Unit 24 · Algebra II: Exponential, Logarithmic and Trigonometric Functions

A video posted Monday has 200 views. Each day the count triples. Tuesday: 200 × 3 = 600. Wednesday: 600 × 3 = 1,800. Thursday: 1,800 × 3 = 5,400. After x days the count is 200 × 3^x, because you have multiplied by 3 a total of x times. After 4 days that is 200 × 81 = 16,200. This is an exponential function: a starting value times a fixed growth factor raised to the number of steps. A linear function adds the same amount each step; an exponential one multiplies by the same amount. That difference is why the count jumps from hundreds to tens of thousands in less than a week.

Most growth in the world is described in percent, so you need to turn a percent into a factor. A town of 12,000 grows 3% each year. Growing by 3% means keeping 100% and adding 3%, so the factor is 1.03. After 10 years the population is 12,000 × 1.03^10. On a calculator 1.03^10 ≈ 1.344, so the town has about 12,000 × 1.344 ≈ 16,127 people. Shrinking works the same way with a factor less than 1. A car worth $20,000 loses 20% of its value each year: it keeps 80%, so the decay factor is 0.80, and after 3 years it is worth 20,000 × 0.8^3 = 20,000 × 0.512 = $10,240.

The common mistake is to add the percents instead of multiplying the factors. Growing 3% a year for 10 years is not 30% total. Ten steps of × 1.03 give × 1.344, which is 34.4% more, because each year’s 3% is taken on a bigger amount. Another mistake is to write the factor as 0.03 or as 3. The rule: factor = 1 + rate for growth and 1 − rate for decay, with the rate written as a decimal. Check any exponential model by computing one or two steps by hand before you trust the formula.

You deposit $1,000 at 6% interest. If the bank pays once a year, after 5 years you have 1,000 × 1.06^5 ≈ $1,338.23. Many banks pay compound interest monthly instead: each month they pay 6% ÷ 12 = 0.5%, so the factor is 1.005 and there are 5 × 12 = 60 payments. Now you have 1,000 × 1.005^60 ≈ $1,348.85. The formula is A = P(1 + r/n)^(nt), where P is the principal, the starting amount, r is the yearly rate as a decimal, n is the number of payments per year and t is the years. Monthly compounding earned about $10.62 more than yearly. Daily compounding (n = 365) gives about $1,349.83. The gains keep shrinking.

Push n higher and higher and the amount creeps toward a ceiling. Take the simplest case: $1 at 100% interest for 1 year, so A = (1 + 1/n)^n. With n = 1 you get 2. With n = 2, (1.5)^2 = 2.25. With n = 12, about 2.613. With n = 365, about 2.7146. With n = 10,000, about 2.7181. The values never pass 2.71828…, the number e. Like π, e never ends and never repeats. Growth that compounds at every instant is continuous growth, and its formula is A = P × e^(rt). For the $1,000 at 6% for 5 years, A = 1,000 × e^0.3 ≈ 1,000 × 1.3499 = $1,349.86, only 3 cents more than daily compounding.

The number e shows up wherever change happens smoothly and constantly: bacteria, cooling coffee, radioactive atoms and money. The e^x key on a calculator computes it. A common error is to type the rate as a percent, e^(6 × 5) instead of e^(0.06 × 5); the first gives a nonsense answer in the trillions. Another is to mix up n and t: with monthly compounding for 5 years the exponent is 60, not 5. Write the formula, label each letter with its value, then compute.

A patient takes 200 mg of a medicine whose half-life in the body is 6 hours. That means every 6 hours, half of what is left is gone. After 6 hours: 100 mg. After 12 hours: 50 mg. After 18 hours: 25 mg. After 24 hours: 12.5 mg. The pattern is 200 × (1/2)^k, where k is the number of half-lives. To get k from clock time, divide the time by the half-life: 24 ÷ 6 = 4 half-lives, and 200 × (1/2)^4 = 200 ÷ 16 = 12.5 mg. In general A = A0 × (1/2)^(t/h), where A0 is the starting amount, t the time and h the half-life. The exponent can be a fraction: after 9 hours, t/h = 1.5, and 200 × (1/2)^1.5 ≈ 200 × 0.354 = 70.7 mg.

The same rule dates old bones and wood. Carbon-14 is a radioactive form of carbon with a half-life of about 5,730 years. A living plant keeps a steady amount; when it dies, the carbon-14 starts to decay with nothing to replace it. A piece of charcoal from an ancient fire holds 1/4 of the carbon-14 a living plant would have. One quarter is (1/2)^2, so two half-lives have passed: 2 × 5,730 = 11,460 years. If it held 1/8, three half-lives have passed, about 17,190 years. This is radiocarbon dating, and it works because the decay factor never changes: half is gone every 5,730 years no matter how much you start with.

Percent decay and half-life describe the same kind of curve. If a quantity loses 12% each year, the factor is 0.88 and the model is A = A0 × 0.88^t. Both forms are exponential decay, with a base below 1. The classic mistake is to think a half-life of 6 hours means everything is gone in 12 hours. Halving twice leaves a quarter, never zero; the curve gets close to zero but never reaches it. Another mistake is to divide the amount by the number of hours. Always count half-lives first, then multiply by 1/2 that many times.

Suppose a colony doubles every hour and you want to know when 1 cell becomes 32. You are asking: 2 to what power is 32? Since 2^5 = 32, the answer is 5. A logarithm is exactly this question written as a function. We write log_2(32) = 5, read ’log base 2 of 32 equals 5.’ The general rule: log_b(y) = x means the same thing as b^x = y. The base b is the growth factor, y is the amount, and the log gives back the exponent. So log_10(1,000) = 3 because 10^3 = 1,000, and log_5(25) = 2 because 5^2 = 25.

Some values follow straight from the definition. log_b(1) = 0 for any base, because b^0 = 1. log_b(b) = 1, because b^1 = b. Negative answers are fine: log_5(1/25) = −2 because 5^(−2) = 1/25. Fractions are fine too: log_4(8) = 1.5 because 4^1.5 = (√4)^3 = 2^3 = 8. What is not fine is a zero or negative input: log_2(0) and log_2(−8) do not exist, because no power of 2 is zero or negative. The domain of a logarithm is the positive numbers only.

Two mistakes come up again and again. The first is dividing instead of asking for an exponent: log_2(8) is not 8 ÷ 2 = 4; it is 3, because 2^3 = 8. The second is switching the base and the input: log_8(2) is 1/3, not 3, since 8^(1/3) = 2. When in doubt, rewrite the log as an exponential sentence and check it. log_3(81) = ? becomes 3^? = 81; since 3 × 3 × 3 × 3 = 81, the answer is 4. That rewriting step is the whole method.

Two bases matter so much that they get their own keys. The common logarithm uses base 10 and is written log(y) with no base shown. It counts powers of ten: log(1,000) = 3, log(1,000,000) = 6, log(0.01) = −2, and log(500) ≈ 2.699 because 500 sits between 10^2 and 10^3, closer to the top. The natural logarithm uses base e and is written ln(y). It undoes e^x: ln(e^2) = 2, ln(e) = 1, ln(1) = 0. Since e ≈ 2.718, ln(10) ≈ 2.303 and ln(2) ≈ 0.693. A calculator’s LOG key is base 10; the LN key is base e.

What if the base is 3? Few calculators have a log_3 key, so use the change-of-base rule: log_b(y) = log(y) ÷ log(b), and the same with ln. Example: log_3(20) = ln(20) ÷ ln(3) ≈ 2.9957 ÷ 1.0986 ≈ 2.727. Check it: 3^2 = 9 and 3^3 = 27, so an exponent of 2.727 lands sensibly between 2 and 3, near the top. Another: log_2(10) = ln(10) ÷ ln(2) ≈ 2.3026 ÷ 0.6931 ≈ 3.322, meaning it takes about 3.3 doublings to multiply by 10. That number explains why 10 doublings is about 1,000 (2^10 = 1,024).

The mistake to avoid is dividing the numbers before taking the log: log_3(20) is not log(20 ÷ 3). The rule says log of the input divided by log of the base, two separate logs. A second mistake is expecting a log to grow fast. Logs grow slowly: log(10) = 1, log(100) = 2, log(1,000) = 3. Multiplying the input by 10 adds only 1 to the output. That slow growth is exactly why scientists use logs to tame huge ranges, from earthquake energy to sound intensity.

Exponents add when you multiply powers: 2^3 × 2^4 = 2^7. Since logs are exponents, logs add when inputs multiply: log_2(8 × 16) = log_2(8) + log_2(16) = 3 + 4 = 7, and indeed 8 × 16 = 128 = 2^7. This is the product rule. The quotient rule follows the same way: log_3(81 ÷ 9) = log_3(81) − log_3(9) = 4 − 2 = 2, and 81 ÷ 9 = 9 = 3^2. The power rule handles exponents inside the log: log(5^3) = 3 × log(5). With log(5) ≈ 0.699, that gives about 2.097, and 5^3 = 125 is indeed a little past 100 = 10^2.

These rules let you build many logs from a few. Take log(2) ≈ 0.301 and log(3) ≈ 0.477. Then log(6) = log(2 × 3) ≈ 0.778. log(12) = log(4 × 3) = 2 log(2) + log(3) ≈ 0.602 + 0.477 = 1.079. log(1.5) = log(3 ÷ 2) ≈ 0.477 − 0.301 = 0.176. log(8) = 3 log(2) ≈ 0.903. Before calculators, people did all their multiplying this way: look up two logs, add them, look up the answer. The power rule is the key to solving equations later, because it pulls an unknown exponent down where you can divide it out: log(2^x) = x × log(2).

The rule students invent that does not exist: log(a + b) = log(a) + log(b). Test it: log(10 + 100) = log(110) ≈ 2.04, but log(10) + log(100) = 1 + 2 = 3. Not equal. Logs turn multiplication into addition, and there is no rule for the log of a sum. Also, log(a) × log(b) is not log(ab), and log(a) ÷ log(b) is not log(a ÷ b); that second one is the change-of-base formula for log_b(a), a different thing. Keep the three true rules on a card and refuse the rest.

A colony of 5 cells doubles every hour. When will it reach 640 cells? Write 5 × 2^x = 640. Divide both sides by 5: 2^x = 128. Now ask the log question: 2 to what power is 128? Since 2^7 = 128, x = 7 hours. When the answer is a whole power of the base, you can match bases and read the exponent off. Another example: 9^x = 27. Write both as powers of 3: (3^2)^x = 3^3, so 3^(2x) = 3^3, so 2x = 3 and x = 1.5. Check: 9^1.5 = (√9)^3 = 3^3 = 27.

Usually the answer is not a whole power. Solve 3^x = 50. No integer works, since 3^3 = 27 and 3^4 = 81. Take the natural log of both sides: ln(3^x) = ln(50). The power rule moves the x out front: x × ln(3) = ln(50). Divide: x = ln(50) ÷ ln(3) ≈ 3.912 ÷ 1.099 ≈ 3.561. Check: 3^3.561 ≈ 50, and 3.561 sits between 3 and 4 as expected. You can use log instead of ln; the ratio comes out the same. This is the standard method for any exponential equation: isolate the power, take a log of both sides, use the power rule, divide.

Money questions work the same way. $1,000 at 5% per year: when does it double? 1,000 × 1.05^t = 2,000, so 1.05^t = 2. Take ln: t × ln(1.05) = ln(2), so t = 0.6931 ÷ 0.04879 ≈ 14.2 years. The error to watch for is taking the log before isolating the power: ln(5 × 2^x) is not 5 × ln(2^x). Divide the 5 away first. A second error is dividing the exponent by the base, writing x = 50 ÷ 3. The exponent comes out only through a log.

Solve log_2(x + 3) = 5. The equation says: 2 to the 5th power is x + 3. So x + 3 = 32 and x = 29. Check by putting 29 back: log_2(32) = 5. True. The method is the definition run backward: whenever a single log equals a number, rewrite it as base^(number) = input. Another: log_3(2x − 1) = 2 becomes 2x − 1 = 9, so x = 5. And ln(x) = 2 becomes x = e^2 ≈ 7.389. If the equation has ln on one side, the base is e.

When two logs are added, combine them first with the product rule. Solve log(x) + log(x − 3) = 1. Combine: log(x(x − 3)) = 1. Rewrite: x(x − 3) = 10^1 = 10. Expand: x^2 − 3x − 10 = 0, which factors as (x − 5)(x + 2) = 0, so x = 5 or x = −2. Now the essential step: check each in the original. x = 5 gives log(5) + log(2) = log(10) = 1. Good. x = −2 gives log(−2), which does not exist. So x = −2 is an extraneous solution, a number the algebra produced that the original equation cannot accept. The only solution is x = 5.

Extraneous solutions appear because combining logs can create an equation that is true more often than the original. A log’s input must be positive, so any candidate that makes an input zero or negative is thrown out. Do not skip the check because the algebra looked fine. Another error is rewriting log(x) + log(x − 3) = 1 as x + (x − 3) = 10; the sum of logs is the log of a product, not the log of a sum. And when both sides are single logs with the same base, as in log_4(3x) = log_4(x + 8), you may set the inputs equal: 3x = x + 8, so x = 4. Check: log_4(12) on both sides.

A city grows 2% a year. How long to double? Set 1.02^t = 2 and take ln: t = ln(2) ÷ ln(1.02) ≈ 0.6931 ÷ 0.0198 ≈ 35 years. For continuous growth at rate r, the model is e^(rt) = 2, so rt = ln(2) and t = ln(2) ÷ r. With r = 0.02, t = 0.6931 ÷ 0.02 ≈ 34.7 years. The two answers are close because 2% is small. Since ln(2) ≈ 0.693, dividing 69.3 by the percent rate gives the doubling time; people round to 70 and call it the rule of 70. At 7% a year, 70 ÷ 7 = 10 years to double. At 1%, about 70 years.

Half-life from a decay rate uses the same steps with 1/2. A medicine leaves the blood at 5% per hour: 0.95^t = 0.5. Take ln: t × ln(0.95) = ln(0.5), so t = (−0.6931) ÷ (−0.0513) ≈ 13.5 hours. Both logs are negative, and the negatives cancel; a positive time is the sign that you did it right. Rule-of-70 style: 70 ÷ 5 = 14 hours, a fine mental estimate. The reverse question also comes up: a population doubles every 12 years, so what is its yearly rate? Solve b^12 = 2: b = 2^(1/12) ≈ 1.0595, about 5.95% per year, nearly 6%.

The rule of 70 fails when the rate is large. At 50% a year, 70 ÷ 50 = 1.4 years, but the exact answer is ln(2) ÷ ln(1.5) ≈ 0.6931 ÷ 0.4055 ≈ 1.71 years. Use the rule for quick estimates and the log for real answers. One more trap: the doubling time does not depend on the starting amount. A town of 5,000 and a city of 5 million growing at 2% both double in about 35 years. Exponential growth is about the factor, never the size.

Earthquake magnitude is a log scale of shaking. A step of 1 in magnitude means the seismograph needle swings 10 times farther. So a magnitude 7 quake shakes 10 times more than a 6 and 10 × 10 = 100 times more than a 5. The difference in magnitude is the exponent: shaking ratio = 10^(difference). Energy grows faster, by about 10^1.5 ≈ 31.6 per step, which is where ’thirty times’ comes from. Energy ratio = 10^(1.5 × difference). From 5 to 7, that is 10^3 = 1,000 times the energy. This is why a change from 5.2 to 6.2 is a very different earthquake.

Sound uses the decibel: dB = 10 × log(I ÷ I0), where I is the sound’s intensity and I0 is the quietest sound a person can hear. Every 10 dB means 10 times the intensity. A 60 dB conversation is 10^3 = 1,000 times the intensity of a 30 dB whisper. Going the other way, a sound with 500 times the base intensity is 10 × log(500) ≈ 10 × 2.7 = 27 dB. Chemists use pH = −log[H+], where [H+] is the hydrogen ion concentration in moles per liter. Pure water has [H+] = 10^(−7), so pH 7. Lemon juice near pH 2 has 10^5 = 100,000 times the hydrogen ion concentration of water. Lower pH means more acid.

Each log scale exists for the same reason: the real quantities span a range far too wide for a normal number line. Earthquake energies differ by factors of billions; the log turns billions into a difference of 6. The trap is to treat the scale as linear. A magnitude 8 is not ’twice’ a 4; it releases about 10^6 = 1,000,000 times the energy. A pH of 4 is not ’a little more acid’ than 6; it is 100 times more. When you read a log scale, subtract to get the exponent, then raise 10 to that power to get the real ratio.

A rumor starts with 10 students in a school of 1,000. At first it spreads fast, nearly doubling each day, because almost everyone who hears it has friends who have not. But the rumor cannot pass 1,000 people. As it nears that ceiling it slows: the students still left to tell are getting rare. The curve rises, bends and flattens into an S shape. This is logistic growth. The ceiling is called the carrying capacity, L. Real populations of deer, bacteria in a dish, or users of an app follow this shape because food, space or new people run out.

The formula is P(t) = L ÷ (1 + C × e^(−kt)). L is the ceiling, k sets the speed, and C is fixed by the starting value. For the rumor, L = 1,000 and P(0) = 10, so 1,000 ÷ (1 + C) = 10, giving 1 + C = 100 and C = 99. With k = 0.8 per day, P(t) = 1,000 ÷ (1 + 99e^(−0.8t)). Day 5: e^(−4) ≈ 0.0183, so 99 × 0.0183 ≈ 1.81, and P ≈ 1,000 ÷ 2.81 ≈ 356 students. Day 10: e^(−8) ≈ 0.000335, 99 × 0.000335 ≈ 0.033, P ≈ 1,000 ÷ 1.033 ≈ 968. Day 15: about 999. The last five days add only about 30 students; the curve has flattened against its limit.

Where does the S bend? The steepest point is at half the ceiling, P = L ÷ 2 = 500 here. Before that point growth speeds up; after it, growth slows. Find that day by solving 99e^(−0.8t) = 1, so e^(−0.8t) = 1/99, so −0.8t = ln(1/99) ≈ −4.595, so t ≈ 5.74 days. The common mistake is to keep using an exponential model past the point where the limit matters; an exponential rumor would ’reach’ 5,000 students in a school of 1,000. When a situation has a hard limit, choose logistic. When the limit is far away, exponential is fine for the early stretch, and the two curves agree there.

Tie a string to a pin, cut it to the circle’s radius, and lay it along the edge of the circle. The angle it spans at the center is one radian. Because the whole edge is 2πr long, a full turn holds 2π radians, about 6.28 of them. Half a turn, 180°, is π radians, so 1 radian is 180° ÷ π ≈ 57.3°. Radians measure angles with the circle’s own ruler, and every formula in physics and calculus gets simpler when angles are in radians. The trade is one conversion: multiply a degree measure by π/180 to get radians, and multiply radians by 180/π to get degrees.

Convert 60°: 60 × π/180 = π/3 radians. Convert 150°: 150 × π/180 = 5π/6. Convert 3π/4 radians: 3π/4 × 180/π = 135°. Convert 2 radians: 2 × 180/π ≈ 114.6°. Exact answers keep π; use decimals only when a measurement needs them. The fractions to know cold are 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2, 180° = π, 270° = 3π/2 and 360° = 2π. Any other angle you can build from these: 210° = 180° + 30° = π + π/6 = 7π/6.

Radians pay off first in arc length. Since one radian spans an arc equal to the radius, an angle of θ radians spans an arc of s = rθ. A bike wheel of radius 30 cm turns through 2π/3 radians: the tire rolls 30 × 2π/3 = 20π ≈ 62.8 cm. With degrees you would need the extra factor of 360, s = (120/360) × 2π × 30, and get the same 62.8 cm the long way. The classic mistake is to plug degrees into s = rθ. If someone writes s = 30 × 120 = 3,600 cm, the units give it away: 36 meters from one third of a wheel turn is impossible.

Draw a circle of radius 1 centered at the origin, the unit circle. Start at the point (1, 0) and walk counterclockwise through an angle θ. Wherever you stop, the x-coordinate is the cosine of θ and the y-coordinate is the sine of θ. That is the whole definition, and it extends the right-triangle ratios to every angle, even past 90° and negative ones. At θ = 0 you are at (1, 0): cos 0 = 1, sin 0 = 0. At θ = π/2 you are at (0, 1): cos(π/2) = 0, sin(π/2) = 1. At θ = π: (−1, 0). At θ = 3π/2: (0, −1).

Between those, a few exact points come from two special triangles. At π/6 (30°) the point is (√3/2, 1/2): so cos(π/6) ≈ 0.866 and sin(π/6) = 0.5. At π/4 (45°) it is (√2/2, √2/2), both about 0.707. At π/3 (60°) it is (1/2, √3/2). Notice sine and cosine trade places between 30° and 60°. Every other special angle is a reflection of these. 5π/6 (150°) sits in the second quadrant, a mirror of π/6 across the y-axis: (−√3/2, 1/2). So cos(5π/6) = −√3/2 and sin(5π/6) = 1/2. 7π/6 (210°) is the mirror through the origin: (−√3/2, −1/2).

Signs follow the quadrants. In quadrant I both coordinates are positive. In quadrant II, x is negative, so cosine is negative and sine positive. In quadrant III both are negative. In quadrant IV cosine is positive and sine negative. The angle back to the nearest part of the x-axis is the reference angle; it tells you the size, and the quadrant tells you the signs. For 4π/3 (240°): reference angle π/3, quadrant III, so the point is (−1/2, −√3/2). The most common mistake is to switch sine and cosine. Remember that x comes first in a point and c comes before s in the alphabet: (cos, sin).

The tangent of an angle is the slope of the line from the origin to its unit-circle point: tan θ = sin θ ÷ cos θ = y ÷ x. At π/4 the point is (√2/2, √2/2), so tan(π/4) = 1: the line rises at 45°. At π/3, tan = (√3/2) ÷ (1/2) = √3 ≈ 1.732. At π/6, tan = (1/2) ÷ (√3/2) = 1/√3 = √3/3 ≈ 0.577. At π/2 the point is (0, 1) and you would divide by zero, so tan(π/2) is undefined; the line is vertical and has no slope. Tangent is positive where sine and cosine share a sign (quadrants I and III) and negative where they differ (II and IV).

Every point on the unit circle satisfies x^2 + y^2 = 1, because that is the equation of the circle. Since x = cos θ and y = sin θ, this gives the Pythagorean identity: sin^2 θ + cos^2 θ = 1, true for every angle. It is the Pythagorean theorem for a triangle whose hypotenuse is 1. Use it to find one function from the other. Suppose 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: cos θ = −4/5. Then tan θ = (3/5) ÷ (−4/5) = −3/4.

The sign step is the one people skip. cos^2 θ = 16/25 has two square roots, 4/5 and −4/5, and only the quadrant can pick. Without a quadrant, the honest answer is cos θ = ±4/5. Another slip is reading sin^2 θ as sin(θ^2); the notation sin^2 θ means (sin θ)^2, the sine squared, not the sine of a squared angle. One more example: if cos θ = 5/13 and θ is in quadrant IV, then sin^2 θ = 1 − 25/169 = 144/169, so sin θ = −12/13 (negative in IV), and tan θ = (−12/13) ÷ (5/13) = −12/5.

Let θ increase from 0 to 2π and plot the height sin θ on a graph, with θ across and sin θ up. The point rises from 0 to 1 at π/2, falls back through 0 at π, drops to −1 at 3π/2 and returns to 0 at 2π. Then the circle repeats and so does the graph. The result is the sine wave. Its period is 2π: after every 2π the pattern starts again, so sin(θ + 2π) = sin θ. Its amplitude is 1: the wave rises 1 above and 1 below its center line, the midline y = 0. The five key points are (0, 0), (π/2, 1), (π, 0), (3π/2, −1) and (2π, 0).

The cosine graph plots the x-coordinate instead. It starts at its top: (0, 1), (π/2, 0), (π, −1), (3π/2, 0), (2π, 1). Same period, same amplitude, same midline, but shifted: the cosine wave is the sine wave slid left by π/2, so cos θ = sin(θ + π/2). If you cover the labels, you cannot tell the two apart except by where they start. Sine starts on the midline heading up; cosine starts at a peak. Each is a periodic function, which means its values repeat after a fixed interval.

To sketch either graph by hand, mark the five key points across one period and connect them with a smooth curve, not straight segments and not sharp corners. The common error is drawing a pointed top like a mountain; a sine wave is rounded at its peaks because the point on the circle slows its up-and-down motion near the top. Another error is making the wave cross the axis at π/2; check the circle: at π/2 the height is 1, the maximum. Once you can name the five key points of sine and cosine without thinking, every other wave in this chapter is a stretched or shifted copy.

A wave rarely has amplitude 1 and period 2π. Take y = 3 sin(2x) + 1. The 3 stretches heights: amplitude 3, so the wave rises 3 above and 3 below its midline. The +1 is a vertical shift that lifts the midline to y = 1, so the maximum is 1 + 3 = 4 and the minimum is 1 − 3 = −2. The 2 inside squeezes the wave sideways: a full cycle now happens when 2x goes from 0 to 2π, that is, when x goes from 0 to π. Period = 2π ÷ B = 2π ÷ 2 = π. So the rule for y = A sin(Bx) + D: amplitude |A|, period 2π/B, midline y = D. A negative A flips the wave upside down, but the amplitude is still |A|.

Going the other way, build the formula from the facts. A Ferris wheel has radius 20 m, its center sits 22 m above the ground, and one turn takes 4 minutes. Height above ground swings from 2 m to 42 m: midline D = 22, amplitude A = 20. The period is 4 minutes, so B = 2π ÷ 4 = π/2. A rider boards at the bottom, so the height starts at its minimum: use a negative cosine. h(t) = 22 − 20 cos(πt/2). Check: t = 0 gives 22 − 20 = 2. t = 2 gives 22 − 20 cos(π) = 22 + 20 = 42, the top, halfway through the turn. t = 1 gives 22 − 20 cos(π/2) = 22 m, the middle.

The two errors here are twins. First, the period is 2π ÷ B, not B: y = sin(4x) has period π/2, not 4, and a period of 12 hours needs B = 2π ÷ 12 = π/6. Second, the amplitude is the distance from the midline to the peak, not from the bottom to the top. A wave that runs from 2 to 42 has amplitude 20, not 40; the 40 is the full swing. Find the midline as the average of the maximum and minimum, (42 + 2) ÷ 2 = 22, and the amplitude as half the difference, (42 − 2) ÷ 2 = 20.

Slide a sine wave to the right by π/3 and its equation becomes y = sin(x − π/3): every feature happens π/3 later. That horizontal slide is the phase shift. In the full form y = A sin(B(x − C)) + D the shift is C, to the right if C is positive and to the left if negative. The catch is that B must be factored out first. y = sin(2x − π/2) looks like a shift of π/2, but factoring gives sin(2(x − π/4)), so the shift is π/4. Divide the inside constant by B, or factor, before you read the shift. For y = 2 sin(3(x − π/6)) + 1: amplitude 2, period 2π/3, shift right π/6, midline 1.

To graph a shifted wave, find the five key points of the plain wave and add C to every x. For y = sin(x − π/3), the plain points x = 0, π/2, π, 3π/2, 2π become π/3, 5π/6, 4π/3, 11π/6, 7π/3, with the same heights 0, 1, 0, −1, 0. A shift of exactly a quarter period turns sine into cosine, which is why both functions exist: cos x = sin(x + π/2). In modeling, the phase shift is where the wave starts; a tide model built on cosine starts at a high tide, so C is the time of a high tide.

The tangent graph is different. Because tan x = sin x ÷ cos x, it blows up wherever cos x = 0, at x = π/2, 3π/2, and every π after: each of those lines is a vertical asymptote, a line the graph approaches but never touches. Between asymptotes the graph climbs from far below through a zero to far above. It passes through (0, 0), (π/4, 1) and (−π/4, −1). Its period is π, not 2π, because the slope of the line through the origin repeats every half turn. For y = tan(3x) the period is π ÷ 3 and the asymptotes are at x = π/6 + kπ/3. The common error is to give tangent the period 2π and to draw it with peaks; it has no maximum or minimum.

Chicago’s daylight runs from about 9.2 hours on the shortest day to about 15.2 hours on the longest. Midline: (15.2 + 9.2) ÷ 2 = 12.2 hours. Amplitude: (15.2 − 9.2) ÷ 2 = 3 hours. Period: 365 days, so B = 2π/365. The daylight crosses its midline heading up near March 21, about day 80 of the year, which is where a sine wave starts. So the model is D(t) = 12.2 + 3 sin(2π(t − 80)/365), with t the day of the year. Test it: June 21 is day 172; (172 − 80) ÷ 365 × 2π ≈ 1.58 radians, about π/2, so sin ≈ 1 and D ≈ 15.2. Good. January 1: (1 − 80) ÷ 365 × 2π ≈ −1.36, sin ≈ −0.98, D ≈ 12.2 − 2.9 = 9.3 hours.

Tides follow the same recipe with a shorter clock. Suppose high tide at a pier is 8 ft at 2:00 a.m. and low tide is 2 ft at 8:00 a.m., six hours later, so the full cycle is 12 hours. Midline: (8 + 2) ÷ 2 = 5 ft. Amplitude: 3 ft. B = 2π ÷ 12 = π/6. A cosine wave starts at its peak, so use cosine with a shift to the peak time: h(t) = 5 + 3 cos(π(t − 2)/6), with t in hours after midnight. Check t = 2: cos(0) = 1, h = 8. Check t = 8: cos(π) = −1, h = 2. Check t = 5: cos(π/2) = 0, h = 5, the midline halfway between. At t = 14 the tide is high again, as it should be 12 hours later.

Real ocean tides repeat about every 12 hours and 25 minutes, because the Moon moves along its orbit as the Earth turns, so a careful model uses that period instead of 12. Lake Michigan has only tiny tides, a few centimeters, so Chicago beaches never notice this cycle. The recipe: midline from the average of max and min, amplitude from half the difference, B from 2π ÷ period, then pick sine or cosine and a shift that matches where the peak or midline crossing happens. The usual mistake is to shift by the peak time while using sine; sine starts at the midline, so a peak-time shift belongs with cosine.

A tuning fork for the note A vibrates 440 times each second. The air pressure at your ear rises and falls as a sine wave with frequency 440 hertz (Hz), which means 440 cycles per second. The period is the time for one cycle: T = 1 ÷ f = 1 ÷ 440 ≈ 0.00227 seconds, about 2.3 thousandths of a second. As a function of time, p(t) = A sin(2π × 440 × t) = A sin(880πt). The B value is 2πf because the period must be 2π ÷ B = 1 ÷ f. Doubling the frequency to 880 Hz gives the same note an octave higher; halving it to 220 Hz drops an octave.

Amplitude A is loudness: a louder A has bigger pressure swings but the same 440 cycles per second, so the pitch does not change. A softer A has the same period and a smaller A. A different note has a different B. Middle C is about 262 Hz, so its wave is p(t) = A sin(2π × 262 × t) with period about 0.0038 s. Most real sounds are sums of several sine waves at different frequencies, which is why a violin and a flute playing the same A sound different: the same main wave, with different smaller waves stacked on top.

The electricity in a U.S. wall outlet is also a sine wave. The voltage swings back and forth 60 times a second, 60 Hz, with a period of 1/60 ≈ 0.0167 s. Its peak is about 170 volts, so v(t) = 170 sin(120πt), where 120π = 2π × 60. The 120 volts printed on appliances is an averaged value, not the peak. When you read any wave in time, translate B into frequency with f = B ÷ 2π and period with T = 1 ÷ f. The common slip is to write sin(440t) for a 440 Hz tone; that wave has period 2π ÷ 440 ≈ 0.0143 s and frequency about 70 Hz, a much lower hum.

Solve sin x = 1/2 for 0 ≤ x < 2π. The reference angle with sine 1/2 is π/6. Sine is positive in quadrants I and II, so the two solutions on one turn are x = π/6 and x = π − π/6 = 5π/6. Check the second: sin(5π/6) is the y-coordinate of (−√3/2, 1/2), which is 1/2. Because sine repeats every 2π, every solution has the form π/6 + 2πk or 5π/6 + 2πk for any integer k. That is the general solution. Most problems ask only for one turn, but ’find all solutions’ means to write those families.

When the equation has extra pieces, isolate the trig function first. 2 cos x + 1 = 0 becomes cos x = −1/2. Reference angle: π/3, since cos(π/3) = 1/2. Cosine is negative in quadrants II and III: x = π − π/3 = 2π/3 and x = π + π/3 = 4π/3. For tan x = 1, the reference angle is π/4 and tangent is positive in I and III: x = π/4 and 5π/4. Since tangent’s period is π, all solutions are π/4 + πk. For a value that is not special, use the inverse function: sin x = 0.3 gives x = arcsin(0.3) ≈ 0.305, and the quadrant II partner is π − 0.305 ≈ 2.837.

The most common mistake is to stop after one answer. A calculator’s arcsin gives only one angle, between −π/2 and π/2, and the second solution on the turn is always π minus that. For cosine, arccos gives an angle between 0 and π, and the partner is 2π minus it. Another mistake is dividing an equation like sin x cos x = 0 by sin x; that throws away the solutions where sin x = 0. Factor instead: sin x = 0 or cos x = 0, giving x = 0, π/2, π, 3π/2. Always test each answer by substituting it back.

A surveyor on the Chicago Riverwalk wants the distance across the water to a doorway on the far bank without crossing. She marks two points A and B on her side, 100 m apart, and calls the doorway C. At A she measures the angle to C as 40°; at B the angle is 60°. The third angle is C = 180° − 40° − 60° = 80°. In any triangle each side is proportional to the sine of the angle across from it: a ÷ sin A = b ÷ sin B = c ÷ sin C, the law of sines. Here c = 100 is the side across from C. So b = c × sin B ÷ sin C = 100 × sin 60° ÷ sin 80° ≈ 100 × 0.866 ÷ 0.985 ≈ 87.9 m, the distance from A to the doorway.

The distance from B: a = 100 × sin 40° ÷ sin 80° ≈ 100 × 0.643 ÷ 0.985 ≈ 65.3 m. Check the sizes: the largest side should sit across from the largest angle, and 100 m is across from 80°, the largest, while 65.3 m is across from 40°, the smallest. It fits. The law works because the triangle’s height, dropped from any vertex, can be written two ways: h = b sin A = a sin B, and dividing gives a ÷ sin A = b ÷ sin B. Any pair of the three ratios can be used; pick the pair where three of the four pieces are known.

A second setup: angle A = 30°, angle C = 45°, side a = 8. Then c = 8 × sin 45° ÷ sin 30° = 8 × 0.7071 ÷ 0.5 ≈ 11.3. Watch two traps. First, the law pairs each side with its own opposite angle; matching a with B gives nonsense. Second, when you know two sides and an angle that is not between them, the sine may point to two possible triangles, because sin θ = sin(180° − θ). And if the problem gives two sides and the angle between them, no side-angle pair is known, so the law of sines cannot start; that is the job of the law of cosines.

Two roads leave a town at an angle of 60°. One car drives 5 miles on the first road, another drives 7 miles on the second. How far apart are they? That triangle has two sides and the angle between them, so use the law of cosines: c^2 = a^2 + b^2 − 2ab cos C, where C is the included angle between sides a and b, and c is across from it. Here c^2 = 5^2 + 7^2 − 2 × 5 × 7 × cos 60° = 25 + 49 − 70 × 0.5 = 74 − 35 = 39, so c = √39 ≈ 6.24 miles. If the angle were 90°, cos 90° = 0 and the formula becomes c^2 = a^2 + b^2, the Pythagorean theorem. The extra term corrects for the angle being smaller or larger than a right angle.

With three sides known, solve for an angle. A triangle has sides 5, 6 and 7. To find the angle C across from the 7: 7^2 = 5^2 + 6^2 − 2 × 5 × 6 × cos C, so 49 = 61 − 60 cos C, so 60 cos C = 12, cos C = 0.2, and C = arccos(0.2) ≈ 78.5°. Another: sides 3, 5, 7, angle across from the 7: 49 = 9 + 25 − 30 cos C, so 30 cos C = −15, cos C = −0.5, C = 120°. A negative cosine means the angle is obtuse, and the largest angle sits across from the largest side, as it should.

Two errors ruin this formula. Dropping the minus sign, writing a^2 + b^2 + 2ab cos C, makes every answer too long; for the two cars it gives c^2 = 109 and 10.4 miles. But at 60° the cars are closer together than they would be at 90°, where the distance would be √74 ≈ 8.6 miles, so 10.4 cannot be right. The second error is to use the wrong angle: the angle in the formula must be the one between the two sides you square. Choosing the law: two angles and a side, law of sines; two sides and the included angle, or three sides, law of cosines.