Course Review

Study Sheet

Course Review

Every definition, rule, and formula from all twelve topics in one place

Angles and Their Measure

Concept
Degrees, radians, and conversion

An angle in standard position has its vertex at the origin and its initial side on the positive xx-axis. One full revolution is 360∘=2π360^\circ=2\pi radians, so a straight angle is 180∘=π180^\circ=\pi radians.

radians=degrees⋅π180∘,degrees=radians⋅180∘π.\text{radians}=\text{degrees}\cdot\frac{\pi}{180^\circ},\qquad \text{degrees}=\text{radians}\cdot\frac{180^\circ}{\pi}.

Coterminal angles share a terminal side; add or subtract full turns: θ+360∘n\theta+360^\circ n (degrees) or θ+2πn\theta+2\pi n (radians), nn an integer. Complementary angles sum to 90∘90^\circ; supplementary angles sum to 180∘180^\circ.

Concept
Arc length, sector area, and speed

For a central angle θ\theta measured in radians in a circle of radius rr:

s=rθ(arc length),A=12r2θ(sector area).s=r\theta\quad(\text{arc length}),\qquad A=\tfrac12 r^2\theta\quad(\text{sector area}).

If a point moves along the circle, its linear speed and angular speed ω\omega are

v=st=rω,ω=θt.v=\frac{s}{t}=r\omega,\qquad \omega=\frac{\theta}{t}.
Example
Worked example: arc length and area

A sector has radius r=6r=6 cm and central angle θ=60∘=π3\theta=60^\circ=\dfrac{\pi}{3}.

s=rθ=6⋅π3=2π≈6.28 cm,A=12r2θ=12(36)π3=6π≈18.85 cm2.s=r\theta=6\cdot\tfrac{\pi}{3}=2\pi\approx 6.28\text{ cm},\quad A=\tfrac12 r^2\theta=\tfrac12(36)\tfrac{\pi}{3}=6\pi\approx 18.85\text{ cm}^2.
Tip

Tip. The formulas s=rθs=r\theta, A=12r2θA=\tfrac12 r^2\theta, and v=rωv=r\omega require θ\theta in radians. Convert degrees first, or the answer will be wrong by a factor of π/180\pi/180.

Right Triangle Trigonometry

Concept
SOH-CAH-TOA and reciprocals

For an acute angle θ\theta in a right triangle,

sin⁡θ=opphyp,cos⁡θ=adjhyp,tan⁡θ=oppadj.\sin\theta=\frac{\text{opp}}{\text{hyp}},\qquad \cos\theta=\frac{\text{adj}}{\text{hyp}},\qquad \tan\theta=\frac{\text{opp}}{\text{adj}}.

The reciprocals are

csc⁡θ=hypopp,sec⁡θ=hypadj,cot⁡θ=adjopp.\csc\theta=\frac{\text{hyp}}{\text{opp}},\qquad \sec\theta=\frac{\text{hyp}}{\text{adj}},\qquad \cot\theta=\frac{\text{adj}}{\text{opp}}.
Concept
Special angles and cofunctions

1.3

Cofunctions of complementary angles are equal:

sin⁡θ=cos⁡(90∘−θ),tan⁡θ=cot⁡(90∘−θ),sec⁡θ=csc⁡(90∘−θ).\sin\theta=\cos(90^\circ-\theta),\quad \tan\theta=\cot(90^\circ-\theta),\quad \sec\theta=\csc(90^\circ-\theta).
Example
Worked example: angle of elevation

From a point 5050 ft from the base of a tree, the angle of elevation to the top is 32∘32^\circ. The height hh satisfies

tan⁡32∘=h50 ⇒ h=50tan⁡32∘≈31.2 ft.\tan 32^\circ=\frac{h}{50}\ \Rightarrow\ h=50\tan 32^\circ\approx 31.2\text{ ft}.
Tip

Tip. An angle of elevation is measured up from the horizontal; an angle of depression is measured down from the horizontal. They are equal (alternate interior angles), so a depression angle can be moved to the object's location.

The Unit Circle and Circular Functions

Concept
Definition from the unit circle

On the unit circle x2+y2=1x^2+y^2=1, if the terminal side of θ\theta meets the circle at (x,y)(x,y) then

cos⁡θ=x,sin⁡θ=y,tan⁡θ=yx,\cos\theta=x,\qquad \sin\theta=y,\qquad \tan\theta=\frac{y}{x},

with sec⁡θ=1x\sec\theta=\frac1x, csc⁡θ=1y\csc\theta=\frac1y, cot⁡θ=xy\cot\theta=\frac xy. For a point (x,y)(x,y) not on the unit circle, let r=x2+y2r=\sqrt{x^2+y^2}; then sin⁡θ=yr\sin\theta=\frac yr, cos⁡θ=xr\cos\theta=\frac xr, tan⁡θ=yx\tan\theta=\frac yx.

Concept
Reference angles, signs, period, and symmetry

The reference angle θ′\theta' is the acute angle to the xx-axis:

Q II: 180∘−θ,Q III: θ−180∘,Q IV: 360∘−θ.\text{Q II: }180^\circ-\theta,\quad \text{Q III: }\theta-180^\circ,\quad \text{Q IV: }360^\circ-\theta.

Signs follow ASTC (“All Students Take Calculus”): in QI all are ++; QII only sin⁡,csc⁡\sin,\csc; QIII only tan⁡,cot⁡\tan,\cot; QIV only cos⁡,sec⁡\cos,\sec. Periods: sin⁡,cos⁡,sec⁡,csc⁡\sin,\cos,\sec,\csc repeat every 2π2\pi; tan⁡,cot⁡\tan,\cot every π\pi. Even: cos⁡,sec⁡\cos,\sec; odd: sin⁡,csc⁡,tan⁡,cot⁡\sin,\csc,\tan,\cot.

Example
Worked example: reference angle then sign

Evaluate sin⁡210∘\sin 210^\circ. It lies in QIII with reference angle 210∘−180∘=30∘210^\circ-180^\circ=30^\circ, and sine is negative in QIII, so

sin⁡210∘=−sin⁡30∘=−12.\sin 210^\circ=-\sin 30^\circ=-\tfrac12.
Tip

Tip. Cosine is the xx-coordinate, sine is the yy-coordinate. At the quadrantal angles a coordinate is 00, making tan⁡,sec⁡\tan,\sec (when x=0x=0) or cot⁡,csc⁡\cot,\csc (when y=0y=0) undefined.

Graphs of Sine and Cosine

Concept
The general sinusoid

For y=asin⁡(b(x−c))+dy=a\sin\big(b(x-c)\big)+d or y=acos⁡(b(x−c))+dy=a\cos\big(b(x-c)\big)+d with b>0b>0:

amplitude=∣a∣,period=2πb,\text{amplitude}=|a|,\qquad \text{period}=\frac{2\pi}{b},
phase shift=c (right if c>0),vertical shift=d.\text{phase shift}=c\ (\text{right if }c>0),\qquad \text{vertical shift}=d.

The midline is y=dy=d; the graph oscillates between y=d−∣a∣y=d-|a| and y=d+∣a∣y=d+|a|. If a<0a<0 the basic shape is reflected across the midline.

Example
Worked example: reading a graph's equation

For y=3cos⁡ ⁣(2x−π)+1=3cos⁡ ⁣(2(x−π2))+1y=3\cos\!\big(2x-\pi\big)+1=3\cos\!\big(2(x-\tfrac{\pi}{2})\big)+1:

∣a∣=3,period=2π2=π,phase shift=π2 right,d=1.|a|=3,\quad \text{period}=\frac{2\pi}{2}=\pi,\quad \text{phase shift}=\frac{\pi}{2}\text{ right},\quad d=1.

Range: [1−3, 1+3]=[−2,4][1-3,\,1+3]=[-2,4].

Tip

Tip. Always factor out bb before reading the phase shift: in sin⁡(bx−c)\sin(bx-c) the shift is c/bc/b, not cc. Divide the period into quarters to plot the five key points (max, zero, min, zero).

Graphs of the Other Trig Functions

Concept
Tangent and cotangent
y=atan⁡(b(x−c)):period=πb,asymptotes where bx=π2+πn.y=a\tan\big(b(x-c)\big):\quad \text{period}=\frac{\pi}{b},\quad \text{asymptotes where } bx=\tfrac{\pi}{2}+\pi n.
y=acot⁡(b(x−c)):period=πb,asymptotes where bx=πn.y=a\cot\big(b(x-c)\big):\quad \text{period}=\frac{\pi}{b},\quad \text{asymptotes where } bx=\pi n.

Both have range all real numbers and no amplitude. Tangent increases on each branch; cotangent decreases.

Concept
Secant and cosecant
y=asec⁡(b(x−c)) and y=acsc⁡(b(x−c)):period=2πb.y=a\sec\big(b(x-c)\big)\text{ and } y=a\csc\big(b(x-c)\big):\quad \text{period}=\frac{2\pi}{b}.

Graph the guide sine/cosine first: sec⁡\sec has vertical asymptotes where its guide cos⁡=0\cos=0; csc⁡\csc where its guide sin⁡=0\sin=0. Range is (−∞,−∣a∣]∪[∣a∣,∞)(-\infty,-|a|]\cup[|a|,\infty); the curve never crosses the midline.

Example
Worked example: period and asymptotes of a tangent

For y=tan⁡ ⁣(12x)y=\tan\!\big(\tfrac12 x\big): period =π1/2=2π=\dfrac{\pi}{1/2}=2\pi, and asymptotes occur where 12x=π2+πn\tfrac12 x=\tfrac{\pi}{2}+\pi n, i.e. x=π+2πnx=\pi+2\pi n.

Tip

Tip. Tangent/cotangent use π/b\pi/b for the period; secant/cosecant use 2π/b2\pi/b. Sketch the reciprocal sine or cosine lightly first --- its zeros become the asymptotes, and its peaks become the U-shaped turning points.

Inverse Trigonometric Functions

Concept
Domains and ranges of the inverses

1.3

sin⁡−1x=y\sin^{-1}x=y means sin⁡y=x\sin y=x with yy in the range above. (Also written arcsin⁡\arcsin, arccos⁡\arccos, arctan⁡\arctan.)

Concept
Composition rules
sin⁡(sin⁡−1x)=x  for x∈[−1,1],sin⁡−1(sin⁡θ)=θ  only if θ∈[−π2,π2].\sin(\sin^{-1}x)=x\ \text{ for } x\in[-1,1],\qquad \sin^{-1}(\sin\theta)=\theta\ \text{ only if } \theta\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right].

For cos⁡(sin⁡−1x)\cos(\sin^{-1}x) and similar mixed compositions, draw a reference triangle: let the inner inverse define an angle, label two sides, find the third by a2+b2=c2a^2+b^2=c^2.

Example
Worked example: a range-restriction trap

Evaluate sin⁡−1 ⁣(sin⁡3π4)\sin^{-1}\!\big(\sin\tfrac{3\pi}{4}\big). Since 3π4∉[−π2,π2]\tfrac{3\pi}{4}\notin[-\tfrac{\pi}{2},\tfrac{\pi}{2}], the answer is not 3π4\tfrac{3\pi}{4}. Compute sin⁡3π4=22\sin\tfrac{3\pi}{4}=\tfrac{\sqrt2}{2}, then take the angle in range: sin⁡−122=π4\sin^{-1}\tfrac{\sqrt2}{2}=\tfrac{\pi}{4}.

Tip

Tip. An inverse trig function returns exactly one angle in its restricted range. cos⁡−1\cos^{-1} never returns a negative angle; sin⁡−1\sin^{-1} and tan⁡−1\tan^{-1} never return an angle outside ±π2\pm\tfrac{\pi}{2}.

Trigonometric Identities

Concept
The fundamental identities

multicols2 Reciprocal:

csc⁡θ=1sin⁡θ, sec⁡θ=1cos⁡θ, cot⁡θ=1tan⁡θ\csc\theta=\frac{1}{\sin\theta},\ \sec\theta=\frac{1}{\cos\theta},\ \cot\theta=\frac{1}{\tan\theta}

Quotient:

tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta},\qquad \cot\theta=\frac{\cos\theta}{\sin\theta}

Pythagorean:

sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta
1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta

multicols Cofunction: sin⁡ ⁣(π2−θ)=cos⁡θ\sin\!\big(\tfrac{\pi}{2}-\theta\big)=\cos\theta, tan⁡ ⁣(π2−θ)=cot⁡θ\tan\!\big(\tfrac{\pi}{2}-\theta\big)=\cot\theta, sec⁡ ⁣(π2−θ)=csc⁡θ\sec\!\big(\tfrac{\pi}{2}-\theta\big)=\csc\theta (and the three partners). Even/odd: cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta, sec⁡(−θ)=sec⁡θ\sec(-\theta)=\sec\theta; sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta, csc⁡(−θ)=−csc⁡θ\csc(-\theta)=-\csc\theta, tan⁡(−θ)=−tan⁡θ\tan(-\theta)=-\tan\theta, cot⁡(−θ)=−cot⁡θ\cot(-\theta)=-\cot\theta.

Example
Worked example: verifying an identity

Verify sec⁡θ−sin⁡θtan⁡θ=cos⁡θ\sec\theta-\sin\theta\tan\theta=\cos\theta.

sec⁡θ−sin⁡θtan⁡θ=1cos⁡θ−sin⁡2θcos⁡θ=1−sin⁡2θcos⁡θ=cos⁡2θcos⁡θ=cos⁡θ.✓\begin{aligned} \sec\theta-\sin\theta\tan\theta &=\frac{1}{\cos\theta}-\frac{\sin^2\theta}{\cos\theta} =\frac{1-\sin^2\theta}{\cos\theta} =\frac{\cos^2\theta}{\cos\theta}=\cos\theta.\quad\checkmark \end{aligned}
Tip

Tip. When stuck, rewrite everything in sin⁡θ\sin\theta and cos⁡θ\cos\theta, combine over a common denominator, and hunt for a Pythagorean pattern such as 1−sin⁡2θ=cos⁡2θ1-\sin^2\theta=\cos^2\theta.

Sum, Difference, and Multiple-Angle Formulas

Concept
Sum and difference
sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B
cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B
tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}

Cosine flips the sign; sine and tangent keep it.

Concept
Double-angle and half-angle

multicols2 Double:

sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A
cos⁡2A=cos⁡2A−sin⁡2A\cos 2A=\cos^2 A-\sin^2 A
cos⁡2A=2cos⁡2A−1=1−2sin⁡2A\phantom{\cos 2A}=2\cos^2 A-1=1-2\sin^2 A
tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A=\frac{2\tan A}{1-\tan^2 A}

Half:

sin⁡A2=±1−cos⁡A2\sin\frac A2=\pm\sqrt{\frac{1-\cos A}{2}}
cos⁡A2=±1+cos⁡A2\cos\frac A2=\pm\sqrt{\frac{1+\cos A}{2}}
tan⁡A2=1−cos⁡Asin⁡A=sin⁡A1+cos⁡A\tan\frac A2=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}

multicols The ±\pm on a half-angle is chosen from the quadrant of A2\tfrac A2.

Concept
Power-reducing, product-to-sum, sum-to-product

Power-reducing:

sin⁡2A=1−cos⁡2A2,cos⁡2A=1+cos⁡2A2,tan⁡2A=1−cos⁡2A1+cos⁡2A.\sin^2 A=\frac{1-\cos 2A}{2},\quad \cos^2 A=\frac{1+\cos 2A}{2},\quad \tan^2 A=\frac{1-\cos 2A}{1+\cos 2A}.

Product-to-sum:

sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]cos⁡Asin⁡B=12[sin⁡(A+B)−sin⁡(A−B)]cos⁡Acos⁡B=12[cos⁡(A−B)+cos⁡(A+B)]sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\begin{aligned} \sin A\cos B&=\tfrac12\big[\sin(A+B)+\sin(A-B)\big]\\ \cos A\sin B&=\tfrac12\big[\sin(A+B)-\sin(A-B)\big]\\ \cos A\cos B&=\tfrac12\big[\cos(A-B)+\cos(A+B)\big]\\ \sin A\sin B&=\tfrac12\big[\cos(A-B)-\cos(A+B)\big] \end{aligned}

Sum-to-product:

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\begin{aligned} \sin A+\sin B&=2\sin\tfrac{A+B}{2}\cos\tfrac{A-B}{2}\\ \sin A-\sin B&=2\cos\tfrac{A+B}{2}\sin\tfrac{A-B}{2}\\ \cos A+\cos B&=2\cos\tfrac{A+B}{2}\cos\tfrac{A-B}{2}\\ \cos A-\cos B&=-2\sin\tfrac{A+B}{2}\sin\tfrac{A-B}{2} \end{aligned}
Example
Worked example: exact value of cos⁡15∘\cos 15^\circ
cos⁡15∘=cos⁡(45∘−30∘)=cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘=22⋅32+22⋅12=6+24.\begin{aligned} \cos 15^\circ&=\cos(45^\circ-30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ\\ &=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12 =\frac{\sqrt6+\sqrt2}{4}. \end{aligned}
Tip

Tip. To find sin⁡\sin or cos⁡\cos of a non-special angle, write it as a sum or difference of 30∘,45∘,60∘30^\circ,45^\circ,60^\circ. Use 2cos⁡2A−12\cos^2A-1 when you know cos⁡A\cos A and 1−2sin⁡2A1-2\sin^2A when you know sin⁡A\sin A.

Trigonometric Equations

Concept
General solutions

Solve for the reference angle, place it in every quadrant the sign allows, then add the period:

sin⁡,cos⁡,sec⁡,csc⁡: add +2πn,tan⁡,cot⁡: add +πn,\sin,\cos,\sec,\csc:\ \text{add } +2\pi n,\qquad \tan,\cot:\ \text{add } +\pi n,

where nn is any integer. Example patterns:

sin⁡θ=12 ⇒ θ=π6+2πn  or  θ=5π6+2πn.\sin\theta=\tfrac12\ \Rightarrow\ \theta=\tfrac{\pi}{6}+2\pi n\ \text{ or }\ \theta=\tfrac{5\pi}{6}+2\pi n.
tan⁡θ=1 ⇒ θ=π4+πn.\tan\theta=1\ \Rightarrow\ \theta=\tfrac{\pi}{4}+\pi n.
Concept
Factoring, quadratic form, and multiple angles
  • [leftmargin=*,itemsep=1pt]
  • Factor and set each factor to zero, e.g. 2sin⁡2θ−sin⁡θ=0⇒sin⁡θ(2sin⁡θ−1)=02\sin^2\theta-\sin\theta=0\Rightarrow\sin\theta(2\sin\theta-1)=0.
  • Quadratic form: let u=sin⁡θu=\sin\theta, solve au2+bu+c=0au^2+bu+c=0, then back-substitute.
  • Multiple angle: for sin⁡2θ=k\sin 2\theta=k, solve for 2θ2\theta over a doubled interval [0,4π)[0,4\pi), then divide each solution by 22.
Example
Worked example: quadratic form on [0,2π)[0,2\pi)

Solve 2cos⁡2θ−cos⁡θ−1=02\cos^2\theta-\cos\theta-1=0. Factor: (2cos⁡θ+1)(cos⁡θ−1)=0(2\cos\theta+1)(\cos\theta-1)=0, so cos⁡θ=−12\cos\theta=-\tfrac12 or cos⁡θ=1\cos\theta=1. Thus

θ=2π3, 4π3, 0.\theta=\tfrac{2\pi}{3},\ \tfrac{4\pi}{3},\ 0.
Tip

Tip. If a solving step multiplies or squares, check for extraneous roots at the end. For multiple-angle equations, expand the interval before solving so you keep every solution in the requested range.

Law of Sines and Law of Cosines

Concept
The two laws

For any triangle with sides a,b,ca,b,c opposite angles A,B,CA,B,C:

Law of Sines:asin⁡A=bsin⁡B=csin⁡C.\textbf{Law of Sines:}\quad \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.
Law of Cosines:c2=a2+b2−2abcos⁡C\textbf{Law of Cosines:}\quad c^2=a^2+b^2-2ab\cos C

(and cyclically for a2a^2 and b2b^2). Solve for an angle as cos⁡C=a2+b2−c22ab\cos C=\dfrac{a^2+b^2-c^2}{2ab}.

Reminder — Law of Sines:  asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
Reminder — Law of Cosines:  c2=a2+b2−2abcos⁡Cc^{2}=a^{2}+b^{2}-2ab\cos C
Concept
Which law, plus area
  • [leftmargin=*,itemsep=1pt]
  • Law of Sines: use for AAS, ASA, or SSA.
  • Law of Cosines: use for SAS or SSS.

Area of a triangle:

Area=12absin⁡C,Heron: Area=s(s−a)(s−b)(s−c),  s=a+b+c2.\text{Area}=\tfrac12 ab\sin C,\qquad \text{Heron: } \text{Area}=\sqrt{s(s-a)(s-b)(s-c)},\ \ s=\tfrac{a+b+c}{2}.
Example
Worked example: Law of Cosines (SAS)

Given a=5a=5, b=7b=7, C=40∘C=40^\circ:

c2=52+72−2(5)(7)cos⁡40∘=74−70cos⁡40∘≈20.37,c^2=5^2+7^2-2(5)(7)\cos 40^\circ=74-70\cos 40^\circ\approx 20.37,

so c≈4.51c\approx 4.51.

Tip

Tip: the ambiguous case (SSA). With two sides and a non-included angle there may be 00, 11, or 22 triangles. After finding one angle from the Law of Sines, test its supplement: if that supplement plus the given angle is still under 180∘180^\circ, a second triangle exists.

Vectors and the Dot Product

Concept
Components, magnitude, and direction

A vector v=⟨a,b⟩\mathbf{v}=\langle a,b\rangle from initial point (x1,y1)(x_1,y_1) to terminal point (x2,y2)(x_2,y_2) has components ⟨x2−x1, y2−y1⟩\langle x_2-x_1,\ y_2-y_1\rangle. Its

magnitude=∥v∥=a2+b2,direction θ=tan⁡−1 ⁣ba.\text{magnitude}=\|\mathbf{v}\|=\sqrt{a^2+b^2},\qquad \text{direction } \theta=\tan^{-1}\!\frac{b}{a}.

In component form v=∥v∥⟨cos⁡θ,sin⁡θ⟩\mathbf{v}=\|\mathbf{v}\|\langle\cos\theta,\sin\theta\rangle. A unit vector is u=v∥v∥\mathbf{u}=\dfrac{\mathbf{v}}{\|\mathbf{v}\|}. Standard basis: i=⟨1,0⟩\mathbf{i}=\langle 1,0\rangle, j=⟨0,1⟩\mathbf{j}=\langle 0,1\rangle.

Concept
Dot product, angle, projection, work

For u=⟨u1,u2⟩\mathbf{u}=\langle u_1,u_2\rangle and v=⟨v1,v2⟩\mathbf{v}=\langle v_1,v_2\rangle:

u⋅v=u1v1+u2v2=∥u∥ ∥v∥cos⁡θ.\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2=\|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta.
cos⁡θ=u⋅v∥u∥ ∥v∥,projvu=u⋅v∥v∥2 v.\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|},\qquad \text{proj}_{\mathbf{v}}\mathbf{u}=\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v}.

Vectors are orthogonal iff u⋅v=0\mathbf{u}\cdot\mathbf{v}=0. Work done by force F\mathbf{F} along displacement d\mathbf{d} is W=F⋅dW=\mathbf{F}\cdot\mathbf{d}.

Example
Worked example: angle between vectors

Let u=⟨3,4⟩\mathbf{u}=\langle 3,4\rangle, v=⟨−1,2⟩\mathbf{v}=\langle -1,2\rangle. Then u⋅v=3(−1)+4(2)=5\mathbf{u}\cdot\mathbf{v}=3(-1)+4(2)=5, ∥u∥=5\|\mathbf{u}\|=5, ∥v∥=5\|\mathbf{v}\|=\sqrt5, so

cos⁡θ=555=15 ⇒ θ≈63.4∘.\cos\theta=\frac{5}{5\sqrt5}=\frac{1}{\sqrt5}\ \Rightarrow\ \theta\approx 63.4^\circ.
Tip

Tip. The dot product is a scalar, not a vector. A positive value means the vectors point in a generally similar direction (θ<90∘\theta<90^\circ); zero means perpendicular; negative means they oppose (θ>90∘\theta>90^\circ).

Polar Coordinates and Complex Numbers

Concept
Polar and rectangular conversion

A point (r,θ)(r,\theta) in polar coordinates relates to rectangular (x,y)(x,y) by

x=rcos⁡θ,y=rsin⁡θ,x=r\cos\theta,\qquad y=r\sin\theta,
r2=x2+y2,tan⁡θ=yx (choose θ by quadrant).r^2=x^2+y^2,\qquad \tan\theta=\frac{y}{x}\ (\text{choose } \theta \text{ by quadrant}).
Concept
Trig form, DeMoivre, and nnth roots

A complex number z=a+biz=a+bi has trig (polar) form

z=r(cos⁡θ+isin⁡θ),r=a2+b2,tan⁡θ=ba.z=r(\cos\theta+i\sin\theta),\quad r=\sqrt{a^2+b^2},\quad \tan\theta=\frac ba.

Product/quotient: multiply/divide moduli, add/subtract arguments. DeMoivre's Theorem:

zn=rn(cos⁡nθ+isin⁡nθ).z^n=r^n\big(\cos n\theta+i\sin n\theta\big).

The nn nnth roots of zz are, for k=0,1,…,n−1k=0,1,\dots,n-1,

zk=r1/n ⁣(cos⁡θ+2πkn+isin⁡θ+2πkn).z_k=r^{1/n}\!\left(\cos\frac{\theta+2\pi k}{n}+i\sin\frac{\theta+2\pi k}{n}\right).
Example
Worked example: DeMoivre's Theorem

Compute (1+i)8\big(1+i\big)^8. Here r=2r=\sqrt2, θ=π4\theta=\tfrac{\pi}{4}, so

(1+i)8=(2)8(cos⁡2π+isin⁡2π)=16(1+0i)=16.(1+i)^8=(\sqrt2)^8\big(\cos 2\pi+i\sin 2\pi\big)=16(1+0i)=16.
Concept
Common polar curves
  • [leftmargin=*,itemsep=1pt]
  • Circles: r=ar=a, r=2acos⁡θr=2a\cos\theta, r=2asin⁡θr=2a\sin\theta.
  • Cardioids/limacons: r=a±bcos⁡θr=a\pm b\cos\theta or r=a±bsin⁡θr=a\pm b\sin\theta (cardioid when a=ba=b).
  • Roses: r=acos⁡nθr=a\cos n\theta or r=asin⁡nθr=a\sin n\theta (nn petals if nn odd, 2n2n if nn even).
  • Lemniscate: r2=a2cos⁡2θr^2=a^2\cos 2\theta.
Tip

Tip. A single point has infinitely many polar names: (r,θ)(r,\theta), (r,θ+2πn)(r,\theta+2\pi n), and (−r,θ+π)(-r,\theta+\pi) all coincide. When converting to polar, always confirm θ\theta lies in the correct quadrant for the signs of xx and yy.