Trigonometry, geometry and periodic models

Learning goals and preparation

We assume Chapter 1's functions, equations and quadratics. This chapter develops angles and periodic functions as quantitative tools, without assuming calculus. You will distinguish degrees and radians, use trigonometric identities with their domains, solve equations over a full interval, and choose the correct branch in triangle and inverse-function problems.

2.1 Radians, the circle, and basic identities

An angle in radians is its signed arc length divided by circle radius. Thus a full turn is $2\pi$ radians and $\theta$ degrees equals $\pi\theta/180$ radians. On the unit circle the coordinate pair is $(\cos\theta,\sin\theta)$, giving $\cos^2\theta+\sin^2\theta=1$.

Sine and cosine are defined for every real angle and repeat after $2\pi$. Tangent is their quotient $\sin\theta/\cos\theta$ and excludes $\theta=\pi/2+k\pi$, where $k\in\mathbb Z$. Its period is $\pi$. Dividing by a trigonometric expression can discard exactly the solutions at which it vanishes, just as dividing an algebraic equation by an unknown can discard zero.

Worked example 1: arc versus chord. On a circle of radius $10$ m, a $60^\circ=\pi/3$ radian angle gives arc length $10\pi/3$ m. The straight chord length is $2(10)\sin(\pi/6)=10$ m. They are different distances even though they connect the same endpoints.

The addition formulas are

$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta,$$ $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$

Proof

Take as a geometric fact that a rotation about the origin preserves distances and orientation and is linear in coordinate components. Let $R_\alpha$ rotate counterclockwise through $\alpha$. On the unit circle, it sends the first and second coordinate basis vectors to

$$R_\alpha e_1=(\cos\alpha,\sin\alpha)^T,\qquad R_\alpha e_2=(-\sin\alpha,\cos\alpha)^T.$$

By linearity, its action on the point at angle $\beta$ is

$$R_\alpha(\cos\beta,\sin\beta)^T =(\cos\alpha\cos\beta-\sin\alpha\sin\beta,\ \sin\alpha\cos\beta+\cos\alpha\sin\beta)^T.$$

Geometrically, successive rotations add their angles, so this same point is $(\cos(\alpha+\beta),\sin(\alpha+\beta))^T$. Comparing coordinates proves both identities without assuming them. Setting $\beta=\alpha=\theta$ gives $\sin2\theta=2\sin\theta\cos\theta$ and $\cos2\theta=\cos^2\theta-\sin^2\theta$. An identity holds for every input in its domain; an equation asks which inputs make a particular equality true.

In particular $\sin2\theta=2\sin\theta\cos\theta$, $\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta$. A trigonometric identity is valid for all permitted inputs; it is not an equation whose solutions should be isolated.

2.2 Equations and principal inverse values

The periodic sine function is not injective on the whole line. Restricting it to $[-\pi/2,\pi/2]$ gives the inverse $\arcsin:[-1,1]\to[-\pi/2,\pi/2]$. Similarly $\arccos$ takes values in $[0,\pi]$, and $\arctan$ in $(-\pi/2,\pi/2)$.

For $c\in[-1,1]$, with $\alpha=\arcsin c$, the full sine solutions are $\theta=\alpha+2k\pi$ or $\theta=\pi-\alpha+2k\pi$, where $k\in\mathbb Z$. An interval restriction selects the allowed members. Principal inverse values are a convenient parameterization, not the complete periodic solution set.

Worked example 2: quadratic in cosine. Solve $2\cos^2\theta-3\cos\theta+1=0$ for $0\le\theta<2\pi$. Factoring gives $(2\cos\theta-1)(\cos\theta-1)=0$. Thus cosine is $1/2$ or $1$. The solutions are $\pi/3$, $5\pi/3$, and $0$. The endpoint $2\pi$ is excluded. Dividing the equation by $\cos\theta-1$ would discard the genuine solution zero.

Worked example 3: an inverse branch trap. $\arcsin(\sin(5\pi/6))=\pi/6$ because the sine value is $1/2$ and the inverse output must lie in $[-\pi/2,\pi/2]$. Returning $5\pi/6$ ignores the function's declared range.

2.3 Triangle geometry and ambiguity

For a triangle with side lengths $a,b,c$ opposite angles $A,B,C$, the cosine rule is

$$a^2=b^2+c^2-2bc\cos A.$$

Place the two sides of lengths $b,c$ as vectors with endpoints $(b,0)$ and $(c\cos A,c\sin A)$. The squared distance between their endpoints is $(b-c\cos A)^2+(c\sin A)^2$, which simplifies to the formula. This derivation uses ordinary coordinate geometry rather than a memorized special case.

The area is $\Delta=bc\sin A/2$. Expressing it using other pairs gives the sine rule $a/\sin A=b/\sin B=c/\sin C$ for nondegenerate triangles.

Worked example 4: two sides and included angle. If $b=5$, $c=7$, and $A=60^\circ$, then $a^2=25+49-70(1/2)=39$, so $a=\sqrt{39}$. The area is $(35/2)(\sqrt3/2)=35\sqrt3/4$. Both calculations retain the positive geometric length or area.

With two sides and a nonincluded angle, there may be two possible triangles. For $a=7$, $b=10$, $A=30^\circ$, the sine rule gives $\sin B=5/7$. Both $B=\arcsin(5/7)$ and $B=\pi-\arcsin(5/7)$ leave a positive third angle after subtracting $A$. A single inverse-sine button cannot choose a unique triangle without further information. If the computed sine exceeded one, there would be no real triangle at all.

2.4 Periodic quantitative models

For $f(t)=m+A\cos(\omega t+\phi)$ with nonzero $A,\omega$, the midpoint is $m$, amplitude $|A|$, and period $T=2\pi/|\omega|$. The output range is $[m-|A|,m+|A|]$. If the modeled quantity cannot be negative, ensure its range respects that restriction or restrict the model's applicable times.

The time variable and angular frequency must use compatible units so the argument is dimensionless. Changing from months to years changes the numerical frequency; it does not merely relabel an unchanged input.

Worked example 5: combine two seasonal components. Let

$$Q(t)=120+20\cos(\pi t/6)-15\sin(\pi t/6),$$

where $t$ is months. Choose $\phi\in(0,\pi/2)$ with $\cos\phi=4/5$, $\sin\phi=3/5$. The addition identity gives $Q(t)=120+25\cos(\pi t/6+\phi)$. Its period is $12$ months, range $[95,145]$, and value at zero is $140$. Its first maximum at nonnegative time is $t=12-6\phi/\pi$, not zero. The phase shifts the timing without changing amplitude or period.

A nonnegative periodic graph is not a probability density without a support and normalization.

2.5 Coordinates and bearings

A directed displacement of length $R$ at angle $\theta$ counterclockwise from the positive horizontal axis has components $(R\cos\theta,R\sin\theta)$. A bearing measured clockwise from north uses a different convention: east and north components are $(R\sin\beta,R\cos\beta)$. State the convention before calculating.

Worked example 6: a survey direction. A path of length $10$ at bearing $30^\circ$ has east component $10\sin30^\circ=5$ and north component $10\cos30^\circ=5\sqrt3$. An angle of $30^\circ$ measured instead from east would exchange those components. The formula changes because the reference direction changes, not because trigonometry is inconsistent.

2.6 Graded exercises

Foundation

1. Convert $150^\circ$ to radians, and $7\pi/6$ radians to degrees.
2. Find arc and chord lengths for a $90^\circ$ angle on a circle of radius $4$.
3. State every real input at which $\tan\theta$ is undefined.

Core

4. Solve $\sin(2x)=0$ for $0\le x<2\pi$.
5. Evaluate $\arccos(\cos(4\pi/3))$ using the principal range.
6. A triangle has sides $3,4$ enclosing a right angle. Find the third side and its area by the general formulas.
7. Find the midpoint, amplitude, range and period of $f(t)=7-2\sin(\pi t/4)$.
8. Express $8\cos t+6\sin t$ as $R\cos(t-\delta)$ with first-quadrant $\delta$, specifying its sine and cosine.

Synthesis

9. In a proposed triangle, $a=3$, $b=8$, and $A=30^\circ$. Use the sine rule to decide whether the triangle exists, explaining why an inverse-sine calculation cannot repair the problem.
10. A path of length $20$ has bearing $120^\circ$ clockwise from north. Find east and north components and check that the resulting component length is $20$.

2.7 Full solutions

1. $150^\circ=5\pi/6$ radians. Conversely $7\pi/6$ radians equals $210^\circ$.
2. The angle is $\pi/2$ radians, so arc length is $4(\pi/2)=2\pi$. The chord is $2(4)\sin(\pi/4)=4\sqrt2$.
3. Tangent is undefined where cosine vanishes: $\theta=\pi/2+k\pi$ for integers $k$.
4. $2x=k\pi$ with $k\in\mathbb Z$, so $x=k\pi/2$. The four permitted values are $0,\pi/2,\pi,3\pi/2$.
5. Cosine at $4\pi/3$ is $-1/2$. The angle with that cosine in $[0,\pi]$ is $2\pi/3$.
6. The cosine rule gives $a^2=3^2+4^2-2(3)(4)\cos(\pi/2)=25$, so the third side is $5$. The area is $(1/2)(3)(4)\sin(\pi/2)=6$.
7. Midpoint is $7$, amplitude $2$, range $[5,9]$, and period $2\pi/(\pi/4)=8$ time units.
8. $R=\sqrt{64+36}=10$, with $\cos\delta=4/5$ and $\sin\delta=3/5$. Expanding $10\cos(t-\delta)$ returns the original expression.
9. The sine rule would require $\sin B=b\sin A/a=8(1/2)/3=4/3>1$. No real angle has that sine, so no such triangle exists. A principal inverse is undefined there because the input lies outside its domain; changing branches cannot create a valid sine value.
10. East component is $20\sin120^\circ=10\sqrt3$; north component is $20\cos120^\circ=-10$. The negative north component means southward motion. The length is $\sqrt{300+100}=20$.

Summary and onward map

Angles need a unit and reference direction. Inverse branches do not eliminate periodic alternatives, and triangle data can be ambiguous or impossible. Identities prove exact relationships; phase, amplitude and frequency describe different features of a periodic model. Component coordinates prepare for the vector and matrix chapter, while radian functions prepare for later differentiation.