions are perpendicular. A zero vector also gives a dot product of 0, but it has no direction.

Worked example. Find the angle between $\vec{u} = \langle 2, 3 \rangle$ and $\vec{v} = \langle 4, -1 \rangle$. The dot product is $8 - 3 = 5$. The magnitudes are $\sqrt{13}$ and $\sqrt{17}$. So $\cos\theta = \frac{5}{\sqrt{221}} \approx 0.336$, which gives $\theta \approx 70.3°$.