Trigonometric functions, identities and non-right triangles
Extend right-triangle ratios to periodic graphs and solve triangles using justified identities and laws.
What you will tackle
- Interpret unit-circle signs and trig graphs
- Use amplitude, period and transformations
- Apply fundamental trig identities
- Use sine and cosine laws with suitable given information
Periodic graphs encode repeated behaviour
For y = A sin(Bx) + D in radians, amplitude is |A|, period is 2π/|B| for B ≠ 0 and midline is D. Unit-circle coordinates give cosine and sine values beyond acute angles; quadrant determines sign. The identity sin²θ + cos²θ = 1 follows from the unit circle, and tan θ = sin θ/cos θ when the denominator is nonzero.
Choose a law from the triangle information
For sides a,b,c opposite angles A,B,C, a/sin A = b/sin B = c/sin C. The cosine law is c² = a² + b² − 2ab cos C, useful for two sides and their included angle. In an SSA sine-law setup, an ambiguous second triangle may be possible; test geometrical conditions rather than assuming uniqueness.
Build your notes, then practise
Work on paper before opening an explanation. Guided tasks provide a starting point; independent tasks remove the scaffolding; application tasks ask you to justify a method or interpretation in context.
Guided practice
Guided 1
For y = 3 sin(2x) − 1 find amplitude, period and midline. If sin θ = 3/5 in quadrant II, find cos θ and tan θ.
Show a starting hint
Use the quadrant before choosing the square-root sign.
Compare with the explained answer
- Amplitude is 3; period is 2π/2 = π; midline is −1.
- The identity gives |cos θ| = 4/5; quadrant II makes cosine negative.
- tan θ = (3/5)/(−4/5) = −3/4.
Answer / response criteria: Amplitude 3, period π, midline −1; cos θ = −4/5 and tan θ = −3/4.
Independent practice
Independent 1
Two sides are 5 and 7 with included angle 60°. Find the opposite side. In another triangle, A = 30°, B = 90° and a = 4; find b.
Compare with the explained answer
- Cosine law gives c² = 25 + 49 − 70(1/2) = 39, so c = √39.
- For the second triangle, b/sin 90° = 4/sin 30°.
- b = 8.
Answer / response criteria: √39; 8.
Application practice
Application 1
A model for water depth is d(t) = 2 + 0.5 cos(πt/6), with t in hours. Find the range, period, d(0) and d(6). Explain a reason the model may fail over a long period.
Compare with the explained answer
- The cosine term ranges from −0.5 to 0.5, so depth is 1.5–2.5.
- Period = 2π/(π/6) = 12 hours.
- d(0) = 2.5 and d(6) = 1.5; weather or long-term changes can alter the cycle.
Answer / response criteria: Range [1.5,2.5]; period 12 h; d(0) = 2.5 and d(6) = 1.5.
What can you explain now?
A model for water depth is d(t) = 2 + 0.5 cos(πt/6), with t in hours. Find the range, period, d(0) and d(6). Explain a reason the model may fail over a long period.
Compare with the explanation
Range [1.5,2.5]; period 12 h; d(0) = 2.5 and d(6) = 1.5.
Explain the reasoning before comparing. This review records your self-assessment rather than an official score or mastery decision.
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Source and review status
Coverage follows the named requirements and examples in the 2026–2027 ACT test guide and subject descriptions. That guide explicitly does not list every possible topic. The broader coded College and Career Readiness Standards are also mapped to conceptual teaching and practice; this does not calibrate a learner to a score band. The separate scientific-background context list and reading-complexity rubric have their own stated limits.
Original teaching and explained self-check practice are available. Specialist review and difficulty calibration remain pending. Essays require self-check or teacher review and are not automatically marked or awarded official scores.
Official source specification / guide