12.3 Trigonometric Graphs and Identities

12.3 Trigonometric Graphs and Identities

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

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.