Module 04: Radians, trigonometry, geometry, and units

8 exercises, each with a complete solution. 1 figure. 1 note from the engine.

On this page

A. Prerequisites and observable goals
B. The central idea: angles are inputs with a geometric scale
C. Essential identities and coordinate formulas
D. Why the point’s height is not the intercept
E. A short decision procedure
F. Worked examples
G. Independent practice
H. Complete solutions: Module 04
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Algebra, function graphs, and basic right-triangle geometry.

After

Use exact unit-circle values, solve simple trigonometric equations, interpret slope with units, and write a line from a point and a slope.

B. The central idea: angles are inputs with a geometric scale#

An angle in radians is the ratio of arc length to radius: $\theta=s/r$ . A full circle has arc length $2\pi r$ , hence angle $2\pi$ radians. Radian measure is dimensionless, but writing “rad” reminds us of the input convention. The derivative formulas in calculus use radians. A calculator in degree mode can produce a plausible-looking but wrong limit.

On the unit circle, the point at angle $\theta$ has coordinates $(\cos\theta,\sin\theta)$ . Therefore sine and cosine describe projections of a rotating radius onto coordinate axes. Their identities are geometric, not merely a table to memorize. They also model oscillation and periodic motion, making them useful for animation as well as standard exams.

One radian. On the unit circle the point at angle

\theta

(\cos\theta,\sin\theta)

, and the arc from the positive

x

-axis to that point has length

\theta

. Here

\theta=1

radian, so the highlighted arc is as long as the radius,

\cos1\approx0.540

, and

\sin1\approx0.841

C. Essential identities and coordinate formulas#

The circle equation gives $\sin^2\theta+\cos^2\theta=1$ . Where denominators are nonzero,

\tan\theta=\frac{\sin\theta}{\cos\theta},\quad \sec\theta=\frac1{\cos\theta},\quad 1+\tan^2\theta=\sec^2\theta.

Sine is odd, cosine even: $\sin(-\theta)=-\sin\theta$ , $\cos(-\theta)=\cos\theta$ . Both have period $2\pi$ . Useful addition identities are

\begin{align*} \sin(u+v)&=\sin u\cos v+\cos u\sin v,\\ \cos(u+v)&=\cos u\cos v-\sin u\sin v. \end{align*}

They give $\sin2u=2\sin u\cos u$ and $\cos2u=1-2\sin^2u=2\cos^2u-1$ .

For distinct horizontal coordinates, the line through $(x_1,y_1)$ and $(x_2,y_2)$ has slope $m=(y_2-y_1)/(x_2-x_1)$ . Point-slope form is

\boxed{y-y_1=m(x-x_1)}.

The distance formula is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . A circle with center $(a,b)$ and radius $r$ satisfies $(x-a)^2+(y-b)^2=r^2$ . Standard geometry and trigonometry definitions: [O1F].

D. Why the point’s height is not the intercept#

The slope compares a change in height to a change in horizontal position. If a line passes through $(2,-1)$ and has slope $3$ , then from that point to any $(x,y)$ on the line,

\frac{y-(-1)}{x-2}=3\quad(x\ne2),

which rearranges to $y+1=3(x-2)$ and extends to $x=2$ . Thus $y=3x-7$ . The vertical-axis intercept is the height at input zero; here it is $-7$ , not $-1$ . Checking $x=2$ gives $-1$ . The incorrect guess $y=3x-1$ gives $5$ there, so it fails the given point.

Units help separate a value from a slope. If $x$ is seconds and $y$ meters, then $y_1$ has units meters while $m$ has units meters per second. Multiplying $m(x-x_1)$ gives meters, allowing the equation’s two sides to have matching units.

E. A short decision procedure#

For angles, convert degrees to radians by multiplying by $\pi/180$ . Use a reference angle and quadrant signs for exact trig values. When solving a trigonometric equation, find all requested angles in the interval, not just the calculator’s principal answer. For a line, identify a point and a slope, use point-slope form, and verify the point. If horizontal change is zero, the line is vertical and has no finite slope; use $x=a$ instead of $y=mx+b$ .

F. Worked examples#

Example 1: exact values. The angle $150^\circ$ is $5\pi/6$ . Its reference angle is $\pi/6$ and it lies in quadrant II, where sine is positive and cosine negative. Hence $\sin(5\pi/6)=1/2$ , $\cos(5\pi/6)=-\sqrt3/2$ , and $\tan(5\pi/6)=-1/\sqrt3$ .

Example 2: geometric units. A wheel of radius $0.4$ m turns through $3\pi/2$ radians without slipping. The travel distance is $s=r\theta=0.6\pi$ m. The units are meters; the angle is not an additional length factor.

Finish the intercept. A line has slope $-2$ through $(3,5)$ . First write $y-5=-2(x-3)$ . Completion: $y=-2x+11$ . The intercept is $11$ because moving left from $x=3$ to $0$ increases height by $6$ .

From the engine: ZeroVectorAn elbow from three side lengths

ZeroVector, the author’s tactical shooter, places a character’s arm so that the hand reaches a target. The upper arm and forearm have fixed lengths, and the distance from shoulder to target is known. Three sides fix a triangle, and the law of cosines, $a^2 = b^2 + c^2 - 2bc\cos A$ for the angle $A$ between sides $b$ and $c$ , turns them into an angle: here the angle at the shoulder between the line to the target and the upper arm.

ZeroVector: two-bone inverse kinematicsC++

// Law of cosines: angle at root between (root->target) and (root->mid).
f32 cosA = (lUpper * lUpper + dEffective * dEffective - lLower * lLower)
         / (2.f * lUpper * dEffective);
cosA = std::clamp(cosA, -1.f, 1.f);
f32 angleA = std::acos(cosA);

// Desired mid direction: dirToTarget rotated by angleA toward the pole.
vec3 desiredDirRootToMid = std::cos(angleA) * dirToTarget
                         + std::sin(angleA) * polePerp;

With an upper arm of 0.30 m, a forearm of 0.25 m, and a target 0.40 m away, $\cos A = (0.30^2 + 0.40^2 - 0.25^2)/(2\cdot 0.30\cdot 0.40) = 0.78125$ , so $A \approx 38.6^\circ$ . The clamp matters: rounding can push the quotient a hair past 1, where $\arccos$ is undefined. The last two lines rotate the unit vector toward the target by $A$ , inside the plane picked out by a second unit vector perpendicular to it. That is the unit-circle picture $(\cos A, \sin A)$ laid into three dimensions.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M04.01FSolution

Convert $225^\circ$ to radians and find its sine and cosine exactly.

M04.02FSolution

Find every solution of $\sin\theta=\sqrt3/2$ in $[0,2\pi)$ .

M04.03CSolution

Simplify $(1-\cos^2x)/\sin x$ and state where the original expression is defined.

M04.04CSolution

Find the line through $(2,-1)$ with slope $3$ , its vertical-axis intercept, and its value at $x=4$ .

M04.05CSolution

Find the slope and equation of the line through $(-1,4)$ and $(3,-2)$ .

M04.06TSolution

A circular path has radius $6$ m. A runner moves through $5\pi/6$ radians in $4$ s. Find arc length and average speed.

M04.07TSolution

The height $H(t)=2+0.5\sin(4t)$ is measured in meters with $t$ in seconds. Give its range, period, and the units of $4$ .

M04.08SSolution

Explain why $\sin^2x=(1-\cos2x)/2$ , and use it to find the average of the two values $\sin^2x$ and $\cos^2x$ for any $x$ .

Optional small hints

For M04.07, a full cycle occurs when the angle increases by $2\pi$ . For M04.08, start with the cosine double-angle identity, not a numerical example.

H. Complete solutions: Module 04#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M04.01Show solution

Reference angle and signs

Multiplying by $\pi/180$ gives $225\pi/180=\boxed{5\pi/4}$ . The reference angle is $\pi/4$ and the angle is in quadrant III, so both coordinates are negative. Thus $\boxed{\sin(5\pi/4)=\cos(5\pi/4)=-\sqrt2/2}$ . Their squared sum is $1$ , as required by the unit circle.

Back to M04.01

M04.02Show solution

All solutions in the requested interval

The positive reference angle with sine $\sqrt3/2$ is $\pi/3$ . Sine is positive in quadrants I and II, giving $\boxed{\pi/3,\,2\pi/3}$ . Adding another full turn leaves $[0,2\pi)$ , so no additional representatives belong to the requested interval.

Back to M04.02

M04.03Show solution

An identity with restrictions

Use $1-\cos^2x=\sin^2x$ . Then $\sin^2x/\sin x=\sin x$ only where $\sin x\ne0$ . The simplified expression is $\boxed{\sin x,\quad x\ne k\pi\ (k\in\mathbb Z)}$ . The simplified sine itself exists at those excluded inputs, but the original quotient does not.

Back to M04.03

M04.04Show solution

Point, slope, and intercept

Point-slope form gives $y+1=3(x-2)$ . Expand and subtract $1$ : $\boxed{y=3x-7}$ . At input zero the height is $-7$ , so the intercept is $(0,-7)$ . At $x=4$ , $y=12-7=5$ . Moving from $x=2$ to $4$ raises the height by $3(2)=6$ , agreeing with $-1+6=5$ .

Back to M04.04

M04.05Show solution

Two points determine a nonvertical line

The slope is $(-2-4)/(3-(-1))=-6/4=-3/2$ . Through $(-1,4)$ ,

y-4=-\frac32(x+1),\qquad\boxed{y=-\frac32x+\frac52}.

Substituting $x=3$ gives $-9/2+5/2=-2$ , checking the second point.

Back to M04.05

M04.06Show solution

Distance and a rate have different units

Arc length is $s=r\theta=6(5\pi/6)=\boxed{5\pi\text{ m}}$ . Average speed is distance divided by elapsed time: $\boxed{5\pi/4\text{ m/s}}$ . The straight-line chord is shorter than the traveled arc and is not the correct distance for speed along the path.

Back to M04.06

M04.07Show solution

Read a periodic model

Since sine ranges from $-1$ to $1$ , $H$ ranges from $2-0.5$ to $2+0.5$ , or $\boxed{[1.5,2.5]\text{ m}}$ . A full cycle satisfies $4T=2\pi$ , so $\boxed{T=\pi/2\text{ s}}$ . The coefficient has units radians per second, conventionally $\mathrm{s}^{-1}$ because radians are dimensionless.

Back to M04.07

M04.08Show solution

Derive rather than memorize

The identity $\cos2x=1-2\sin^2x$ rearranges to $2\sin^2x=1-\cos2x$ ; dividing by $2$ gives the requested half-angle identity. The arithmetic mean of $\sin^2x$ and $\cos^2x$ is $(\sin^2x+\cos^2x)/2=\boxed{1/2}$ . This is an exact all-input identity, not an approximation or an average over an interval.

Back to M04.08

I. Retention and readiness#

Return to M02.06 and M03.07 to connect transformations and units. Before starting derivatives, solve M04.04 from a blank page and explain why the $-1$ is not the intercept. For trigonometry, independently retrieve the unit-circle values at $0,\pi/6,\pi/4,\pi/3,\pi/2$ and determine quadrant signs.