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5.2 Trigonometric Functions of Real Numbers

Use sine, cosine, tangent, and reciprocal ratios.

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Lesson brief and rules

5.2 Trigonometric Functions of Real Numbers

$\tan \theta = \frac{\sin \theta }{\cos \theta }$ where $\cos \theta \ne 0$ ; $\csc \theta = \frac{1}{\sin \theta }$ where $\sin \theta \ne 0$ .

Remember
Tangent is undefined when cosine is zero.

Worked example

1

01 / Define all six trigonometric functions using a unit-circle terminal point. Give their global domains, then complete the six-function rows at

\pi

\dfrac{5 \pi }{6}

. Use column order sine, cosine, tangent, cosecant, secant, cotangent, and explain every undefined entry. Show the requested reasoning; written work and sketches require teacher review. Also construct the complete first-quadrant special-value table for

0

\dfrac{\pi }{6}

\dfrac{\pi }{4}

\dfrac{\pi }{3}

\dfrac{\pi }{2}

, including all six functions and every undefined cell. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

02 /

\sin t=y,\ \cos t=x,\ \tan t=\dfrac{y}{x},\ \csc t=\dfrac1y,\ \sec t=\dfrac1x,\ \cot t=\dfrac{x}{y}

03 / Sine and cosine allow all real inputs. Tangent and secant exclude

t=\dfrac\pi2+k\pi

; cotangent and cosecant exclude

t=k\pi

, for integers

k

. A zero numerator gives zero only when the denominator is nonzero.

04 /

t=\pi

gives

\left(-1,0\right)

Sine	Cosine	Tangent	Cosecant	Secant	Cotangent
$0$	$-1$	$0$	undefined	$-1$	undefined

05 /

t=\dfrac{5 \pi }{6}

gives

\left(- \dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)

Sine	Cosine	Tangent	Cosecant	Secant	Cotangent
$\dfrac{1}{2}$	$- \dfrac{\sqrt{3}}{2}$	$- \dfrac{\sqrt{3}}{3}$	$2$	$- \dfrac{2\,\sqrt{3}}{3}$	$- \sqrt{3}$

06 /

Input	Sine	Cosine	Tangent	Cosecant	Secant	Cotangent
$0$	$0$	$1$	$0$	undefined	$1$	undefined
$\dfrac{\pi }{6}$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{3}}{3}$	$2$	$\dfrac{2\,\sqrt{3}}{3}$	$\sqrt{3}$
$\dfrac{\pi }{4}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$	$1$	$\sqrt{2}$	$\sqrt{2}$	$1$
$\dfrac{\pi }{3}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$	$\dfrac{2\,\sqrt{3}}{3}$	$2$	$\dfrac{\sqrt{3}}{3}$
$\dfrac{\pi }{2}$	$1$	$0$	undefined	$1$	undefined	$0$

Worked example 2 · Coordinate definitions and domains

Define all six trigonometric functions using a unit-circle terminal point. Give their global domains, then complete the six-function rows at $\dfrac{3 \pi }{2}$ , $\dfrac{11 \pi }{6}$ . Use column order sine, cosine, tangent, cosecant, secant, cotangent, and explain every undefined entry. Show the requested reasoning; written work and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

$\sin t=y,\ \cos t=x,\ \tan t=\dfrac{y}{x},\ \csc t=\dfrac1y,\ \sec t=\dfrac1x,\ \cot t=\dfrac{x}{y}$ .
Sine and cosine allow all real inputs. Tangent and secant exclude $t=\dfrac\pi2+k\pi$ ; cotangent and cosecant exclude $t=k\pi$ , for integers $k$ . A zero numerator gives zero only when the denominator is nonzero.
$t=\dfrac{3 \pi }{2}$ gives $\left(0,-1\right)$ .

Sine	Cosine	Tangent	Cosecant	Secant	Cotangent
$-1$	$0$	undefined	$-1$	undefined	$0$

t=\dfrac{11 \pi }{6}

gives

\left(\dfrac{\sqrt{3}}{2},- \dfrac{1}{2}\right)

Sine	Cosine	Tangent	Cosecant	Secant	Cotangent
$- \dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$- \dfrac{\sqrt{3}}{3}$	$-2$	$\dfrac{2\,\sqrt{3}}{3}$	$- \sqrt{3}$

Written-work review:

Use the stated definitions and the complete allowed domain.
Show all intermediate values and justify signs, restrictions and conclusions.
Give every requested result in the stated order and units.

Read the graphs

Exact circular evaluation

For

t=3 \pi

, reduce full turns, find the reference number and signed terminal point, and calculate all six functions in the order sine, cosine, tangent, cosecant, secant, cotangent. Explain which functions are undefined, if any.

Show the requested reasoning; written work and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

$3 \pi \equiv \pi \pmod{2\pi }$ . Reference number $0$ .
Coordinate magnitudes come from the special first-quadrant point; the quadrant or axis fixes their signs. The terminal point is $\left(-1,0\right)$ .
$\sin t=y,\ \cos t=x,\ \tan t=\dfrac{y}{x},\ \csc t=\dfrac1y,\ \sec t=\dfrac1x,\ \cot t=\dfrac{x}{y}$ .
Substitution gives $0$ , $-1$ , $0$ , undefined, $-1$ , undefined. Sine and cosine allow all real inputs. Tangent and secant exclude $t=\dfrac\pi2+k\pi$ ; cotangent and cosecant exclude $t=k\pi$ , for integers $k$ . A zero numerator gives zero only when the denominator is nonzero.
Written-work review:

Use the stated definitions and the complete allowed domain.
Show all intermediate values and justify signs, restrictions and conclusions.
Give every requested result in the stated order and units.

Answer: $0$ , $-1$ , $0$ , undefined, $-1$ , undefined

Diagram versus calculator evaluation

Estimate

\sin(\dfrac{67}{100})

from the plotted terminal point A, first reading both coordinates to the nearest tenth

(

use hundredths if rounding would make a required denominator zero

)

. Then evaluate in radian mode to six decimals and compare the absolute difference, rounded to three decimals. Report estimate, calculator value and absolute difference.

The projections give approximately $\left(0.80,0.60\right)$ . Use $\sin t=y,\ \cos t=x,\ \tan t=\dfrac{y}{x},\ \csc t=\dfrac1y,\ \sec t=\dfrac1x,\ \cot t=\dfrac{x}{y}$ to form the requested coordinate or ratio.
The diagram estimate is $0.600$ . Do not treat an inverse-trig key as a reciprocal.
Radian-mode evaluation gives $0.620986$ . The absolute discrepancy is $0.021$ . Rounding coordinates before division can magnify tangent or reciprocal error.
Written-work review:

Use the stated definitions and the complete allowed domain.
Show all intermediate values and justify signs, restrictions and conclusions.
Give every requested result in the stated order and units.

Answer: $0.600$ , $0.620986$ , $0.021$

Geometric reduction formulas

A terminal point is

\left(\dfrac{4}{5},\dfrac{3}{5}\right)

. Using actual geometry, prove the half-turn and positive quarter-turn reduction formulas for sine, cosine and tangent, with domains. Then apply them to find the three values at

t + \pi

in sine, cosine, tangent order.

In the figure, $B$ and $E$ are opposite endpoints of a diameter with midpoint $O$ . A half-turn maps the radius $OB$ onto $OE$ and sends $(x,y)$ to $(-x,-y)$ . Therefore $\sin(t+\pi )=-\sin t,\quad\cos(t+\pi )=-\cos t$ and $\tan(t+\pi )=\tan t$ whenever $\cos t\ne0$ .
For a quarter-turn, form the two right triangles using the horizontal projections of the original and rotated radii. Match vertex $O$ of triangle $OAB$ to $D$ of triangle $DCO$ , A to $C$ , and $B$ to $O$ : the acute angles at $O$ and $D$ are equal, and the acute angles at $B$ and $O$ are complementary to them. The included hypotenuses $OB$ and $DO$ both have length $1$ . ASA congruence therefore swaps the leg lengths. The new horizontal leg points left while the new vertical leg points up, giving $(-y,x)$ .
Thus $\sin(t+\dfrac\pi2)=\cos t,\quad\cos(t+\dfrac\pi2)=-\sin t,\quad\tan(t+\dfrac\pi2)=-\cot t$ . The tangent identity requires $\sin t\ne0$ on both sides. Axis and other-quadrant cases follow by the same rotation and signed coordinates.
Reverse the quarter-turn for a negative shift, or add a full turn for $\dfrac{3 \pi }{2}$ . The requested point is $\left(- \dfrac{4}{5},- \dfrac{3}{5}\right)$ ; the sine, cosine and tangent values are $- \dfrac{3}{5}$ , $- \dfrac{4}{5}$ , $\dfrac{3}{4}$ .
Written-work review:

Identify the congruent right triangles and the signed coordinate swap.
Derive all six stated reduction identities, not only the numerical answer.
Keep tangent and cotangent denominator restrictions and evaluate the requested shifted values.

Answer: $- \dfrac{3}{5}$ , $- \dfrac{4}{5}$ , $\dfrac{3}{4}$

Angles versus real inputs

A right triangle has adjacent leg

16

, opposite leg

12

and hypotenuse

20

. Normalize it to the unit circle, use similarity to reconcile triangle ratios with real-input sine and cosine, and find the acute angle in radians to five decimals. Find the subtended arc on the original-radius circle to three decimals. Report unit-circle horizontal coordinate, vertical coordinate, tangent, angle and original arc length.

Divide every length by the hypotenuse $20$ . Similarity gives the unit point $\left(\dfrac{4}{5},\dfrac{3}{5}\right)$ .
Opposite over hypotenuse is $\dfrac{3}{5}$ and adjacent over hypotenuse is $\dfrac{4}{5}$ , exactly the sine and cosine coordinate definitions. Tangent is $\dfrac{3}{4}$ .
$\theta=\arctan(\dfrac{3}{4})\approx 0.64350$ radians. On radius $1$ , radian measure equals oriented arc length, so the real input equals this angle.
$s=r\theta=20\arctan(\dfrac{3}{4})\approx 12.870$ . Arc length equals angle only on a unit circle.
Written-work review:

Use the stated definitions and the complete allowed domain.
Show all intermediate values and justify signs, restrictions and conclusions.
Give every requested result in the stated order and units.

Answer: $\dfrac{4}{5}$ , $\dfrac{3}{5}$ , $\dfrac{3}{4}$ , $0.64350$ , $12.870$

15 questions with a worked answer for each.Start practice ↗

Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.

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5.2 Trigonometric Functions of Real Numbers

Ideas in this lesson

5.2 Trigonometric Functions of Real Numbers

Read the graphs

Exact circular evaluation

Diagram versus calculator evaluation

Geometric reduction formulas

Angles versus real inputs