A Cumulative Workflow for Mixed Trig Problems
When trigonometry feels messy, it is usually because the problem is mixing representations: an angle might be described by a rotation, a triangle might be implied by a situation, and an expression might hide an identity. Use one consistent workflow to keep control.
The 5-step workflow
- (1) Choose a representation: triangle, unit circle, or graph. Pick the one that makes the given information easiest to “see.”
- (2) Identify knowns and unknowns: list what you know (angle measure, side length, coordinate, quadrant, etc.) and what you need (a trig value, a length, an angle, a simplified form).
- (3) Select tools: ratios, special angles, identities, inverse trig, or algebra. Choose the smallest toolset that can finish the job.
- (4) Compute: do the algebra carefully, keep exact values exact, and delay rounding until the end.
- (5) Interpret: check sign (quadrant), reasonableness (size), and units (meters, degrees, etc.).
Representation Choice: Quick Decision Guide
- Unit circle: best for exact values, quadrant/sign, and reference angles.
- Right triangle: best for real contexts (heights, distances, slopes) and when a diagram is implied.
- Graph: best for interpreting periodic behavior, intercepts, and where sine/cosine are positive or negative.
Mixed Skill Set 1: Exact Values via the Unit Circle
These problems are about choosing the unit circle representation, then using quadrant and reference angle to get exact values.
Example 1: Find exact values at a non-standard angle
Problem: Find exact values of sin(210°), cos(210°), and tan(210°).
(1) Representation: unit circle.
(2) Knowns/unknowns: angle 210°; need sin, cos, tan.
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(3) Tools: reference angle and quadrant signs; tan = sin/cos.
(4) Compute: 210° = 180° + 30°, so the reference angle is 30° and the angle is in Quadrant III. In Quadrant III, sin and cos are negative. Using the 30° values: |sin| = 1/2 and |cos| = √3/2. Therefore sin(210°) = -1/2, cos(210°) = -√3/2. Then tan(210°) = (-1/2)/(-√3/2) = 1/√3 = √3/3.
(5) Interpret: Quadrant III makes tan positive, which matches √3/3.
Example 2: Use a coordinate on the unit circle
Problem: A point on the unit circle is P = (−√2/2, √2/2). Find the angle(s) θ in [0, 2π) that match, and compute tan(θ).
(1) Representation: unit circle coordinate.
(2) Knowns/unknowns: cos(θ) = −√2/2, sin(θ) = √2/2; need θ and tan(θ).
(3) Tools: special-angle coordinates; tan = sin/cos.
(4) Compute: The magnitudes match 45° (π/4). Signs are (cos negative, sin positive) so Quadrant II. Thus θ = 3π/4. Then tan(θ) = (√2/2)/(−√2/2) = −1.
(5) Interpret: Quadrant II implies tan negative, consistent with −1.
Practice set A (exact values)
- Find exact values: sin(330°), cos(330°), tan(330°).
- Find exact values: sin(5π/6), cos(5π/6), tan(5π/6).
- Given cos(θ) = 1/2 and θ is in Quadrant IV, find sin(θ) and tan(θ).
- Evaluate exactly: sin(−2π/3), cos(−2π/3), tan(−2π/3).
Mixed Skill Set 2: Right Triangles from Real Contexts (with Visual Reasoning)
In applied problems, the hardest part is often drawing the triangle and labeling it correctly. Your workflow should force a diagram early.
Example 3: Angle of elevation with a horizontal distance
Problem: From a point on level ground 48 m from the base of a building, the angle of elevation to the top is 35°. Find the building’s height to the nearest tenth of a meter.
(1) Representation: right triangle (ground is horizontal, building is vertical).
(2) Knowns/unknowns: adjacent side = 48 m, angle = 35°, unknown opposite side = height h.
(3) Tools: tangent relates opposite/adjacent.
(4) Compute: tan(35°) = h/48, so h = 48 tan(35°). Using a calculator, tan(35°) ≈ 0.7002, so h ≈ 48(0.7002) = 33.6 m.
(5) Interpret: Height is less than 48 m because the angle is well below 45°, which matches 33.6 m. Units: meters.
Example 4: Ladder problem with a required safety angle
Problem: A ladder must reach a window 6.2 m above the ground. Safety guidelines require the ladder to make a 75° angle with the ground. How long must the ladder be (nearest hundredth of a meter)?
(1) Representation: right triangle (ladder is hypotenuse).
(2) Knowns/unknowns: opposite side = 6.2 m, angle at ground = 75°, unknown hypotenuse L.
(3) Tools: sine relates opposite/hypotenuse.
(4) Compute: sin(75°) = 6.2/L, so L = 6.2/sin(75°). With sin(75°) ≈ 0.9659, L ≈ 6.2/0.9659 = 6.42 m.
(5) Interpret: Ladder length slightly longer than height, reasonable. Units: meters.
Example 5: Two-step diagram problem (horizontal + vertical components)
Problem: A drone travels 120 m at an angle of 28° above the horizontal. How far horizontally did it travel, and how much did it gain in altitude (nearest tenth)?
(1) Representation: right triangle with hypotenuse 120 m and angle 28° from horizontal.
(2) Knowns/unknowns: hypotenuse = 120 m, angle = 28°, unknown adjacent (horizontal) x and opposite (vertical) y.
(3) Tools: cosine for adjacent, sine for opposite.
(4) Compute: x = 120 cos(28°) ≈ 120(0.8829) = 105.9 m. y = 120 sin(28°) ≈ 120(0.4695) = 56.3 m.
(5) Interpret: Because 28° is small, horizontal distance should be larger than vertical gain; 105.9 m vs 56.3 m fits. Units: meters.
Practice set B (right-triangle modeling)
- A wheelchair ramp rises 0.85 m over a horizontal run of 7.5 m. Find the incline angle (nearest tenth of a degree).
- A guy-wire is attached to a pole 9.0 m high and anchored 4.2 m from the base. Find the wire length (nearest hundredth of a meter).
- A boat heads 2.4 km at 18° north of east. Find the eastward and northward components (nearest hundredth of a km).
- A sign is viewed from 30 m away at an angle of elevation of 12°. Find the sign’s height above the viewer’s eye level (nearest tenth of a meter).
Mixed Skill Set 3: Simplifying Expressions Using Identities (with Strategy)
Identity problems are about recognizing structure. Before manipulating, decide what form you want: all sine and cosine, or a single trig function, or a constant. Keep the workflow: choose a representation (often “algebraic identity space”), list what you want to eliminate, then select identities that reduce complexity.
Strategy checklist for identity simplification
- Look for a common denominator when fractions involve sin and cos.
- Convert tan and sec into sin/cos if that creates cancellation.
- Use Pythagorean identities to replace 1 − sin² or 1 − cos², or to convert 1 + tan² into sec² (and similarly for cot/csc).
- Factor when you see a difference of squares or a common factor.
- State domain restrictions when you divide by an expression that could be zero (for example, dividing by cos θ assumes cos θ ≠ 0).
Example 6: Simplify to a single trig function
Problem: Simplify (1 − cos²θ)/sinθ.
(1) Representation: identity/algebra.
(2) Knowns/unknowns: expression in sin and cos; want simpler form.
(3) Tools: Pythagorean identity: 1 − cos²θ = sin²θ.
(4) Compute: (1 − cos²θ)/sinθ = sin²θ/sinθ = sinθ, provided sinθ ≠ 0.
(5) Interpret: The simplified form matches the original wherever defined; note the original expression is undefined when sinθ = 0.
Example 7: Simplify a complex fraction
Problem: Simplify (sinθ)/(1 + cosθ) + (1 + cosθ)/(sinθ).
(1) Representation: identity/algebra.
(2) Knowns/unknowns: sum of two fractions; want a simpler combined form.
(3) Tools: common denominator; Pythagorean identity.
(4) Compute: Common denominator is sinθ(1 + cosθ). Combine: [sin²θ + (1 + cosθ)²]/[sinθ(1 + cosθ)]. Expand numerator: sin²θ + (1 + 2cosθ + cos²θ) = (sin²θ + cos²θ) + 1 + 2cosθ = 1 + 1 + 2cosθ = 2(1 + cosθ). So the expression becomes 2(1 + cosθ)/[sinθ(1 + cosθ)] = 2/sinθ, provided sinθ ≠ 0 and 1 + cosθ ≠ 0.
(5) Interpret: Final form is 2cscθ with domain restrictions from the original denominators.
Example 8: Prove an identity by transforming one side
Problem: Verify that (sec²θ − 1)/tanθ = tanθ for angles where both sides are defined.
(1) Representation: identity/algebra.
(2) Knowns/unknowns: show left side equals right side.
(3) Tools: identity sec²θ − 1 = tan²θ.
(4) Compute: (sec²θ − 1)/tanθ = tan²θ/tanθ = tanθ, provided tanθ ≠ 0.
(5) Interpret: Equality holds on the domain where the expressions are defined (tanθ ≠ 0 and cosθ ≠ 0).
Practice set C (identity simplification)
- Simplify: (1 − sin²θ)/cosθ.
- Simplify: (tanθ + cotθ) in terms of sinθ and cosθ, then simplify as much as possible.
- Simplify: (secθ − cosθ)/tanθ.
- Verify: (1 + tan²θ)cos²θ = 1 for angles where defined.
Mini-Assessment: Putting It All Together
Instructions: For each item, follow the workflow. Start by choosing a representation and sketching a diagram when appropriate. Give exact values when asked; otherwise round as directed. Include units for applied problems.
Part A: Diagram interpretation (describe what you “see”)
- A point on the unit circle is shown in Quadrant IV with x-coordinate √3/2. Without computing yet, state the sign of sin(θ), cos(θ), and tan(θ). Then find sin(θ) and tan(θ) exactly.
- A right triangle diagram shows angle A at the left, a right angle at the bottom right, hypotenuse labeled 15, and the side adjacent to angle A labeled 9. Identify which trig ratio directly gives sin(A), then compute sin(A) and cos(A) (exact radical form if possible).
- A sine graph crosses the x-axis at x = 0 and is increasing there. State whether sin(x) is positive or negative just to the right of 0, and whether cos(0) is positive, negative, or zero.
Part B: Short computation (exact when possible)
- Compute exactly: cos(7π/6) and sin(7π/6).
- Compute exactly: tan(315°).
- If sin(θ) = −3/5 and θ is in Quadrant III, find cos(θ) and tan(θ).
- Simplify: (1 − cos²θ)/(1 + cosθ).
- Simplify: (sinθ/cosθ)·(1/sinθ).
Part C: Multi-step word problems (diagram + unit-aware answers)
- Observation tower: An observer is 72 m from the base of a tower. The angle of elevation to the top is 41°. (a) Draw and label a right triangle. (b) Find the tower height to the nearest tenth of a meter. (c) If the observer’s eye level is 1.6 m above the ground, find the tower height from ground to top (nearest tenth).
- Roof pitch: A roof rises 3.4 m over a horizontal run of 9.8 m. (a) Find the roof angle relative to the horizontal (nearest tenth of a degree). (b) Find the length of the sloped roof segment (nearest hundredth of a meter).
- Rescue cable: A cable runs from the top of a 18 m cliff down to a point on the ground 25 m from the base of the cliff. (a) Find the cable length (nearest tenth of a meter). (b) Find the angle the cable makes with the ground (nearest tenth of a degree). (c) Explain briefly how you know your angle is acute.
- Two-leg walk: A hiker walks 1.8 km at 20° north of east, then 1.2 km at 35° south of east. (a) Draw a component diagram for each leg. (b) Compute the total eastward displacement and total northward displacement (nearest hundredth of a km). (c) Find the magnitude of the net displacement and its direction relative to east (nearest hundredth and nearest tenth of a degree).
