Unit Circle and Trig Graphs

• Convert between degrees and radians (180° = π radians)
• Recall sin, cos, and tan of the unit-circle benchmark angles (0, π/6, π/4, π/3, π/2)
• Apply the Pythagorean identity sin²θ + cos²θ = 1 to simplify expressions
• Identify the period and amplitude of y = A sin(Bx) and y = A cos(Bx)
• Recognize the shape of a tangent graph (period π, vertical asymptotes) and distinguish it from sin/cos

Radians measure angles as arc length on a unit circle, where 180° = π radians. Convert by multiplying: degrees × (π/180) gives radians, and radians × (180/π) gives degrees. The common angles you should know cold: 0 rad = 0°, π/6 = 30°, π/4 = 45°, π/3 = 60°, π/2 = 90°, π = 180°, 3π/2 = 270°, 2π = 360°. On the unit circle (radius 1, centered at origin), the point at angle θ is (cos θ, sin θ). Memorize these values: at 0, (1, 0); at π/6, (√3/2, 1/2); at π/4, (√2/2, √2/2); at π/3, (1/2, √3/2); at π/2, (0, 1). So sin(π/6) = 1/2 and cos(π/4) = √2/2. The Pythagorean identity sin²θ + cos²θ = 1 is derived from the unit-circle equation x² + y² = 1. It works for every angle and is the most tested identity on the ACT. The ACT also tests tan θ = sin θ / cos θ. For graphs, y = A sin(Bx) has amplitude |A| (max height from centerline) and period 2π/|B| (length of one full cycle). So y = 3 sin(2x) has amplitude 3 and period π. Cosine has the same shape as sine, but the cosine graph is the sine graph shifted left by π/2, that is cos(x) = sin(x + π/2). Cosine starts at its maximum; sine starts at 0. Tangent is different: its period is π (not 2π), and it has vertical asymptotes wherever cos θ = 0 (at π/2, 3π/2, etc.). The ACT asks you to identify a graph from an equation or vice versa. Always check amplitude and period first. Harder graph questions use y = A sin(B(x − C)) + D: the midline is y = D, the maximum is D + |A|, the minimum is D − |A|, the period is 2π/|B|, and C shifts the graph sideways without changing any of those values. The reciprocal functions come from the basic three: csc θ = 1/sin θ, sec θ = 1/cos θ, and cot θ = 1/tan θ. An equation such as sin θ = 1/2 has two solutions from 0 to 2π (π/6 and 5π/6) because sine is positive in two quadrants: find the reference angle, then place it in every quadrant where the sign matches (sine is positive in Quadrants I and II, cosine in I and IV, tangent in I and III).

• π radians = 180°. Write this on your scratch paper at the start of the section.
• If you forget a unit-circle value, draw the 30-60-90 or 45-45-90 triangle and read it off.
• sin²θ + cos²θ = 1 can be rearranged as sin²θ = 1 − cos²θ. This is how most ACT identity questions are disguised.
• For y = A sin(Bx): amplitude = |A|, period = 2π/|B|. Do not confuse the two.
• Tangent graphs have vertical asymptotes and repeat every π. Sine and cosine graphs never have asymptotes, so a curve with asymptotes cannot be a sine or cosine graph.