1.3 Trigonometric Functions

Big idea

Trigonometry enters calculus wearing a different hat than it wore in geometry. There, sine and cosine were ratios of sides in a right triangle, defined only for acute angles. Here they are functions of a real number, defined for every input, positive or negative, and their defining picture is a point travelling around the unit circle rather than a triangle. That shift is what lets you write $\sin x$ next to $x^2$ and treat both as functions to be differentiated.

Radians come with the shift, and they are not a matter of taste. An angle in radians is the arc length it cuts on a circle of radius $1$, so the angle and the arc it subtends are the same number. Every derivative formula for a trigonometric function in this course assumes radians; in degrees the derivative of $\sin x$ picks up a stray factor of $\pi/180$. Convert to radians once, at the start of a problem, and never think about it again.

The unit circle is the lookup table. Cosine reads off the horizontal coordinate of the point at angle $\theta$ and sine reads off the vertical coordinate, which explains in one stroke why both stay between $-1$ and $1$, why $\cos^2\theta + \sin^2\theta = 1$, why cosine is even and sine is odd, and why both repeat every $2\pi$. Knowing the handful of special angles cold means you can evaluate later answers exactly instead of reaching for a decimal.

The last theme is sinusoidal modelling. Anything that oscillates steadily - a tide, a current, a vibrating string - is described by a sine or cosine adjusted by four numbers: how tall, how fast, where it starts, and what it oscillates around. These are the transformations of the previous section applied to one particular base graph, and they are worth being fast at, because such functions are the standard test cases for periodic behavior of derivatives and integrals.

Decoder

If a ray from the origin makes an angle $\theta$ with the positive $x$-axis and meets the unit circle at the point $(x, y)$, then $\cos\theta = x$ and $\sin\theta = y$.

Everything in the section follows from that sentence. The angle is measured counterclockwise from the positive $x$-axis, so a negative $\theta$ means clockwise. Since the point lies on the circle of radius $1$, its coordinates satisfy $x^2 + y^2 = 1$, which is the Pythagorean identity written in disguise. Since the point comes back to itself after a full turn, both functions repeat with period $2\pi$.

The definition also settles signs without memorization. In the second quadrant the horizontal coordinate is negative and the vertical one is positive, so cosine is negative and sine is positive there. Reflecting the point across the $x$-axis sends $\theta$ to $-\theta$, keeps $x$, and negates $y$, giving $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$.

The classic mistake is mixing units inside one problem, usually by leaving a calculator in degree mode while working with $\pi$-flavored inputs, or by reading a right-triangle value for an angle whose terminal point is in a quadrant where the sign flips. Decide the quadrant first, take the reference angle second, and attach the sign last.

Definitions and results

Radian measure. An angle of $\theta$ radians subtends an arc of length $\theta$ on a circle of radius $1$. A full turn is $2\pi$ radians, so $180$ degrees equals $\pi$ radians. Convert by multiplying by $\pi/180$ to go from degrees to radians, and by $180/\pi$ to go back.

Arc length and sector area. On a circle of radius $r$, an angle of $\theta$ radians cuts an arc and a wedge of size

$$ s = r\theta, \qquad A = \tfrac{1}{2} r^2 \theta $$

Both formulas are false in degrees, which is the most practical reason radians are the default.

The six functions. With $\cos\theta$ and $\sin\theta$ defined by the unit circle, the rest are quotients: $\tan\theta = \sin\theta/\cos\theta$, $\cot\theta = \cos\theta/\sin\theta$, $\sec\theta = 1/\cos\theta$, $\csc\theta = 1/\sin\theta$. Each is undefined wherever its denominator vanishes, so tangent and secant are undefined at odd multiples of $\pi/2$.

Special values. The angles $\pi/6$, $\pi/4$ and $\pi/3$ have coordinates $(\sqrt{3}/2, 1/2)$, $(\sqrt{2}/2, \sqrt{2}/2)$ and $(1/2, \sqrt{3}/2)$. Every other exact value in this course is one of these with a sign attached by the quadrant, or a quadrantal angle where a coordinate is $0$ or $\pm 1$.

Reference angles. The reference angle of $\theta$ is the acute angle between its terminal ray and the $x$-axis. Sine and cosine of $\theta$ equal those of the reference angle up to sign, and the quadrant supplies the sign.

Pythagorean identities. From $x^2 + y^2 = 1$ comes $\cos^2\theta + \sin^2\theta = 1$. Dividing that equation by $\cos^2\theta$ gives $1 + \tan^2\theta = \sec^2\theta$, and dividing by $\sin^2\theta$ gives $1 + \cot^2\theta = \csc^2\theta$.

Periodicity and symmetry. Sine and cosine have period $2\pi$; tangent has period $\pi$. Cosine is even and sine, tangent, cosecant and cotangent are odd.

Sinusoidal form. The graph of

$$ y = A\sin\big(B(x - C)\big) + D $$

has amplitude $|A|$, period $2\pi/|B|$, horizontal shift $C$ and midline $y = D$, so its values run from $D - |A|$ to $D + |A|$. Factor $B$ out before reading the shift.

Worked examples

Converting, then measuring an arc

An angle measures $135$ degrees. Multiply by $\pi/180$:

$$ 135 \cdot \frac{\pi}{180} = \frac{3\pi}{4} $$

On a circle of radius $8$, that angle cuts an arc of length $s = 8 \cdot 3\pi/4 = 6\pi$ and a sector of area $A = \tfrac{1}{2}(64)(3\pi/4) = 24\pi$.

Check the arc against the whole circle. The full circumference is $2\pi(8) = 16\pi$, and $135$ degrees is three eighths of a turn; three eighths of $16\pi$ is $6\pi$. The sector check matches too: three eighths of $\pi(64) = 64\pi$ is $24\pi$.

Exact values from the circle

Evaluate the three main functions at $\theta = 5\pi/6$.

This angle sits in the second quadrant, and its reference angle is $\pi - 5\pi/6 = \pi/6$. The reference coordinates are $(\sqrt{3}/2, 1/2)$; in the second quadrant the horizontal coordinate turns negative. So

$$ \cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}, \qquad \sin\frac{5\pi}{6} = \frac{1}{2}, \qquad \tan\frac{5\pi}{6} = \frac{1/2}{-\sqrt{3}/2} = -\frac{\sqrt{3}}{3} $$

Verify with the Pythagorean identity: $3/4 + 1/4 = 1$.

Repeat at $\theta = 4\pi/3$, which lies in the third quadrant with reference angle $\pi/3$. Both coordinates are negative there, so $\cos = -1/2$, $\sin = -\sqrt{3}/2$, and the tangent is the quotient of two negatives, $\tan(4\pi/3) = \sqrt{3}$.

Reading a sinusoid

Describe $y = 3\sin(2x - \pi/2) + 1$.

Factor the $2$ out of the argument before reading anything: $y = 3\sin\big(2(x - \pi/4)\big) + 1$. Now amplitude $3$, period $2\pi/2 = \pi$, shift right by $\pi/4$, midline $y = 1$. The graph oscillates between $1 - 3 = -2$ and $1 + 3 = 4$.

Locate the first maximum. Sine peaks when its argument is $\pi/2$, so $2(x - \pi/4) = \pi/2$ gives $x = \pi/2$. Check by substituting: $3\sin(\pi - \pi/2) + 1 = 3\sin(\pi/2) + 1 = 4$, the predicted maximum.

Solving a basic equation

Solve $2\cos\theta + \sqrt{3} = 0$ for $\theta$ in $[0, 2\pi)$.

Isolate the cosine: $\cos\theta = -\sqrt{3}/2$. The reference angle with cosine $\sqrt{3}/2$ is $\pi/6$, and cosine is negative in the second and third quadrants, so

$$ \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}, \qquad \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} $$

Both were read off the circle rather than guessed, and both check against the values computed above.

Practice

Start with units. Convert between degrees and radians in both directions, and use radian measure to compute arc lengths and sector areas.

Next, exact values. Place the angle, take its reference angle, and let the quadrant decide the sign.

Last, the four numbers that shape a wave. Factor the coefficient out of the argument before reading amplitude, period, shift and midline.

Quiz

Six items on radian and degree measure, exact unit circle values, and the amplitude, period and shift of a sinusoid.