12.3 Trigonometric Graphs and Identities

12.3 Trigonometric Graphs and Identities

Table 11’s trigonometry cluster asks you to evaluate equivalent trigonometric functions and to graph trig relationships. Those two skills sit on top of the unit-circle coordinates from Unit Circle, Radians, and Arc Length: once (cos θ, sin θ) is a point on the circle, plotting y against θ produces the sine and cosine waves, and the identity sin^2 θ + cos^2 θ = 1 is just x^2 + y^2 = 1 rewritten. Skills Insight 276–300 is the band that expects this fluency. Work FREE mixed AAF items at /practice/accuplacer-advanced-algebra.

Parent graph y = sin x

The sine graph is the y-coordinate of the unit circle, unrolled along a horizontal axis named x (the same letter now means the angle, in radians).

• Period 2π: the pattern repeats every full turn. sin(x + 2π) = sin x.
• Amplitude 1: the wave peaks at 1 and troughs at −1. Range is [−1, 1].
• Domain all real numbers.
• Zeros at x = nπ for integer n — wherever the unit-circle point sits on the x-axis.
• Maximum 1 at x = π/2 + 2πn.
• Minimum −1 at x = 3π/2 + 2πn.
• Odd function: sin(−x) = −sin x. The graph is origin-symmetric and passes through (0, 0).

Five-point sketch on [0, 2π]: (0, 0), (π/2, 1), (π, 0), (3π/2, −1), (2π, 0). Mid-quadrant values match the special-angle table: sin(π/6) = 1/2, sin(π/3) = √3/2.

Parent graph y = cos x

The cosine graph is the x-coordinate of the unit circle, unrolled the same way.

• Period 2π, amplitude 1, range [−1, 1], domain all reals — same envelope as sine.
• Starts at a peak: cos 0 = 1, so the graph crosses (0, 1), not the origin.
• Zeros at x = π/2 + nπ.
• Maximum 1 at x = 2πn.
• Minimum −1 at x = π + 2πn.
• Even function: cos(−x) = cos x. The graph is symmetric about the y-axis.

Five-point sketch on [0, 2π]: (0, 1), (π/2, 0), (π, −1), (3π/2, 0), (2π, 1). Cosine is a sine wave shifted left by π/2: cos x = sin(x + π/2). That is already an equivalent-function identity.

Amplitude and period for y = A sin(Bx) and y = A cos(Bx)

AAF will change the letters. For y = A sin(Bx) or y = A cos(Bx) with B ≠ 0:

• Amplitude = |A|. The factor A is a vertical stretch. If A < 0, the graph also reflects across the x-axis.
• Period = 2π / |B|. The factor B is a horizontal compression. y = sin(2x) finishes a full wave on [0, π] instead of [0, 2π].

Worked: y = 4 sin x has amplitude 4 and period 2π. It still zeros at nπ, but now the peak is 4 at π/2.

Worked: y = −3 cos(2x) has amplitude 3 (absolute value), period 2π/2 = π, and a reflection, so at x = 0 the value is −3 rather than +3.

Worked: y = (1/2) sin(x/3) has amplitude 1/2 and period 2π / (1/3) = 6π. A slower wave. Do not report the period as 3 or as 2π/3 — divide 2π by |B|, and here |B| = 1/3.

A vertical shift y = A sin(Bx) + D moves the midline from y = 0 to y = D and the range to [D − |A|, D + |A|]. A horizontal (phase) shift y = A sin(B(x − C)) moves the graph right by C. Table 11’s graphing language is satisfied if you can read amplitude, period, intercepts, and the parent shape; phase-shift items still use the same substitution you already use on other function families.

Parent graph y = tan x

Tangent is sin x / cos x, so it inherits zeros from sine and vertical asymptotes from cosine’s zeros.

• Period π, not 2π. tan(x + π) = tan x. One copy of the graph lives on (−π/2, π/2).
• Vertical asymptotes at x = π/2 + nπ, where cos x = 0.
• Zeros at x = nπ.
• Range all real numbers — no amplitude in the sine/cosine sense, because the graph is unbounded.
• Odd function: tan(−x) = −tan x. Through the origin, with tan(π/4) = 1 and tan(−π/4) = −1.

For y = A tan(Bx) the period is π / |B|. Worked: y = tan(2x) has period π/2 and asymptotes where 2x = π/2 + nπ, i.e. x = π/4 + nπ/2. The first positive asymptote is at π/4, not at π/2. That is the classic trap of copying the parent asymptote without dividing by B.

Pythagorean identity and equivalent values

The unit-circle equation is the Pythagorean identity:

sin^2 θ + cos^2 θ = 1

Divide through by cos^2 θ (where defined) to get tan^2 θ + 1 = sec^2 θ. Divide through by sin^2 θ to get 1 + cot^2 θ = csc^2 θ. AAF most often uses the sine-cosine form.

Worked, Quadrant II: sin θ = 3/5 and θ is in Quadrant II. Then cos^2 θ = 1 − 9/25 = 16/25, so cos θ = ±4/5. Quadrant II makes cosine negative, therefore cos θ = −4/5 and tan θ = (3/5) / (−4/5) = −3/4. The unsigned 3-4-5 triple is the same right-triangle work as 12.1; the unit circle only added the sign.

Worked, Quadrant IV: cos θ = 8/17 and θ is in Quadrant IV. Then sin^2 θ = 1 − 64/289 = 225/289, so sin θ = ±15/17. Quadrant IV makes sine negative, therefore sin θ = −15/17. Equivalent forms of that cosine include sin(π/2 − θ) and cos(−θ).

Even/odd and cofunction equivalents

These rewrites are the Table 11 “equivalent trigonometric functions” skill:

Worked: sin(150°) = sin(180° − 30°) = sin 30° = 1/2. Equivalently sin(5π/6) = 1/2.

Worked: cos(−π/3) = cos(π/3) = 1/2 because cosine is even. sin(−π/3) = −sin(π/3) = −√3/2 because sine is odd.

Worked: Which expression equals cos θ? Options of the form sin(π/2 − θ), sin(θ − π/2), −sin θ, and cos(π − θ) are not all equivalent. sin(π/2 − θ) equals cos θ. cos(π − θ) equals −cos θ. sin(θ − π/2) equals −cos θ as well, because sin(θ − π/2) = −sin(π/2 − θ) = −cos θ.

Reading a trig graph without a pretty picture

AAF may describe a graph in words: “a cosine curve with amplitude 2 and period π.” Translate:

1. Cosine parent, so a peak at the origin unless a phase shift is named.
2. Amplitude 2 means |A| = 2.
3. Period π = 2π / |B| forces |B| = 2.
4. A matching equation is y = 2 cos(2x) or y = −2 cos(2x) depending on whether it opens from a high point or a low point at x = 0.

If the description is “the graph of y = sin x reflected across the x-axis and stretched vertically by 5,” write y = −5 sin x. Reflection is the negative amplitude, not a period change.

If two graphs are offered as equivalent, check one convenient x-value. y = cos x and y = sin(x + π/2) both equal 1 at x = 0 and both equal 0 at x = π/2. y = sin(x − π/2) equals −1 at x = 0 and is a different graph — it is actually −cos x.

AAF traps for graphs and identities

• Period of tangent. y = tan x has period π. Copying 2π from sine is the most common miss.
• Amplitude of a negative coefficient. Amplitude is |A|, so y = −4 sin x still has amplitude 4. The leading minus is a reflection.
• Period formula backwards. Period is 2π/|B|, not |B|/2π. For sin(x/2), B = 1/2 and the period is 4π.
• Missing the second identity form. sin^2 θ = 1 − cos^2 θ is the Pythagorean identity solved for sine. A stem that gives cosine and asks for sine still needs a quadrant to pick the sign.
• Even/odd swap. Cosine is the even one. cos(−θ) = cos θ; sin(−θ) is not sin θ.
• Asymptotes of tan(Bx). Solve Bx = π/2 + nπ. Do not leave the parent’s π/2 in place.

Graphing and identities feed the last section: once you know sine is 1/2 at both π/6 and 5π/6, you can solve sin θ = 1/2 on [0, 2π). That solving step, plus the law of sines and the law of cosines, is Solving Trig Equations; Law of Sines and Cosines.