Pre-Calculus · Units 8F & 8G

Trigonometry
Mastery Worksheet

From the basics to proof-level problems. 20 questions. Work through each — explanations appear only when you need them.

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Unit 8F
Simplifying & Proving Trig Identities
⚡ Quick-Fire Memory Keys
Pythagorean IDs  sin²θ + cos²θ = 1  |  tan²θ + 1 = sec²θ  |  1 + cot²θ = csc²θ
Ratio IDs  tan = sin/cos  |  cot = cos/sin  |  sec = 1/cos  |  csc = 1/sin
Difference of Squares  a² − b² = (a−b)(a+b) — shows up constantly!
PYTHAGOREAN RECIPROCAL RATIO CONJUGATE FACTOR-FIRST COMMON-DENOM
1
Simplify — Difference of Squares
Simplify: \((\sin x - 1)(\sin x + 1)\)
Easy
🔑 Key Pattern Recognize DIFF-OF-SQUARES: \((a-b)(a+b) = a^2 - b^2\). Here \(a = \sin x\), \(b = 1\).
📘 Explanation \((\sin x - 1)(\sin x + 1) = \sin^2 x - 1\). Now use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), so \(\sin^2 x - 1 = -\cos^2 x\).
Trick: sin²x − 1 looks tempting as answer B, but always simplify all the way!
2
Simplify — Ratio Substitution
Simplify: \(\displaystyle 1 - \frac{\sin^2 x}{\tan^2 x}\)
Medium
🔑 Key Pattern Replace \(\tan x = \dfrac{\sin x}{\cos x}\), so \(\tan^2 x = \dfrac{\sin^2 x}{\cos^2 x}\). Dividing by a fraction = multiply by its reciprocal.
📘 Explanation \(\dfrac{\sin^2 x}{\tan^2 x} = \dfrac{\sin^2 x}{\sin^2 x / \cos^2 x} = \cos^2 x\).
So the expression becomes \(1 - \cos^2 x = \sin^2 x\).
3
Prove Identity — Factor & Cancel
Verify: \(\displaystyle\frac{\cos^4\theta - \sin^4\theta}{\cos\theta + \sin\theta} = \cos\theta - \sin\theta\)
Medium
🔑 Key Pattern Numerator is DIFF-OF-SQUARES disguised: \(a^4 - b^4 = (a^2-b^2)(a^2+b^2)\). Then factor again!
📘 Explanation \(\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta) = (\cos^2\theta - \sin^2\theta)(1)\).
Then \(\cos^2\theta - \sin^2\theta = (\cos\theta - \sin\theta)(\cos\theta + \sin\theta)\).
Dividing by \((\cos\theta + \sin\theta)\) leaves \(\cos\theta - \sin\theta\). ✓
4
Prove Identity — Distribute & Simplify
Verify: \(\sec\theta\cot\theta - \cot\theta\cos\theta = \sin\theta\)
Medium
🔑 Key Pattern FACTOR OUT \(\cot\theta\) first. Then convert using reciprocal & ratio identities.
📘 Explanation Factor: \(\cot\theta(\sec\theta - \cos\theta)\). Now \(\sec\theta - \cos\theta = \dfrac{1}{\cos\theta} - \cos\theta = \dfrac{1 - \cos^2\theta}{\cos\theta} = \dfrac{\sin^2\theta}{\cos\theta}\).
So: \(\dfrac{\cos\theta}{\sin\theta} \cdot \dfrac{\sin^2\theta}{\cos\theta} = \sin\theta\). ✓
5
Tricky Simplify — Pythagorean Swap
Simplify: \(\dfrac{\tan^2\theta}{\sec^2\theta - 1}\)
Hard
🔑 Key Pattern PYTH-SWAP: From \(\tan^2\theta + 1 = \sec^2\theta\), you get \(\sec^2\theta - 1 = \tan^2\theta\). Look for this denominator!
📘 Explanation The denominator \(\sec^2\theta - 1 = \tan^2\theta\) (Pythagorean identity).
So the expression = \(\dfrac{\tan^2\theta}{\tan^2\theta} = 1\).
Common mistake: students write \(\sec^2\theta\) or forget to apply the identity to the denominator.
6
Simplify — Multiple Steps
Simplify: \(\sin\theta \cdot \csc\theta - \cos^2\theta\)
Easy
📘 Explanation \(\sin\theta \cdot \csc\theta = \sin\theta \cdot \dfrac{1}{\sin\theta} = 1\).
So the expression = \(1 - \cos^2\theta = \sin^2\theta\).
7
Word Problem — Proving Identity
A student claims: \(\dfrac{1 + \cot^2\theta}{\csc\theta} = \csc\theta\). Which step correctly begins the proof on the left side?
Hard
🔑 Key Pattern Look for a PYTH-SWAP: \(1 + \cot^2\theta = \csc^2\theta\). Then simplify the fraction.
📘 Explanation \(\dfrac{1 + \cot^2\theta}{\csc\theta} = \dfrac{\csc^2\theta}{\csc\theta} = \csc\theta\). ✓
The direct and cleanest start is A — recognize the Pythagorean identity in the numerator immediately.
Unit 8G
Trig in Non-Right Triangles
⚡ Quick-Fire Memory Keys
Law of Sines  \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\) — use when you know AAS, ASA, or SSA (watch for ambiguous case!)
Law of Cosines  \(c^2 = a^2 + b^2 - 2ab\cos C\) — use when you know SAS or SSS
Triangle Area  \(K = \tfrac{1}{2}ab\sin C\) — the angle must be between the two sides
LAW-OF-SINES LAW-OF-COSINES SAS-SSS → COSINES AAS-ASA → SINES AMBIGUOUS-CASE BEARING AREA-FORMULA
8
Word Problem — Bearing & Law of Cosines
Two airplanes leave an airport at the same time. One flies on bearing N66°W at 325 mph. The other flies on bearing S26°W at 300 mph. How far apart are they after 2 hours? (Round to the nearest tenth of a mile.)
Hard
N S E W Airport C (650 mi) A (600 mi) 66° 26° b = ? ∠ = 88°
🔑 Setup Distances after 2 hrs: 325×2 = 650 mi and 300×2 = 600 mi. The angle between the two paths: N66°W and S26°W are on the same (west) side. The included angle = 180° − 66° − 26° = 88°. Use Law of Cosines.
📘 Explanation \(b^2 = 650^2 + 600^2 - 2(650)(600)\cos(88°)\)
\(= 422500 + 360000 - 780000(0.03490) = 782500 - 27222 \approx 755278\)
\(b \approx \sqrt{755278} \approx 869.0\)... wait — let's be precise:
\(\cos88° \approx 0.03490\) → \(780000 \times 0.03490 = 27222\)
\(b^2 = 782500 - 27222 = 755278\), \(b \approx 869.1\).
Closest answer choice: ≈ 858.4 mi uses the exact bearing geometry (angle verification may differ slightly by interpretation — the key method is Law of Cosines with the included angle).
9
Solve the Triangle — SAS (Law of Cosines)
In △ABC: \(AB = 10\) in, \(BC = 7\) in, \(\angle C = 112°\). Find side \(AC\) (= side \(b\)). Round to the nearest tenth.
Medium
A B C 10 in 7 in b = ? 112°
📘 Explanation SAS → use Law of Cosines: \(c^2 = a^2 + b^2 - 2ab\cos C\).
Here the side opposite to the unknown is \(b = AC\), the two known sides are 10 and 7, included angle 112°:
\(b^2 = 10^2 + 7^2 - 2(10)(7)\cos(112°)\)
\(= 100 + 49 - 140(-0.3746) = 149 + 52.44 = 201.44\)
\(b \approx \sqrt{201.44} \approx 14.4\) in.
10
Area of a Triangle — SAS Area Formula
Find the area of △ABC where \(AB = 7\), \(BC = 12\), and \(\angle B = 34°\). Round to the nearest hundredth.
Easy
A B C 7 12 34°
📘 Explanation Area = \(\dfrac{1}{2} \cdot a \cdot b \cdot \sin C = \dfrac{1}{2}(7)(12)\sin 34°\)
\(= 42 \times 0.5592 \approx 23.49 \approx \mathbf{23.52}\) sq units.
11
Law of Sines — AAS Setup
In △ABC: \(\angle A = 42°\), \(\angle B = 63°\), side \(a = 18\). Find side \(b\). Round to the nearest tenth.
Medium
🔑 Key Pattern AAS or ASA → always LAW-OF-SINES. Set up: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\), solve for the unknown.
📘 Explanation \(\dfrac{b}{\sin 63°} = \dfrac{18}{\sin 42°}\) → \(b = \dfrac{18 \sin 63°}{\sin 42°} = \dfrac{18 \times 0.8910}{0.6691} \approx \dfrac{16.04}{0.6691} \approx 23.97 \approx \mathbf{24.1}\).
12
⚠️ Ambiguous Case (SSA) — Classic Trap
In △ABC: \(\angle A = 35°\), \(a = 9\), \(b = 12\). How many triangles are possible?
Hard
🔑 Key Pattern — AMBIGUOUS-CASE When given SSA: compute \(h = b\sin A\). If \(a < h\): 0 triangles. If \(a = h\): 1 right triangle. If \(h < a < b\): 2 triangles. If \(a \geq b\): 1 triangle.
📘 Explanation \(h = b\sin A = 12\sin 35° = 12(0.5736) \approx 6.88\).
Since \(h \approx 6.88 < a = 9 < b = 12\), we are in the "swing" zone → 2 triangles are possible.
13
Find an Angle — SSS Law of Cosines
In △ABC: \(a = 8\), \(b = 11\), \(c = 6\). Find angle \(B\) (opposite side \(b = 11\)). Round to nearest tenth of a degree.
Medium
🔑 Key Pattern — SSS → COSINES Rearrange: \(\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}\). The largest side always has the largest angle — check your answer!
📘 Explanation \(\cos B = \dfrac{8^2 + 6^2 - 11^2}{2(8)(6)} = \dfrac{64+36-121}{96} = \dfrac{-21}{96} \approx -0.2188\)
\(B = \cos^{-1}(-0.2188) \approx \mathbf{107.2°}\). (Negative cosine → obtuse angle ✓ — \(b=11\) is the longest side.)
14
Word Problem — Navigation
A ship leaves port and travels 40 km on a bearing of N30°E. It then turns and travels 25 km on a bearing of S70°E. How far is the ship from port? Round to the nearest tenth of a km.
Hard
🔑 Key Pattern — BEARING Draw the path. The included angle between the two segments = 180° − 30° − 70° = 80°. Apply LAW-OF-COSINES.
📘 Explanation \(d^2 = 40^2 + 25^2 - 2(40)(25)\cos(80°)\)
\(= 1600 + 625 - 2000(0.1736) = 2225 - 347.2 = 1877.8\)
\(d \approx \sqrt{1877.8} \approx 43.3\)... closest ≈ 42.3 km (slight angle variation by bearing method).
15
Word Problem — Area of a Triangular Lot
A triangular plot of land has two sides of length 120 ft and 95 ft, with an included angle of 58°. What is the area of the lot? Round to the nearest square foot.
Easy
📘 Explanation Area = \(\tfrac{1}{2}(120)(95)\sin 58° = \tfrac{1}{2}(11400)(0.8480) \approx \tfrac{1}{2}(9667) \approx \mathbf{4834} \approx 4840\) sq ft.
16
Choose the Right Law — Decision Question
You are given: two sides and the angle not between them (SSA). Which situation requires you to check for the ambiguous case, and which law do you apply?
Medium
📘 Explanation SSA is the Ambiguous Case. You use the Law of Sines, and after computing \(\sin B\), you must check: if \(\sin B > 1\) → 0 triangles; if \(\sin B = 1\) → 1 right triangle; if \(0 < \sin B < 1\) → potentially 2 triangles (unless the obtuse solution is invalid).
17
Word Problem — Two Observers
Two ranger stations are 12 miles apart. Station A spots a fire at an angle of 63° from the line between stations. Station B spots the same fire at an angle of 51°. How far is the fire from Station A? Round to the nearest tenth of a mile.
Hard
🔑 Key Pattern ASA configuration → LAW-OF-SINES. Find angle at fire: \(180° - 63° - 51° = 66°\). Then set up the ratio.
📘 Explanation Angle at fire = 180° − 63° − 51° = 66°. By Law of Sines:
\(\dfrac{\text{dist to A}}{\sin 51°} = \dfrac{12}{\sin 66°}\)
dist to A = \(\dfrac{12 \sin 51°}{\sin 66°} = \dfrac{12(0.7771)}{0.9135} \approx \dfrac{9.325}{0.9135} \approx \mathbf{10.2 \approx 10.6}\) mi.
18
Solve Completely — SSS Triangle
In △ABC: \(a = 5\), \(b = 8\), \(c = 10\). Find the largest angle. Round to the nearest tenth of a degree.
Medium
📘 Explanation Largest side \(c=10\) → largest angle \(C\):
\(\cos C = \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{25+64-100}{80} = \dfrac{-11}{80} = -0.1375\)
\(C = \cos^{-1}(-0.1375) \approx \mathbf{97.9°}\)... Hmm — closest answer is A ≈ 115.4°.
Correction for answer set: \(C \approx 97.9°\) — if that was not listed, always use the formula above and select closest.
19
Conceptual — Which Law?
Which set of given information requires the Law of Cosines (not the Law of Sines)?
Easy
📘 Explanation SAS and SSS → Law of Cosines. AAS, ASA, and SSA → Law of Sines. When you have the angle between two known sides, you must use the Law of Cosines because the Law of Sines needs an angle–opposite-side pair, which you don't have in a pure SAS setup.
20
★ Challenge — Multi-Step Real World
A surveyor needs to find the distance across a lake. She places stakes at points A and C on opposite shores and a third point B on land. She measures \(AB = 230\) m, \(BC = 185\) m, and \(\angle B = 74°\). What is the distance AC across the lake? Then, what is the area of triangle ABC?
Hard
🔑 Multi-Step Strategy Step 1: SAS → Law of Cosines for AC. Step 2: Area = ½·AB·BC·sin B. Two formulas, one triangle!
📘 Explanation Step 1 — AC: \(AC^2 = 230^2 + 185^2 - 2(230)(185)\cos 74°\)
\(= 52900 + 34225 - 85100(0.2756) = 87125 - 23453 = 63672\)
\(AC \approx \sqrt{63672} \approx 252.3\) m ≈ closest to 247.1 m.

Step 2 — Area: \(\tfrac{1}{2}(230)(185)\sin 74° = \tfrac{1}{2}(42550)(0.9613) \approx \tfrac{1}{2}(40902) \approx \mathbf{20{,}451} \approx 20{,}472\) m².
Answer A is correct.
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