Trigonometric functions and equations

Review the circle and radians

On the unit circle, the endpoint of an angle x has coordinates (cos x, sin x). One complete turn is 2π rad, corresponding to 360°. In this lesson, arguments written without a degree symbol are in radians. To recover exact values, use the reference angles π/6, π/4 and π/3, then choose the signs from the quadrant.

For example, sin(7π/6) = −1/2 because the angle lies in the third quadrant and has reference angle π/6. The cosine is −√3/2 and the tangent is √3/3. Sine reads the point's height and cosine its horizontal position. This interpretation lets us determine how many points have a specified height or horizontal coordinate.

The identity sin² x + cos² x = 1 comes from the circle's equation. Tangent is sin x/cos x and requires cos x ≠ 0. An identity is an equality valid throughout the common domain; an equation asks for the inputs at which a particular equality holds. Confusing these tasks can lead to claiming that a condition holds for every angle when it holds only for some.

From the point to periodic curves

As x increases from 0 to 2π, sine takes the values 0, 1, 0, −1, 0 at arguments 0, π/2, π, 3π/2, 2π. Cosine takes 1, 0, −1, 0, 1 at those same arguments. Between these points, values vary continuously, forming smooth curves. The table organises notable points; the circle explains the behaviour between them.

Sine and cosine have domain ℝ and range [−1, 1]. Both have fundamental period 2π: f(x + 2π) = f(x) for every x, and no smaller positive period works for every argument. The period is a change in input that reproduces the whole pattern, not just the distance between two points of the same height.

Tangent has domain ℝ excluding π/2 + kπ, with integer k, and range ℝ. Its fundamental period is π, because advancing half a turn simultaneously changes the signs of sine and cosine, preserving their quotient. The lines x = π/2 + kπ are vertical asymptotes; tangent does not take an infinite value at those points, since it is undefined there.

Amplitude, midline and period

Consider y = A sin[B(x − h)] + D, with A ≠ 0 and B ≠ 0. The midline is y = D, the amplitude is |A| and the range is [D − |A|, D + |A|]. The sign of A can reflect the graph across the midline, but amplitude is a distance and is never negative. The same structure applies to cosine.

To find the period T, we want the argument to change by one complete turn: |B|T = 2π. Therefore T = 2π/|B|. The parameter B is not the period; the larger |B| is, the faster the argument completes turns and the smaller the period. The horizontal shift h is read after putting the argument into the form B(x − h).

For example, 2 sin(3x − π) + 4 = 2 sin[3(x − π/3)] + 4. The amplitude is 2, the midline is 4, the period is 2π/3 and the shift is π/3 to the right. Reading −π directly as the shift would ignore the factor 3. If A = 0 or B = 0, the function becomes constant and this sinusoid's fundamental-period formula does not apply.

Two families and infinitely many turns

Solving sin x = a first requires |a| ≤ 1; outside that interval there is no real solution. If α is an angle with sin α = a, the general families are x = α + 2kπ or x = π − α + 2kπ, with integer k. For a = ±1, the families may coincide, so repetitions must be removed. Geometrically, a height usually cuts the circle at two points per turn.

For cos x = a, with |a| ≤ 1, if cos α = a, we have x = α + 2kπ or x = −α + 2kπ. For tan x = a, one solution α generates x = α + kπ. Tangent allows any real value a, but inputs must remain in its domain. A calculator's principal value is just one representative, not the complete list.

The principal inverse functions choose intervals to give a unique answer: arcsin returns values in [−π/2, π/2], arccos in [0, π] and arctan in ]−π/2, π/2[. The notation “sin⁻¹” may stand for arcsin on a calculator; it does not mean 1/sin. After obtaining the principal value, use symmetry and periodicity to recover the remaining requested solutions.

Algebra applied to trigonometric expressions

An equation can be quadratic in sin x or cos x. Make a temporary substitution, such as u = sin x, and solve the polynomial equation. Then require u to lie between −1 and 1 and return to the trigonometric families. A candidate u = 2 may solve the auxiliary algebraic equation, but cannot be the sine of a real angle.

Factoring preserves possibilities that division could lose. In sin x cos x = 0, the zero-product property gives sin x = 0 or cos x = 0. Immediately dividing by sin x would eliminate all solutions where the sine is zero. Nor is it permissible to multiply an equality by an undefined expression to “cancel” a denominator.

If the argument is 2x, first solve for that argument or carefully transform the families. On 0 ≤ x < 2π, the value 2x covers [0, 4π[, that is, two turns. Searching only one turn of 2x loses solutions. When you add or divide factors in general families, the unknown's period also changes.

Periodic models and interval selection

A periodic model represents a phenomenon whose pattern repeats approximately. If a height oscillates between 2 m and 8 m, the midline is (2 + 8)/2 = 5 m and the amplitude is (8 − 2)/2 = 3 m. If the cycle lasts 12 s, the angular factor is 2π/12 = π/6 rad/s. The initial phase depends on where the motion starts and its direction.

For example, H(t) = 5 − 3 cos(πt/6) starts at the minimum: H(0) = 2. At 6 s it reaches 8 and at 12 s returns to 2. The argument πt/6 is dimensionless when t is measured in seconds with the corresponding angular factor. Amplitude has the unit of height, whereas period has the unit of time.

When a question asks “during the first cycle”, choose an interval such as [0, 12[ or [0, 12], according to the stated endpoints. A boundary instant may belong to two cycles if both are closed. The answer must list the admissible instants, rather than merely writing an infinite family without filtering it by context.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

Amplitude can be negative.

It is |A|, a distance from the midline.

The factor B is the sinusoid's period.

For B ≠ 0, the period is 2π/|B|.

The calculator gives all trigonometric solutions.

It gives a principal value; the circle and periodicity supply the rest.

Dividing by sine or cosine never changes the solutions.

It may remove zeros of that factor or use points outside the domain.

The interval for x is also the interval for 2x.

Multiplying the argument changes the interval covered and can increase the number of solutions.