Trigonometry

Right-triangle trig, laws of sines and cosines, the unit circle, identities and graphs.

Worked Examples

<details> <summary><b>Example 1: Find both legs from the hypotenuse and an angle</b></summary>

Question: In right triangle PQR, the right angle is at R, hypotenuse PQ = 18, and angle P measures 28°. Find QR and PR.

Solution:

1. Stand at angle P. QR does not touch P, so it is opposite. PR touches P, so it is adjacent.
2. Opposite with hypotenuse → sine: $\sin 28^\circ = \frac{QR}{18}$, so $QR = 18\sin 28^\circ \approx 18(0.4695) \approx 8.45$.
3. Adjacent with hypotenuse → cosine: $PR = 18\cos 28^\circ \approx 18(0.8829) \approx 15.89$.
4. Check: both legs are shorter than 18, and $8.45^2 + 15.89^2 \approx 324 = 18^2$. ✓

</details> <details> <summary><b>Example 2: Find an angle with inverse trig</b></summary>

Question: A wheelchair ramp rises 3 feet over a horizontal distance of 10 feet. What angle does the ramp make with the ground?

Solution:

1. At the ground angle, the 3-foot rise is opposite and the 10-foot run is adjacent. No hypotenuse is involved → tangent.
2. $\tan\theta = \frac{3}{10} = 0.3$, so $\theta = \tan^{-1}(0.3) \approx 16.7^\circ$.
3. If the question asks for an expression, the answer is simply $\tan^{-1}\left(\frac{3}{10}\right)$. ✓

</details> <details> <summary><b>Example 3: Build the triangle from one ratio, then switch angles</b></summary>

Question: In right triangle XYZ, the right angle is at Y and $\tan X = \frac{5}{12}$. Find $\sin X$ and $\cos Z$.

Solution:

1. Opposite X is YZ = 5k; adjacent to X is XY = 12k. Hypotenuse XZ = 13k (a 5-12-13 triple).
2. $\sin X = \frac{5}{13}$.
3. Now stand at Z: the adjacent leg is YZ = 5k, so $\cos Z = \frac{5}{13}$, the same as $\sin X$, exactly as the complementary-angle rule predicts. ✓

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Worked Examples

<details> <summary><b>Example 1: Angle of depression</b></summary>

Question: From the top of a 120-foot cliff, the angle of depression to a boat is 18°. How far is the boat from the base of the cliff?

Solution:

1. Move the 18° angle down to the boat (alternate interior angles).
2. At the boat, the 120-foot cliff is opposite and the distance $d$ is adjacent → tangent.
3. $\tan 18^\circ = \frac{120}{d}$, so $d = \frac{120}{\tan 18^\circ} \approx \frac{120}{0.3249} \approx 369$ feet. ✓

</details> <details> <summary><b>Example 2: Law of Sines (AAS)</b></summary>

Question: In triangle ABC, angle A = 50°, angle B = 65°, and side a = 10. Find side b.

Solution:

1. You have a matched pair (A with a) → Law of Sines.
2. $\frac{10}{\sin 50^\circ} = \frac{b}{\sin 65^\circ}$, so $b = \frac{10\sin 65^\circ}{\sin 50^\circ} \approx \frac{10(0.9063)}{0.7660} \approx 11.8$.
3. Check: the larger angle (65°) is opposite the longer side (11.8 > 10). ✓

</details> <details> <summary><b>Example 3: Law of Cosines (SAS and SSS)</b></summary>

SAS: Two sides are 6 and 10 with an included angle of 120°. Find the third side.

$c^2 = 36 + 100 - 2(6)(10)\cos 120^\circ = 136 - 120\left(-\tfrac{1}{2}\right) = 196, \quad c = 14$

Because the angle is obtuse, the cosine is negative and the third side comes out longer than it would in a right triangle.

SSS: A triangle has sides 3, 5, and 7. Find its largest angle (opposite 7).

$49 = 9 + 25 - 2(3)(5)\cos C \implies 15 = -30\cos C \implies \cos C = -\tfrac{1}{2} \implies C = 120^\circ$ ✓

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Worked Examples

<details> <summary><b>Example 1: Convert in both directions</b></summary>

• $240^\circ \cdot \frac{\pi}{180} = \frac{240\pi}{180} = \frac{4\pi}{3}$
• $\frac{5\pi}{6} \cdot \frac{180}{\pi} = 150^\circ$
• $2 \text{ radians} \cdot \frac{180}{\pi} \approx 114.6^\circ$ (no π, so the answer is not a "nice" angle) ✓

</details> <details> <summary><b>Example 2: Evaluate with quadrant + reference angle + sign</b></summary>

• $\sin 315^\circ$: Quadrant IV, reference 45°, sine negative → $-\frac{\sqrt{2}}{2}$.
• $\cos\frac{2\pi}{3}$: that is 120°, Quadrant II, reference 60°, cosine negative → $-\frac{1}{2}$.
• $\tan 210^\circ$: Quadrant III, reference 30°, tangent positive → $\frac{\sqrt{3}}{3}$. ✓

</details> <details> <summary><b>Example 3: One value plus a quadrant gives the rest</b></summary>

Question: $\sin\theta = -\frac{5}{13}$ and θ is in Quadrant III. Find $\cos\theta$ and $\tan\theta$.

Solution:

1. Reference triangle: opposite 5, hypotenuse 13, so the other leg is 12.
2. In Quadrant III, x is negative: $\cos\theta = -\frac{12}{13}$.
3. Tangent is positive in Quadrant III: $\tan\theta = \frac{-5/13}{-12/13} = \frac{5}{12}$. ✓

</details> <details> <summary><b>Example 4: A point off the unit circle</b></summary>

Question: The terminal side of θ passes through $(-6, 8)$. Find sin θ, cos θ, and tan θ.

Solution: $r = \sqrt{36 + 64} = 10$. So $\sin\theta = \frac{8}{10} = \frac{4}{5}$, $\cos\theta = -\frac{6}{10} = -\frac{3}{5}$, $\tan\theta = \frac{8}{-6} = -\frac{4}{3}$. The point is in Quadrant II, and only sine is positive, as ASTC predicts. ✓

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Worked Examples

<details> <summary><b>Example 1: Simplify with a Pythagorean pattern</b></summary>

Question: Simplify $\dfrac{1 - \cos^2\theta}{\sin\theta\cos\theta}$.

Solution:

1. Replace $1 - \cos^2\theta$ with $\sin^2\theta$: $\dfrac{\sin^2\theta}{\sin\theta\cos\theta}$.
2. Cancel one $\sin\theta$: $\dfrac{\sin\theta}{\cos\theta} = \tan\theta$. ✓

</details> <details> <summary><b>Example 2: Convert to sines and cosines</b></summary>

Question: Simplify $\sec\theta\cot\theta$.

Solution: $\dfrac{1}{\cos\theta} \cdot \dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta} = \csc\theta$. ✓

</details> <details> <summary><b>Example 3: Double angle from one given ratio</b></summary>

Question: θ is acute and $\cos\theta = \frac{3}{5}$. Find $\sin 2\theta$ and $\cos 2\theta$.

Solution:

1. 3-4-5 triangle: $\sin\theta = \frac{4}{5}$.
2. $\sin 2\theta = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$.
3. $\cos 2\theta = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}$. (Negative is fine: θ ≈ 53°, so 2θ ≈ 106° is in Quadrant II.) ✓

</details> <details> <summary><b>Example 4: Answer in terms of a variable</b></summary>

Question: $\cos x = k$ for an acute angle $x$. Express $\tan x$ in terms of $k$.

Solution: Adjacent $k$, hypotenuse 1, opposite $\sqrt{1 - k^2}$. So $\tan x = \dfrac{\sqrt{1 - k^2}}{k}$. ✓

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Worked Examples

<details> <summary><b>Example 1: Read every feature from an equation</b></summary>

Question: For $y = -2\cos(3x) + 1$, find the amplitude, period, midline, maximum, minimum, and the starting point at x = 0.

Solution:

• Amplitude $\lvert -2\rvert = 2$; period $\frac{2\pi}{3}$; midline $y = 1$.
• Maximum $1 + 2 = 3$; minimum $1 - 2 = -1$.
• At x = 0, $\cos 0 = 1$, so $y = -2(1) + 1 = -1$: the graph starts at its minimum because A is negative. ✓

</details> <details> <summary><b>Example 2: Find the first maximum after a phase shift</b></summary>

Question: For $y = 4\sin(2x - \pi) + 3$, where is the first maximum with x > 0?

Solution:

1. Phase shift: $-\frac{C}{B} = \frac{\pi}{2}$ to the right.
2. Sine peaks when its input is $\frac{\pi}{2}$: $2x - \pi = \frac{\pi}{2} \implies 2x = \frac{3\pi}{2} \implies x = \frac{3\pi}{4}$.
3. Height $= A + D = 4 + 3 = 7$. First maximum: $\left(\frac{3\pi}{4}, 7\right)$.
4. Check: $y = 4\sin(2x)$ peaks at $\frac{\pi}{4}$, and $\frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$. ✓

</details> <details> <summary><b>Example 3: Write the equation from a graph</b></summary>

Question: A graph of the form $y = A\cos(Bx) + D$ has a maximum at $(0, 6)$, and the next minimum is at $\left(\frac{\pi}{2}, -2\right)$. Find the equation.

Solution:

1. Amplitude $= \frac{6 - (-2)}{2} = 4$; midline $D = \frac{6 + (-2)}{2} = 2$.
2. Max to next min is half a cycle, so the period is $2 \cdot \frac{\pi}{2} = \pi$, and $B = \frac{2\pi}{\pi} = 2$.
3. A maximum at x = 0 means positive cosine: $y = 4\cos(2x) + 2$. ✓

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Worked Examples

<details> <summary><b>Example 1: Build a Ferris wheel model from a description</b></summary>

Question: A Ferris wheel is 60 feet in diameter, its lowest point is 4 feet above the ground, and it makes one revolution every 8 minutes. A rider boards at the lowest point at t = 0. Write h(t), the rider's height in feet after t minutes, and find the height at t = 2 and t = 4.

Solution:

1. Radius $r = 30$; center $c = 4 + 30 = 34$; $B = \frac{2\pi}{8} = \frac{\pi}{4}$.
2. Starting at the minimum → negative cosine: $h(t) = -30\cos\left(\frac{\pi}{4}t\right) + 34$.
3. $h(2) = -30\cos\frac{\pi}{2} + 34 = 0 + 34 = 34$ feet (a quarter turn: level with the center).
4. $h(4) = -30\cos\pi + 34 = 30 + 34 = 64$ feet (the top). ✓

</details> <details> <summary><b>Example 2: Build a tide model</b></summary>

Question: At a dock, high tide of 11 feet occurs at midnight and the next low tide, 3 feet, occurs at 6 a.m. Write the depth d(t), t hours after midnight, and find the depth at 2 a.m.

Solution:

1. $A = \frac{11 - 3}{2} = 4$, $D = \frac{11 + 3}{2} = 7$.
2. High to low is half a cycle: period $= 12$ hours, so $B = \frac{2\pi}{12} = \frac{\pi}{6}$.
3. Maximum at t = 0 → cosine: $d(t) = 4\cos\left(\frac{\pi}{6}t\right) + 7$.
4. $d(2) = 4\cos\frac{\pi}{3} + 7 = 4\left(\frac{1}{2}\right) + 7 = 9$ feet. The next high tide is at noon. ✓

</details> <details> <summary><b>Example 3: Interpret a shifted model</b></summary>

Question: A city's average monthly temperature, in °F, is modeled by $T(m) = 20\sin\left(\frac{\pi}{6}(m - 4)\right) + 55$, where m = 1 is January. In which month is it hottest, and what is that temperature?

Solution: Sine peaks when its input is $\frac{\pi}{2}$: $\frac{\pi}{6}(m - 4) = \frac{\pi}{2} \implies m - 4 = 3 \implies m = 7$ (July). The maximum is $55 + 20 = 75$°F. ✓

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Worked Examples

<details> <summary><b>Example 1: Point on the terminal side, then a double angle</b></summary>

Question: The terminal side of θ passes through $(-8, 15)$. Find $\sin 2\theta$.

Solution:

1. $r = \sqrt{64 + 225} = 17$.
2. $\sin\theta = \frac{15}{17}$ and $\cos\theta = -\frac{8}{17}$ (Quadrant II).
3. $\sin 2\theta = 2 \cdot \frac{15}{17} \cdot \left(-\frac{8}{17}\right) = -\frac{240}{289}$. ✓

</details> <details> <summary><b>Example 2: Phase shift and the first maximum</b></summary>

Question: For $y = 5\sin\left(2x - \frac{\pi}{2}\right) + 2$, find the first maximum with x > 0.

Solution:

1. Set the inside equal to $\frac{\pi}{2}$: $2x - \frac{\pi}{2} = \frac{\pi}{2} \implies x = \frac{\pi}{2}$.
2. Height $= 5 + 2 = 7$. First maximum: $\left(\frac{\pi}{2}, 7\right)$.
3. Check with the shift: phase shift $= \frac{\pi/2}{2} = \frac{\pi}{4}$ right; the unshifted peak at $\frac{\pi}{4}$ moves to $\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$. ✓

</details> <details> <summary><b>Example 3: Law of Cosines with an obtuse angle</b></summary>

Question: A triangle has sides 7, 8, and 13. Find the angle opposite the side of length 13.

Solution: $169 = 49 + 64 - 2(7)(8)\cos C \implies 56 = -112\cos C \implies \cos C = -\frac{1}{2} \implies C = 120^\circ$. The negative cosine confirms the angle is obtuse, as $169 > 113$ predicted. ✓

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