AP Precalculus: Trigonometric Functions

Setting β = α in the sum identities gives the double-angle identities: sin 2α = 2 sin α cos α and cos 2α = cos2 α − sin2 α = 1 − 2 sin2 α = 2 cos2 α − 1. The three forms of cosine let you choose the one that uses only the function you already know. When an equation mixes different angles (such as 2x and x) or different functions (such as sin and cos), use an identity first to rewrite it in terms of a single function and a single angle. Then factor or use the quadratic formula, solve for the trig function, and find the angles in the given interval. Never divide both sides by a trig function, since that can lose solutions; factor it out instead. Finally, check each candidate value: a quadratic in sin x can produce a value larger than 1 in absolute value, which has no angle solution.

1. Periodic Phenomena: Period, Midline, and Amplitude

Trigonometry begins with things that repeat: ocean tides, the seasons, a swinging pendulum, a rider on a Ferris wheel. A function is periodic if its outputs repeat over equal-length intervals of input. The length of the smallest such interval is the period, so f(x + period) = f(x) for every x. Three quantities describe a wave-like periodic function. The midline is the horizontal line halfway between the maximum and minimum: y = (max + min)/2. The amplitude is the distance from the midline to the maximum, which equals (max − min)/2. The period is the horizontal length of one full cycle, for example from one peak to the next. These three numbers are exactly the ones you will later read off a sine or cosine equation, so it pays to be fluent with them from a graph.

Worked Example: A Ferris wheel has a diameter of 30 m, its lowest point is 2 m above the ground, and it makes one full turn every 12 minutes. Describe the rider’s height over time.

The highest point is 2 + 30 = 32 m and the lowest is 2 m. Midline = (32 + 2)/2 = 17 m. Amplitude = (32 − 2)/2 = 15 m. The period is 12 minutes, so the rider is again at the bottom at t = 12, 24, … and at the top at t = 6, 18, … Because the pattern repeats, the height at t = 30 equals the height at t = 6 (30 − 24 = 6), which is 32 m.

midline 17, amplitude 15, period 12; h(30) = 32

61218245101520253035xymin 2max 32
The height h(t) = 17 − 15 cos(πt/6) of a rider on a Ferris wheel. The wave repeats every 12 minutes (the period), swings 15 m above and below the midline y = 17 (the amplitude), and runs from 2 m to 32 m.

1. A periodic function has a maximum value of 14 and a minimum value of −6. What are its midline and amplitude?

Explanation: Midline = (14 + (−6))/2 = 4. Amplitude = (14 − (−6))/2 = 10. The amplitude is half the total vertical spread, not the whole spread (20).

2. A periodic function has period 6, f(2) = 3 and f(5) = 8. What is f(14)?

Explanation: Subtract whole periods: 14 − 12 = 2, so f(14) = f(2) = 3. Two periods of 6 equal 12.

3. Which quantity is the best example of a periodic phenomenon?

Explanation: Daylight hours rise and fall in the same pattern every year. The other quantities do not repeat in equal intervals.

4. A tide gauge records a high of 9 ft and a low of 1 ft, and the pattern repeats every 12 hours. What is the amplitude?

Explanation: Amplitude = (9 − 1)/2 = 4 ft. The 5 ft is the midline height, 12 is the period, and 8 is the full spread.

2. Angles, Radians, and Arc Length

An angle in standard position has its vertex at the origin and its initial side on the positive x-axis; a counterclockwise rotation is positive and a clockwise rotation is negative. One radian is the measure of a central angle whose arc has the same length as the radius, so for a circle of radius r, the angle θ in radians satisfies θ = s/r, or s = rθ for arc length s. A full turn is 360° = 2π radians, which gives the conversions radians = degrees × π/180 and degrees = radians × 180/π. Angles that end on the same terminal side are coterminal; add or subtract multiples of 360° or 2π to find them. In calculus and in most of AP Precalculus, angles are in radians unless a degree sign is shown, and the calculator must be in radian mode.

Worked Example: (a) Convert 150° to radians. (b) Convert 7π/12 to degrees. (c) Find an angle in [0, 2π) coterminal with 13π/6. (d) Find the arc length on a circle of radius 8 cm cut off by a central angle of 3π/4.

(a) 150 × π/180 = 5π/6.

(b) (7π/12) × (180/π) = 105°.

(c) 13π/6 − 2π = 13π/6 − 12π/6 = π/6.

(d) s = rθ = 8 × 3π/4 = 6π cm ≈ 18.85 cm.

5π/6, 105°, π/6, 6π cm

1. Convert 240° to radians.

Explanation: 240 × π/180 = 4π/3. Reduce 240/180 = 4/3.

2. Convert 11π/6 radians to degrees.

Explanation: (11π/6)(180/π) = 11 × 30 = 330°.

3. Which angle in [0, 2π) is coterminal with −π/3?

Explanation: Add 2π: −π/3 + 6π/3 = 5π/3. The terminal side is one-third of a turn clockwise from the positive x-axis.

4. A circle has radius 12 cm. What is the arc length for a central angle of 75°, to the nearest tenth of a centimeter?

Explanation: Convert first: 75° = 5π/12 radians. Then s = rθ = 12 × 5π/12 = 5π ≈ 15.7 cm. Multiplying 12 by 75 without converting gives 900, which is wrong because the formula needs radians.

3. The Unit Circle and the Trigonometric Functions

Let P(x, y) be any point on the terminal side of an angle θ in standard position, and let r = √(x2 + y2) be its distance from the origin, always positive. Then sin θ = y/r, cos θ = x/r, and tan θ = y/x = sin θ/cos θ (undefined when x = 0). On the unit circle, r = 1, so the point is simply (cos θ, sin θ), and x2 + y2 = 1 becomes cos2 θ + sin2 θ = 1. The signs follow the quadrant: in Quadrant I all three are positive, in II only sine is positive, in III only tangent is positive, and in IV only cosine is positive. The point does not have to lie on the unit circle; you only need to divide by r.

Worked Example: The point P(−5, 12) is on the terminal side of θ. Find sin θ, cos θ, and tan θ.

First r = √((−5)2 + 122) = √169 = 13. Then sin θ = 12/13, cos θ = −5/13, and tan θ = 12/(−5) = −12/5. The point is in Quadrant II, where sine is positive and cosine and tangent are negative, which matches.

sin θ = 12/13, cos θ = −5/13, tan θ = −12/5

-1.5-1-0.50.511.5-1.5-1-0.50.511.5xy(cos θ, sin θ)θ = 5π/6
On the unit circle the point at angle θ = 5π/6 is (cos θ, sin θ) = (−√3/2, 1/2). Its coordinates are cosine and sine directly because r = 1.

1. The point P(8, −6) is on the terminal side of θ. What is sin θ?

Explanation: r = √(64 + 36) = 10, so sin θ = y/r = −6/10 = −3/5. The value −3/4 is tan θ, not sin θ.

2. In which quadrant is sin θ < 0 and cos θ > 0?

Explanation: Sine is the y-coordinate (negative below the x-axis) and cosine is the x-coordinate (positive to the right of the y-axis). That is Quadrant IV.

3. The point P(√5, −2) is on the terminal side of θ. What is tan θ?

Explanation: tan θ = y/x = −2/√5 = −2√5/5 after rationalizing. (Here r = √(5 + 4) = 3, so sin θ = −2/3, which is not the tangent.)

4. If sin θ = 3/5 and θ is in Quadrant II, what is cos θ?

Explanation: Use the legs 3 and 4 with hypotenuse 5. In Quadrant II the x-coordinate is negative, so cos θ = −4/5.

4. Exact Values of Trigonometric Functions

You should be able to produce the sine, cosine, and tangent of the special angles 0, π/6, π/4, π/3, π/2 and their multiples without a calculator. The coordinates (cos θ, sin θ) are (1, 0), (√3/2, 1/2), (√2/2, √2/2), (1/2, √3/2), (0, 1). For an angle in another quadrant, find its reference angle (the acute angle between the terminal side and the x-axis), use that angle’s values, and then attach the sign for the quadrant. For example, 7π/6 is π/6 past π, so it is in Quadrant III with reference angle π/6. Tangent is sine divided by cosine, and it is undefined at π/2 and 3π/2 where cosine is 0.

Worked Example: Find (a) sin(5π/4), (b) cos(2π/3), (c) tan(7π/6), (d) sin(−π/2).

(a) 5π/4 is in Quadrant III with reference angle π/4. Sine is negative there, so sin(5π/4) = −√2/2.

(b) 2π/3 is in Quadrant II with reference angle π/3. Cosine is negative, so cos(2π/3) = −1/2.

(c) 7π/6 is in Quadrant III with reference angle π/6. Tangent is positive there: tan = (−1/2)/(−√3/2) = 1/√3 = √3/3.

(d) −π/2 points straight down to (0, −1), so sin(−π/2) = −1.

-1.5-1-0.50.511.5-1.5-1-0.50.511.5xyπ/6π/4π/3π/2 (0, 1)0 (1, 0)
The first-quadrant points: π/6 → (√3/2, 1/2), π/4 → (√2/2, √2/2), π/3 → (1/2, √3/2). Every other special angle is a reflection of one of these.

1. What is cos(5π/6)?

Explanation: 5π/6 is in Quadrant II with reference angle π/6. cos(π/6) = √3/2, and cosine is negative in Quadrant II.

2. What is tan(3π/4)?

Explanation: The reference angle is π/4, where sine and cosine have equal size. In Quadrant II they have opposite signs, so tan = −1.

3. What is sin(11π/6)?

Explanation: 11π/6 is in Quadrant IV with reference angle π/6. sin(π/6) = 1/2, and sine is negative in Quadrant IV.

4. What is cos(7π/3)?

Explanation: 7π/3 − 2π = π/3, so cos(7π/3) = cos(π/3) = 1/2. Subtract full turns first.

5. Graphs of the Sine and Cosine Functions

The graphs of f(θ) = sin θ and g(θ) = cos θ are smooth waves. Both have domain all real numbers, range [−1, 1], amplitude 1, midline y = 0, and period 2π. Sine starts at the origin and rises to its maximum at π/2; it has zeros at every multiple of π and is an odd function, sin(−θ) = −sin θ, so it is symmetric about the origin. Cosine starts at its maximum 1, crosses zero at π/2, and is an even function, cos(−θ) = cos θ, so it is symmetric about the y-axis. Dividing one period into four equal parts of length π/2 gives the five key points that locate a maximum, a minimum, and three midline crossings, and they are the backbone of every sketch. The graph is concave down where the function is above the midline and concave up where it is below, so the points where the curve crosses the midline are inflection points.

Worked Example: Use the graph to answer: on [0, 2π], (a) where is sin x increasing, (b) where is cos x = 0, and (c) how many solutions does sin x = 0.4 have?

(a) Sine rises from its minimum to its maximum, so it is increasing on [0, π/2] and on [3π/2, 2π].

(b) Cosine crosses the x-axis at π/2 and 3π/2.

(c) The horizontal line y = 0.4 cuts the sine curve once on the way up and once on the way down within one period, so there are 2 solutions.

π/2π3π/22π-1.5-1-0.50.511.5xyy = sin xy = cos x
One period of y = sin x and y = cos x. Both oscillate between −1 and 1 with period 2π; the cosine curve is the sine curve shifted π/2 units to the left.

1. On [0, 2π], where does sin x = 0?

Explanation: Sine is the y-coordinate on the unit circle, which is 0 at the points (1, 0) and (−1, 0): angles 0, π, and 2π.

2. Which function is even?

Explanation: cos(−x) = cos x, so the cosine graph is symmetric about the y-axis. Sine is odd.

3. On [0, 2π], on which interval is cos x decreasing?

Explanation: Cosine falls from its maximum 1 at x = 0 to its minimum −1 at x = π, then rises back up. So it is decreasing on (0, π).

4. How many solutions does cos x = 0.3 have on [0, 4π]?

Explanation: Each period of length 2π contains exactly 2 solutions (one with the curve falling and one rising). Two periods give 4 solutions.

6. Sinusoidal Functions: Amplitude, Period, Phase Shift, and Midline

A sinusoidal function is any function of the form y = a sin b(x − c) + d or y = a cos b(x − c) + d. Each constant has a geometric meaning. The amplitude is |a|; if a < 0 the graph is reflected over its midline. The period is 2π/|b|, and the frequency is its reciprocal |b|/(2π). The phase shift c moves the graph c units right (left if c is negative). The vertical shift d gives the midline y = d, so the maximum is d + |a| and the minimum is d − |a|. Be careful when the equation is written as sin(bx − k): factor out b first, because sin(bx − k) = sin b(x − k/b), so the phase shift is k/b and not k. A transformation of the parent graph can be done one step at a time: stretch, reflect, shift horizontally, shift vertically.

Worked Example: Describe y = −2 sin(4x − π) + 3.

Factor out the 4 from the input: −2 sin 4(x − π/4) + 3. So a = −2, b = 4, c = π/4, d = 3.

Amplitude = 2 (and the graph is reflected). Period = 2π/4 = π/2. Phase shift = π/4 to the right. Midline y = 3, so the maximum is 5 and the minimum is 1. The x-scale is (π/2)/4 = π/8, so the key points are spaced π/8 apart starting at x = π/4.

amplitude 2, period π/2, shift right π/4, midline y = 3, range [1, 5]

π/4π/23π/4π123456xymidline y = 3
y = −2 sin 4(x − π/4) + 3. The amplitude is 2, the period is π/2 (two full cycles on [0, π]), the midline is y = 3, and the curve is flipped so it first dips below the midline after the shifted start at x = π/4.

1. What is the period of y = 5 sin(πx/3)?

Explanation: Period = 2π/b with b = π/3, so 2π ÷ (π/3) = 6. The 5 is the amplitude and does not change the period.

2. What is the maximum value of y = −3 cos x + 2?

Explanation: The midline is 2 and the amplitude is 3, so the values run from 2 − 3 = −1 to 2 + 3 = 5. The negative sign flips the graph but does not change the maximum value 5.

3. What is the phase shift of y = sin(2x − π/2)?

Explanation: Factor: sin(2x − π/2) = sin 2(x − π/4). The phase shift is π/4 to the right, not π/2.

4. What is the range of y = 2 sin x − 5?

Explanation: The midline is −5 and the amplitude is 2, so y ranges from −5 − 2 = −7 to −5 + 2 = −3.

7. Graphing by Hand and Writing an Equation from a Graph

Both skills use the same six-question checklist. (1) Amplitude: half the distance from maximum to minimum, a = (max − min)/2. (2) Period: the length of one cycle, which gives b = 2π/period. (3) x-scale: period/4, the spacing of the key points. (4) Midline: d = (max + min)/2. (5) Phase shift: pick a starting point. A sine curve begins at the midline going up; a cosine curve begins at a maximum. The x-value where your chosen curve starts is c. (6) Reflection: if the curve starts the opposite way, make a negative. To graph, plot the five key points of one cycle by moving along the x-scale and then repeat. To write an equation, read the maximum, minimum and one full cycle from the graph, and choose either sine or cosine. Both are correct if the shifts are consistent, and cosine is often easier when the graph has an obvious peak on the y-axis.

Worked Example: A sinusoidal curve has a maximum of 7 at x = 0, a minimum of −1, and its next maximum is at x = π. Write an equation.

(1) a = (7 − (−1))/2 = 4. (2) The period is π, so b = 2π/π = 2. (3) The x-scale is π/4. (4) d = (7 + (−1))/2 = 3. (5) The curve starts at a maximum when x = 0, so use cosine with no phase shift, c = 0. (6) A cosine curve starts at a maximum and falls, which is what this curve does, so there is no reflection.

y = 4 cos 2x + 3

π/2π3π/22π-2-112345678xy
y = 4 cos 2x + 3. The maximum 7 and minimum −1 are marked. One cycle takes π units, so the curve peaks at 0, π, and 2π.

1. A sinusoid has maximum 10 and minimum 2, passes through the midline going up at x = 0, and has period 8. Which equation fits?

Explanation: a = (10 − 2)/2 = 4, d = (10 + 2)/2 = 6, b = 2π/8 = π/4. Starting at the midline and going up is a sine curve.

2. For y = 3 sin 2x, at what value of x > 0 does the first maximum occur?

Explanation: The period is π, so the x-scale is π/4. A sine curve reaches its maximum one quarter of the way through the cycle: x = π/4.

3. A cosine-type curve has its minimum −5 at x = 0, maximum 1, and period 6. Which equation fits?

Explanation: a = 3 and d = (1 + (−5))/2 = −2, b = 2π/6 = π/3. A cosine curve starts at a maximum, but this one starts at a minimum, so it is reflected: a = −3.

4. A sine curve crosses its midline going upward at x = π/3 and has amplitude 1, period 2π, and midline y = 0. Which equation fits?

Explanation: A regular sine curve crosses the midline upward at x = 0. Shifting that start to x = π/3 means a shift right by π/3: sin(x − π/3).

8. Sinusoidal Function Context and Data Modeling

Any data that rise and fall in a regular cycle can be modeled with a sinusoid. The method is the six-step checklist applied to a table or scatterplot. Find the maximum and minimum values of the data to get a and d. Decide the period from the context (12 months for a yearly pattern, 24 hours for a daily one) and use b = 2π/period. Then pick a data point at or near the midline on the way up (for sine) or at the maximum (for cosine) to get the phase shift c. Real data are rarely perfect, so the hand model is an approximation; a graphing calculator’s sinusoidal regression (SinReg on a TI-84) finds the best-fitting values, usually in the form a sin(bx + k) + d, which you can factor to compare with your c. Once you have a model, you can substitute a value to make a prediction, or set the model equal to a number and solve to find when something occurs. Always set the calculator to radian mode and comment on whether a prediction is reasonable, since a model is only as good as its fit to the data.

Worked Example: The mean monthly high temperature in a city is 6°C in January (m = 1) and 30°C in July (m = 7). Model it with a sine function and predict October (m = 10).

a = (30 − 6)/2 = 12. d = (30 + 6)/2 = 18. The pattern repeats yearly, so the period is 12 and b = 2π/12 = π/6. The midline value 18 occurs halfway between the low in January and the high in July, in April (m = 4), on the way up, so a sine curve starts there: c = 4. No reflection is needed.

T(m) = 12 sin(π/6 (m − 4)) + 18. For October, T(10) = 12 sin(π/6 × 6) + 18 = 12 sin(π) + 18 = 18°C. October is on the way back down through the midline, which is reasonable.

12345678910111261218243036xymax 30 (July)min 6 (Jan)
The model T(m) = 12 sin(π/6 (m − 4)) + 18 for monthly high temperature (°C). It crosses the midline T = 18 going up at m = 4 (April), peaks at 30 in July (m = 7), and returns to the low of 6 in January.

1. The tide height at a harbor is modeled by h(t) = 5 sin(πt/6) meters, t hours after midnight. What is the height at 4 a.m., to the nearest hundredth?

Explanation: h(4) = 5 sin(4π/6) = 5 sin(2π/3) = 5(√3/2) ≈ 4.33 m. Use radian mode; in degree mode the calculator gives the wrong answer 0.18.

2. A quantity repeats every 24 hours and is modeled with y = a sin(bt) + d. What is b?

Explanation: b = 2π/period = 2π/24 = π/12.

3. A rider’s height is h(t) = 17 − 15 cos(πt/6) meters. What is h(2)?

Explanation: h(2) = 17 − 15 cos(π/3) = 17 − 15(1/2) = 9.5 m.

4. Monthly data have a maximum of 30 and a minimum of −4. What midline should the model use?

Explanation: The midline is the average of the extremes: (30 + (−4))/2 = 13. The amplitude is 17, which is half of 34.

9. The Tangent Function and Its Transformations

Since tan θ = sin θ/cos θ, the tangent graph has vertical asymptotes wherever cos θ = 0, which is at the odd multiples of π/2, that is, θ = π/2 + kπ. Its key features are: period π (one full cycle between consecutive asymptotes), domain all reals except π/2 + kπ, range all real numbers, zeros at every multiple of π, no amplitude and no maximum or minimum, and the function is always increasing on each branch. The graph changes from concave down to concave up at each x-intercept, which is an inflection point. Transformations work as they do for other functions. For y = a tan b(x − c) + d, a is a vertical dilation by |a| (negative a reflects over the x-axis), the period is π/|b|, c is the horizontal shift, and d is the vertical shift. Notice that a does not create an amplitude; it controls how steeply the branches rise. The asymptotes move with the horizontal shift and the period: for the center point (c, d), they sit half a period to each side.

−π−π/2π/2π-4-3-2-11234xy
y = tan x on [−π, π]. The dashed lines are the vertical asymptotes at odd multiples of π/2, the zeros are at multiples of π, and the points (π/4, 1) and (−π/4, −1) are marked. The pattern repeats every π.

Worked Example: Describe y = 3 tan 2(x − π/4) + 1 and find its asymptotes.

Here a = 3, b = 2, c = π/4, d = 1. The graph is a vertical dilation by 3, the period is π/2, it is shifted right π/4 and up 1. The center point of one branch is (π/4, 1). The asymptotes are half a period (π/4) on each side of the center: x = 0 and x = π/2, then every π/2 after that. Algebraically, set the inside equal to π/2 + kπ: 2(x − π/4) = π/2 + kπ gives x = π/2 + kπ/2.

period π/2, center (π/4, 1), asymptotes x = kπ/2

π/4π/23π/4π-8-6-4-2246810xy
y = 3 tan 2(x − π/4) + 1. The period is π/2 and the center point of each branch is on the line y = 1 (green points). The dashed lines are asymptotes at 0, π/2, and π.

1. What is the period of y = tan(3x)?

Explanation: For tangent, period = π/|b| = π/3. Do not use 2π/b, which is the period for sine and cosine.

2. Which x-value is not in the domain of y = tan x?

Explanation: Tangent is undefined where cosine is 0, at odd multiples of π/2. 5π/2 = π/2 + 2π.

3. What is the smallest positive vertical asymptote of y = tan(x + π/3)?

Explanation: Set x + π/3 = π/2, so x = π/2 − π/3 = π/6. The graph of tan x is shifted left π/3.

4. Which statement is true about y = −tan x on (−π/2, π/2)?

Explanation: The negative sign reflects tan x over the x-axis, turning an increasing function into a decreasing one. Tangent has no amplitude and its range is all real numbers.

10. Inverse Trigonometric Functions

Sine, cosine, and tangent are not one-to-one, so none has an inverse until the domain is restricted so the graph passes the horizontal line test. The standard restrictions give: y = arcsin x (also written sin−1 x), the angle in [−π/2, π/2] whose sine is x, with domain [−1, 1]; y = arccos x, the angle in [0, π] whose cosine is x, with domain [−1, 1]; and y = arctan x, the angle in (−π/2, π/2) whose tangent is x, with domain all real numbers. The input of an inverse trig function is a ratio and the output is an angle. Be careful: sin−1 x means the inverse function, not 1/sin x. To evaluate one, ask “which angle in the restricted range has this value?” using the unit circle. The restricted range matters: sin−1(sin(7π/6)) is not 7π/6, because 7π/6 is outside [−π/2, π/2]. When a value is not a special ratio, use a calculator in radian mode. Right-triangle sketches help with composites: if θ = arccos x with x > 0, then sin θ = √(1 − x2).

Worked Example: Evaluate (a) arcsin(−√2/2), (b) arccos(−1/2), (c) arctan(√3), (d) sin−1(sin(7π/6)), (e) sin(arccos(3/5)), and (f) arccos(2/9) to the nearest thousandth.

(a) The angle in [−π/2, π/2] with sine −√2/2 is −π/4.

(b) The angle in [0, π] with cosine −1/2 is 2π/3 (Quadrant II).

(c) tan(π/3) = √3, so the answer is π/3.

(d) sin(7π/6) = −1/2, and the angle in [−π/2, π/2] with sine −1/2 is −π/6.

(e) Let θ = arccos(3/5) so the adjacent side is 3 and the hypotenuse 5. The opposite side is 4, so sin θ = 4/5.

(f) 2/9 is not a special value, so use a calculator in radian mode: arccos(2/9) ≈ 1.347.

-1.5-1-0.50.511.5−π/2π/2xy(−1, −π/2)(1, π/2)
y = arcsin x has domain [−1, 1] and range [−π/2, π/2]. It is the reflection of sin x on [−π/2, π/2] over the line y = x.

1. What is arccos(√3/2)?

Explanation: cos(π/6) = √3/2 and π/6 is in the arccosine range [0, π].

2. What is arctan(−1)?

Explanation: The arctangent range is (−π/2, π/2), and tan(−π/4) = −1. Angles such as 3π/4 have the right tangent but are outside the range.

3. What is sin−1(sin(5π/6))?

Explanation: sin(5π/6) = 1/2. The angle in [−π/2, π/2] with sine 1/2 is π/6, so the answer is π/6, not 5π/6.

4. What is cos(arcsin(5/13))?

Explanation: Let θ = arcsin(5/13): opposite 5, hypotenuse 13, adjacent 12. Since arcsin is in [−π/2, π/2], cosine is positive: 12/13.

11. Trigonometric Equations and Inequalities

Solve a trigonometric equation in two stages. First, isolate the trigonometric function (for example, sin x = −1/2). Second, find the angles, remembering that the function takes each value in more than one quadrant and that trig functions are periodic. Use the unit circle to get the reference angle (the Quadrant I angle with the same size of value), and place it in every quadrant where the sign is right: sine is negative in III and IV, cosine is negative in II and III, and tangent is negative in II and IV. If the problem gives an interval such as [0, 2π], list only solutions in it. If it gives no interval, add the full period: k·2π for sine and cosine, kπ for tangent, where k is an integer. If the equation is a quadratic in a trig function, factor or use the quadratic formula first, and discard any value outside [−1, 1] for sine and cosine. For inequalities, find the crossing points, then use the graph to see where the curve is above or below the line. If a graphing calculator gives a reference angle, always evaluate it with a positive input so the angle is in Quadrant I, then place it in the correct quadrants yourself.

Worked Example: (a) Solve 2 sin x + 1 = 0 on [0, 2π]. (b) Solve 2 sin2 x − sin x − 1 = 0 on [0, 2π]. (c) Solve sin x > 1/2 on [0, 2π].

(a) sin x = −1/2. The reference angle is π/6, and sine is negative in Quadrants III and IV, so x = π + π/6 = 7π/6 and x = 2π − π/6 = 11π/6.

(b) Factor: (2 sin x + 1)(sin x − 1) = 0. So sin x = −1/2 gives 7π/6 and 11π/6, and sin x = 1 gives π/2. Solutions: π/2, 7π/6, 11π/6.

(c) sin x = 1/2 at π/6 and 5π/6. The curve is above the line between them, so π/6 < x < 5π/6.

With a calculator: sin x = 0.3 has reference angle sin−1(0.3) ≈ 0.305, so x ≈ 0.305 and x ≈ π − 0.305 ≈ 2.837 on [0, 2π].

π/2π3π/22π-1.5-1-0.50.511.5xyy = 1/2
Sine is above the line y = 1/2 between the crossing points π/6 and 5π/6, so sin x > 1/2 on (π/6, 5π/6) in this period. The green segment marks that interval.

1. Solve 2 sin x = √2 on [0, 2π].

Explanation: sin x = √2/2, reference angle π/4, and sine is positive in Quadrants I and II: x = π/4 and π − π/4 = 3π/4.

2. Solve tan x = −1 on [0, 2π].

Explanation: The reference angle is π/4 and tangent is negative in Quadrants II and IV: π − π/4 = 3π/4 and 2π − π/4 = 7π/4.

3. Solve 2 cos2 x + cos x − 1 = 0 on [0, 2π].

Explanation: Factor: (2 cos x − 1)(cos x + 1) = 0. cos x = 1/2 gives π/3 and 5π/3; cos x = −1 gives π.

4. Solve cos x < −1/2 on [0, 2π].

Explanation: cos x = −1/2 at 2π/3 and 4π/3. Cosine is below −1/2 on the stretch around π between these angles.

12. The Secant, Cosecant, and Cotangent Functions

The three remaining trig functions are reciprocals: sec θ = 1/cos θ, csc θ = 1/sin θ, and cot θ = 1/tan θ = cos θ/sin θ. Each is undefined wherever its denominator is zero, and those input values are vertical asymptotes. Secant is undefined where cos θ = 0 (odd multiples of π/2), cosecant where sin θ = 0 (multiples of π), and cotangent where sin θ = 0 (multiples of π). Secant and cosecant have period 2π and range (−∞, −1] ∪ [1, ∞): wherever the cosine or sine graph has a maximum of 1, the reciprocal has a local minimum of 1, and wherever it has a minimum of −1, the reciprocal has a local maximum of −1. Cotangent has period π, no amplitude, and is decreasing between consecutive asymptotes. To evaluate any of these, find the value of the matching sine, cosine, or tangent first, then take the reciprocal.

Worked Example: Find the exact value of (a) sec(π/3), (b) csc(7π/6), (c) cot(2π/3), and (d) cos−1(csc(π/2)).

(a) cos(π/3) = 1/2, so sec(π/3) = 2.

(b) sin(7π/6) = −1/2, so csc(7π/6) = −2.

(c) cos(2π/3) = −1/2 and sin(2π/3) = √3/2, so cot = (−1/2)/(√3/2) = −1/√3 = −√3/3.

(d) csc(π/2) = 1/1 = 1, and cos−1(1) = 0.

π/2π3π/22π-4-3-2-11234xy
y = sec x with its reciprocal y = cos x on [0, 2π]. Where cos x = 0 there are vertical asymptotes. Where cos x = ±1, sec x = ±1 (green points). The range of sec x is (−∞, −1] ∪ [1, ∞).

1. What is sec(π/3)?

Explanation: sec(π/3) = 1/cos(π/3) = 1/(1/2) = 2.

2. What is csc(7π/6)?

Explanation: sin(7π/6) = −1/2, so csc(7π/6) = 1/(−1/2) = −2.

3. What is cot(2π/3)?

Explanation: cot = cos/sin = (−1/2)/(√3/2) = −1/√3 = −√3/3.

4. What is the range of y = csc x?

Explanation: Since |sin x| ≤ 1, |csc x| = 1/|sin x| ≥ 1. So csc x ≤ −1 or csc x ≥ 1.

13. Pythagorean Identities and Proving Identities

An identity is an equation that is true for every input where both sides are defined. Because (cos θ, sin θ) lies on the unit circle, x2 + y2 = 1 gives the Pythagorean identity sin2 θ + cos2 θ = 1. Dividing every term by cos2 θ gives tan2 θ + 1 = sec2 θ, and dividing by sin2 θ gives 1 + cot2 θ = csc2 θ. Each can be rearranged, for example sin2 θ = 1 − cos2 θ. These let you find the other trig values from one: if cos θ is known, then sin θ = ±√(1 − cos2 θ) with the sign chosen from the quadrant. To prove an identity, work on one side (usually the more complicated one) until it matches the other, and never move terms across the equals sign. Useful strategies: rewrite everything in sine and cosine, factor, combine fractions over a common denominator, and substitute a known identity. If one side alone does not work, simplify both sides to a common expression.

Worked Example: Prove that csc θ − sin θ = cos θ cot θ.

Start with the left side and write it with sine and cosine: 1/sin θ − sin θ. Combine over a common denominator: (1 − sin2 θ)/sin θ. Use the Pythagorean identity 1 − sin2 θ = cos2 θ: cos2 θ/sin θ. Split it as cos θ · (cos θ/sin θ) = cos θ cot θ, which is the right side.

Both sides are equal, so the identity is proved.

1. Which expression is equivalent to sec2 θ − 1?

Explanation: From tan2 θ + 1 = sec2 θ, subtract 1 to get sec2 θ − 1 = tan2 θ.

2. If cos θ = −3/5 and θ is in Quadrant III, what is sin θ?

Explanation: sin2 θ = 1 − 9/25 = 16/25, so sin θ = ±4/5. Sine is negative in Quadrant III: −4/5.

3. Simplify sin θ(csc θ − sin θ).

Explanation: Distribute: sin θ csc θ − sin2 θ = 1 − sin2 θ = cos2 θ.

4. Simplify (1 + cot2 θ) sin2 θ.

Explanation: 1 + cot2 θ = csc2 θ, and csc2 θ · sin2 θ = 1.

14. Sum and Difference Identities

The sum identities are sin(α + β) = sin α cos β + cos α sin β and cos(α + β) = cos α cos β − sin α sin β. The difference identities follow by replacing β with −β and using that sine is odd and cosine is even: sin(α − β) = sin α cos β − cos α sin β and cos(α − β) = cos α cos β + sin α sin β. Notice that cosine changes the sign between the two terms while sine keeps it. They have two main uses. First, they give exact values for angles such as π/12 or 7π/12 by writing them as a sum or difference of special angles (for example π/12 = π/3 − π/4). Second, they let you collapse a long expression into a single trig value when you recognize the pattern. Remember that sin(α + β) is not sin α + sin β.

Worked Example: (a) Find the exact value of cos(π/12). (b) Simplify sin(5π/18)cos(π/18) + cos(5π/18)sin(π/18).

(a) π/12 = π/3 − π/4, so cos(π/12) = cos(π/3)cos(π/4) + sin(π/3)sin(π/4) = (1/2)(√2/2) + (√3/2)(√2/2) = (√2 + √6)/4.

(b) The expression matches sin(α + β) with α = 5π/18 and β = π/18. So it equals sin(6π/18) = sin(π/3) = √3/2.

1. What is the exact value of cos(5π/12)?

Explanation: 5π/12 = π/4 + π/6. cos(π/4 + π/6) = (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4.

2. Simplify cos(2π/9)cos(π/9) − sin(2π/9)sin(π/9).

Explanation: This is cos(α + β) with α = 2π/9 and β = π/9, so it equals cos(3π/9) = cos(π/3) = 1/2.

3. If sin α = 3/5 and cos β = 5/13, with both angles in Quadrant I, what is sin(α + β)?

Explanation: cos α = 4/5 and sin β = 12/13. sin(α + β) = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65.

4. What is the exact value of cos(11π/12)?

Explanation: 11π/12 = 2π/3 + π/4. cos = (−1/2)(√2/2) − (√3/2)(√2/2) = −√2/4 − √6/4 = −(√2 + √6)/4.

15. Double-Angle Identities and Equations with Identities

Setting β = α in the sum identities gives the double-angle identities: sin 2α = 2 sin α cos α and cos 2α = cos2 α − sin2 α = 1 − 2 sin2 α = 2 cos2 α − 1. The three forms of cosine let you choose the one that uses only the function you already know. When an equation mixes different angles (such as 2x and x) or different functions (such as sin and cos), use an identity first to rewrite it in terms of a single function and a single angle. Then factor or use the quadratic formula, solve for the trig function, and find the angles in the given interval. Never divide both sides by a trig function, since that can lose solutions; factor it out instead. Finally, check each candidate value: a quadratic in sin x can produce a value larger than 1 in absolute value, which has no angle solution.

Worked Example: (a) If sin θ = 3/5 and θ is in Quadrant II, find sin 2θ and cos 2θ. (b) Solve sin 2x = sin x on [0, 2π].

(a) cos θ = −4/5 (Quadrant II). sin 2θ = 2(3/5)(−4/5) = −24/25. cos 2θ = 1 − 2(3/5)2 = 1 − 18/25 = 7/25.

(b) Replace sin 2x: 2 sin x cos x = sin x. Move everything to one side and factor (do not divide by sin x): sin x (2 cos x − 1) = 0. Then sin x = 0 gives x = 0, π, 2π, and cos x = 1/2 gives x = π/3, 5π/3.

x = 0, π/3, π, 5π/3, 2π

1. If sin x = 2/3, what is cos 2x?

Explanation: Use cos 2x = 1 − 2 sin2 x = 1 − 2(4/9) = 1 − 8/9 = 1/9.

2. If cos θ = 3/5 and θ is in Quadrant IV, what is sin 2θ?

Explanation: sin θ = −4/5 in Quadrant IV. sin 2θ = 2(−4/5)(3/5) = −24/25.

3. Which values solve sin 2x = sin x on [0, 2π)?

Explanation: sin x(2 cos x − 1) = 0 gives sin x = 0 (0 and π) and cos x = 1/2 (π/3 and 5π/3). Dividing by sin x would lose 0 and π.

4. How many solutions does 2 sin2 x − 5 sin x + 1 = 0 have on [0, 2π]?

Explanation: The quadratic formula gives sin x = (5 ± √17)/4. The plus root is about 2.28, which is greater than 1 and has no solution. The minus root is about 0.219, which gives two angles (Quadrants I and II).

Common Mistakes to Avoid

  • Calculating in degree mode. Check that the calculator is in radian mode whenever an angle has π or the problem is in radians.
  • Using the period 2π/b for tangent. For tan and cot the period is π/|b|; for sine, cosine, secant, and cosecant it is 2π/|b|.
  • Reading the phase shift from sin(bx − k) as k. Factor out b first: the shift is k/b. For sin(2x − π/2) the shift is π/4, not π/2.
  • Confusing amplitude with the full height. Amplitude is half the distance from maximum to minimum, and tangent, secant, cosecant, and cotangent do not have an amplitude at all.
  • Treating sin−1 x as 1/sin x. The superscript −1 on a function name means inverse (arcsine), and the reciprocal is csc x.
  • Giving an angle outside the range of an inverse function, such as writing sin−1(sin(7π/6)) = 7π/6. The output of arcsine must be in [−π/2, π/2], arccosine in [0, π], arctangent in (−π/2, π/2).
  • Finding only one solution to an equation like sin x = 1/2. Look in every quadrant where the sign matches, and add the period (k·2π or kπ) when no interval is given.
  • Dividing both sides of an equation by a trig function such as sin x, which drops the solutions where it equals 0. Factor it out instead.
  • Writing sin(α + β) = sin α + sin β. Trig functions are not linear; use the sum identity.
  • Moving terms across the equals sign when proving an identity. Work on one side until it matches the other.

Frequently Asked Questions

What is the difference between degrees and radians?

Both measure angles. A full turn is 360° or 2π radians. Radians measure an angle by the arc it cuts off compared with the radius (θ = s/r), which makes formulas for arc length and calculus much cleaner. To convert, multiply degrees by π/180, or radians by 180/π.

How do I remember the exact values on the unit circle?

Memorize the three Quadrant I points for π/6, π/4, and π/3 (the numerators of the coordinates are 1, √2, √3 over 2 in order for sine, and reversed for cosine). Every other special angle is a reflection of one of these, so only the signs change by quadrant.

How do I decide between a sine and a cosine model?

Both work. If the graph starts at its maximum or minimum on the y-axis, cosine needs no phase shift. If it starts on the midline going up, sine needs no shift. Otherwise pick either and find the horizontal shift to the nearest start of that curve.

Why does tangent have no amplitude?

Tangent has no maximum or minimum. Near each vertical asymptote it grows without bound, so its range is all real numbers and amplitude is not defined.

Which trig identities do I need to memorize for AP Precalculus?

Know the Pythagorean identities, the reciprocal and quotient identities, the sum and difference identities for sine and cosine, and the double-angle identities. These are the ones used to rewrite expressions and solve equations.

Where can I practice more problems like these?

The AP Precalculus Trigonometric and Polar Functions quiz in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.