Trigonometric Functions

Study Sheet

Trigonometric Functions

Angles, the unit circle, graphs, and inverse trig functions

Angles: Degree and Radian Measure

Concept
Measuring Angles

An angle in standard position has its vertex at the origin and initial side on the positive xx-axis. A positive angle rotates counterclockwise; a negative angle rotates clockwise.

  • Degrees: one full revolution =360∘=360^\circ.
  • Radians: one full revolution =2π=2\pi. One radian is the central angle subtending an arc equal in length to the radius.

Conversions:  radians=degrees×π180∘\text{radians}=\text{degrees}\times\dfrac{\pi}{180^\circ},   degrees=radians×180∘π\text{degrees}=\text{radians}\times\dfrac{180^\circ}{\pi}.

Precalculus revisits the trig functions with an eye toward calculus: radians throughout, the unit-circle definitions, exact values, graphs with their transformations, and the inverse functions with their restricted ranges. The emphasis is on functions — periodicity, symmetry, and behavior — more than on triangles.

Example
Converting between degrees and radians
xy5π/6xy7π/4
150∘=5π6150^\circ=\tfrac{5\pi}{6} (left) and 7π4=315∘\tfrac{7\pi}{4}=315^\circ (right).

Convert 150∘150^\circ to radians:  150∘⋅π180∘=150π180=5π6150^\circ\cdot\dfrac{\pi}{180^\circ}=\dfrac{150\pi}{180}=\dfrac{5\pi}{6}. Convert 7π4\dfrac{7\pi}{4} to degrees:  7π4⋅180∘π=7⋅180∘4=315∘\dfrac{7\pi}{4}\cdot\dfrac{180^\circ}{\pi}=\dfrac{7\cdot180^\circ}{4}=315^\circ.

Side note
Where it goes wrong

Confusing sin⁡−1x\sin^{-1}x (the inverse function, arcsine) with 1sin⁡x\tfrac{1}{\sin x} (the reciprocal, cosecant). The notation is unfortunate; the meanings are different.

Concept
Coterminal, Complementary, Supplementary
  • Coterminal angles share a terminal side: add or subtract 360∘360^\circ (or 2π2\pi) any whole number of times.
  • Complementary angles sum to 90∘90^\circ (or π2\dfrac{\pi}{2}).
  • Supplementary angles sum to 180∘180^\circ (or π\pi).
Example
Coterminal and complements
xy−π/35π/3
Clockwise π3\tfrac{\pi}{3} and counterclockwise 5π3\tfrac{5\pi}{3} end on the same ray.

A positive coterminal angle for −π3-\dfrac{\pi}{3}:  −π3+2π=5π3-\dfrac{\pi}{3}+2\pi=\dfrac{5\pi}{3}. The complement of 40∘40^\circ is 50∘50^\circ; the supplement of 40∘40^\circ is 140∘140^\circ.

Tip

Tip: Complements and supplements only make sense for angles between the required bounds, but coterminal angles always exist. To find the smallest nonnegative coterminal angle, add/subtract full turns until you land in [0∘,360∘)[0^\circ,360^\circ) or [0,2π)[0,2\pi).

Arc Length, Sector Area, and Speed

Concept
Circular Arcs and Sectors

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

Arc length: s=rθSector area: A=12r2θ.\text{Arc length: } s=r\theta \qquad \text{Sector area: } A=\tfrac{1}{2}r^2\theta.

For an object moving along the circle:

Angular speed: ω=θtLinear speed: v=st=rω.\text{Angular speed: } \omega=\frac{\theta}{t}\qquad \text{Linear speed: } v=\frac{s}{t}=r\omega.

A central angle of θ\theta radians is the fraction θ2π\dfrac{\theta}{2\pi} of a full turn, so the arc is that fraction of the circumference 2πr2\pi r, giving rθr\theta, and the sector is that fraction of the area πr2\pi r^2, giving 12r2θ\tfrac12r^2\theta. Dividing s=rθs=r\theta by time turns it into v=rωv=r\omega.

Example
Arc length and sector area
π/36 cms = 2π
A sixth of the circle: arc 2π2\pi cm, area 6π6\pi cm2^2.

A circle has radius r=6r=6 cm and central angle θ=π3\theta=\dfrac{\pi}{3}.

s=rθ=6⋅π3=2π cm,A=12r2θ=12(36)π3=6π cm2.s=r\theta=6\cdot\tfrac{\pi}{3}=2\pi\text{ cm}, \qquad A=\tfrac12 r^2\theta=\tfrac12(36)\tfrac{\pi}{3}=6\pi\text{ cm}^2.
Example
Angular and linear speed

A wheel of radius 0.50.5 m turns at 120120 rev/min. In radians per second:

ω=120⋅2π rad/min=240π rad/min=240π60=4π rad/s.\omega=120\cdot 2\pi \text{ rad/min}=240\pi\text{ rad/min}=\frac{240\pi}{60}=4\pi\text{ rad/s}.

Linear speed: v=rω=0.5⋅4π=2π≈6.28v=r\omega=0.5\cdot 4\pi=2\pi\approx 6.28 m/s.

Tip

Tip: s=rθs=r\theta and A=12r2θA=\tfrac12 r^2\theta require radians. Convert first if the angle is in degrees. One revolution =2π=2\pi radians.

Try it
Quick check

(a) Convert 7π6\dfrac{7\pi}{6} to degrees and −225∘-225^\circ to radians. (b) Find the angle in [0,2π)[0,2\pi) coterminal with 17π4\dfrac{17\pi}{4}.

Show the answer

(a) 7π6⋅180∘π=210∘\dfrac{7\pi}{6}\cdot\dfrac{180^\circ}{\pi}=210^\circ and −225⋅π180=−5π4-225\cdot\dfrac{\pi}{180}=-\dfrac{5\pi}{4}. (b) π4\dfrac{\pi}{4}. Two full turns are 16π4\dfrac{16\pi}{4}, and 17π4−16π4=π4\dfrac{17\pi}{4}-\dfrac{16\pi}{4}=\dfrac{\pi}{4}.

The Unit Circle and the Six Trig Functions

Concept
Definition on the Unit Circle

On the unit circle (x2+y2=1x^2+y^2=1), the terminal side of angle θ\theta meets the circle at (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta). For any angle with terminal point (x,y)(x,y) at radius r=x2+y2r=\sqrt{x^2+y^2}:

sin⁡θ=yr,cos⁡θ=xr,tan⁡θ=yx,\sin\theta=\frac{y}{r},\quad \cos\theta=\frac{x}{r},\quad \tan\theta=\frac{y}{x},
csc⁡θ=ry,sec⁡θ=rx,cot⁡θ=xy.\csc\theta=\frac{r}{y},\quad \sec\theta=\frac{r}{x},\quad \cot\theta=\frac{x}{y}.

The point where the terminal side meets a circle of radius rr is the unit-circle point scaled by rr, so dividing its coordinates by rr gives (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta) for every angle, not just acute ones. On the unit circle itself r=1r=1 and the coordinates are the values.

Concept
Exact Values at Special Angles
θ0π6 (30∘)π4 (45∘)π3 (60∘)sin⁡θ0122232cos⁡θ1322212tan⁡θ03313θsin⁡/cos⁡/tan⁡π21/ 0/ undef.\begin{array}{c|cccc} \theta & 0 & \tfrac{\pi}{6}\,(30^\circ) & \tfrac{\pi}{4}\,(45^\circ) & \tfrac{\pi}{3}\,(60^\circ) \\\hline \sin\theta & 0 & \tfrac12 & \tfrac{\sqrt2}{2} & \tfrac{\sqrt3}{2} \\[3pt] \cos\theta & 1 & \tfrac{\sqrt3}{2} & \tfrac{\sqrt2}{2} & \tfrac12 \\[3pt] \tan\theta & 0 & \tfrac{\sqrt3}{3} & 1 & \sqrt3 \end{array} \qquad \begin{array}{c|c} \theta & \sin/\cos/\tan \\\hline \tfrac{\pi}{2} & 1/\,0/\,\text{undef.} \end{array}
Example
Reading the unit circle
(√2/2, √2/2)
At π4\tfrac{\pi}{4} the two coordinates are equal, so tan⁡π4=1\tan\tfrac{\pi}{4}=1.

Find the six trig functions of θ=π4\theta=\dfrac{\pi}{4}. The terminal point is (22,22)\left(\dfrac{\sqrt2}{2},\dfrac{\sqrt2}{2}\right), so

sin⁡π4=cos⁡π4=22,tan⁡π4=1,\sin\tfrac{\pi}{4}=\cos\tfrac{\pi}{4}=\tfrac{\sqrt2}{2},\quad \tan\tfrac{\pi}{4}=1,
csc⁡π4=sec⁡π4=2,cot⁡π4=1.\csc\tfrac{\pi}{4}=\sec\tfrac{\pi}{4}=\sqrt2,\quad \cot\tfrac{\pi}{4}=1.
Tip

Tip (“n/2\sqrt{n}/2” pattern): For sin⁡\sin at 0,30,45,60,900,30,45,60,90 read 02,12,22,32,42\dfrac{\sqrt0}{2},\dfrac{\sqrt1}{2},\dfrac{\sqrt2}{2},\dfrac{\sqrt3}{2},\dfrac{\sqrt4}{2}. Cosine runs the same list backward.

Try it
Quick check

Find sin⁡5π3\sin\dfrac{5\pi}{3}, cos⁡5π3\cos\dfrac{5\pi}{3} and tan⁡5π3\tan\dfrac{5\pi}{3}.

Show the answer

Answer: −32-\dfrac{\sqrt3}{2}, 12\dfrac12, −3-\sqrt3. 5π3\dfrac{5\pi}{3} is the π3\dfrac{\pi}{3} point reflected into Quadrant IV: (12,−32)\left(\tfrac12,-\tfrac{\sqrt3}{2}\right). The tangent is −3/21/2=−3\dfrac{-\sqrt3/2}{1/2}=-\sqrt3.

Right-Triangle Trig and Fundamental Identities

Concept
SOH-CAH-TOA

For an acute angle θ\theta in a right triangle with the opposite leg, adjacent leg, and hypotenuse:

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

The reciprocals are csc⁡θ,sec⁡θ,cot⁡θ\csc\theta,\sec\theta,\cot\theta.

Example
Using the triangle
345θ
Opposite 33, adjacent 44, hypotenuse 55.

For the 33-44-55 triangle shown, with θ\theta opposite the side 33, sin⁡θ=35\sin\theta=\dfrac35, cos⁡θ=45\cos\theta=\dfrac45, tan⁡θ=34\tan\theta=\dfrac34, and sec⁡θ=54\sec\theta=\dfrac54.

Concept
Fundamental Identities

Reciprocal:  csc⁡θ=1sin⁡θ\csc\theta=\dfrac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta=\dfrac{1}{\cos\theta}, cot⁡θ=1tan⁡θ\cot\theta=\dfrac{1}{\tan\theta}. Quotient:  tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta},   cot⁡θ=cos⁡θsin⁡θ\cot\theta=\dfrac{\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.

Example
Applying an identity

If sin⁡θ=35\sin\theta=\dfrac35 and θ\theta is in Quadrant I, then cos⁡θ=1−925=1625=45\cos\theta=\sqrt{1-\tfrac{9}{25}}=\sqrt{\tfrac{16}{25}}=\dfrac45, so tan⁡θ=3/54/5=34\tan\theta=\dfrac{3/5}{4/5}=\dfrac34.

Tip

Tip: From sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 divide by cos⁡2θ\cos^2\theta to get 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta; divide by sin⁡2θ\sin^2\theta to get 1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta. You only need to memorize one.

Try it
Quick check

θ\theta is acute and tan⁡θ=724\tan\theta=\dfrac{7}{24}. Find sin⁡θ\sin\theta and cos⁡θ\cos\theta.

Show the answer

Answer: sin⁡θ=725\sin\theta=\dfrac{7}{25}, cos⁡θ=2425\cos\theta=\dfrac{24}{25}. Take opposite 77 and adjacent 2424; the hypotenuse is 49+576=625=25\sqrt{49+576}=\sqrt{625}=25.

Reference Angles and Signs by Quadrant

Concept
Reference Angle

The reference angle θ′\theta' is the acute angle between the terminal side and the xx-axis. The value of a trig function at θ\theta equals its value at θ′\theta' up to a sign.

Quadrant of θReference angle θ′IθII180∘−θ  (π−θ)IIIθ−180∘  (θ−π)IV360∘−θ  (2π−θ)\begin{array}{c|c} \text{Quadrant of }\theta & \text{Reference angle }\theta' \\\hline \text{I} & \theta \\ \text{II} & 180^\circ-\theta \ \ (\pi-\theta)\\ \text{III} & \theta-180^\circ \ \ (\theta-\pi)\\ \text{IV} & 360^\circ-\theta \ \ (2\pi-\theta) \end{array}
Concept
Signs by Quadrant (“All Students Take Calculus”)
  • QI: all positive.   QII: only sin⁡\sin (and csc⁡\csc) positive.
  • QIII: only tan⁡\tan (and cot⁡\cot) positive.   QIV: only cos⁡\cos (and sec⁡\sec) positive.
Example
Evaluating any angle
30°(−√3/2, −1/2)
The 210∘210^\circ point is left of the yy-axis: cosine negative.

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

cos⁡210∘=−cos⁡30∘=−32.\cos 210^\circ=-\cos 30^\circ=-\tfrac{\sqrt3}{2}.
Example
A radian example

Evaluate sin⁡5π6\sin\dfrac{5\pi}{6}. QII, reference angle π−5π6=π6\pi-\tfrac{5\pi}{6}=\tfrac{\pi}{6}. Sine is positive in QII, so sin⁡5π6=sin⁡π6=12\sin\tfrac{5\pi}{6}=\sin\tfrac{\pi}{6}=\tfrac12.

Tip

Tip: Steps to evaluate any angle: (1) find a coterminal angle in [0,2π)[0,2\pi); (2) identify the quadrant; (3) find the reference angle; (4) evaluate at the reference angle; (5) attach the correct sign.

Try it
Quick check

Evaluate tan⁡300∘\tan 300^\circ and give the reference angle of 11π6\dfrac{11\pi}{6}.

Show the answer

Answer: tan⁡300∘=−3\tan300^\circ=-\sqrt3; the reference angle of 11π6\dfrac{11\pi}{6} is π6\dfrac{\pi}{6}. 300∘300^\circ is in Quadrant IV with reference angle 60∘60^\circ, and tangent is negative there. 11π6\dfrac{11\pi}{6} is in Quadrant IV too: 2π−11π6=π62\pi-\dfrac{11\pi}{6}=\dfrac{\pi}{6}.

Graphs of Sine and Cosine

Concept
Sinusoidal Form

For y=asin⁡(b(x−c))+dy=a\sin\big(b(x-c)\big)+d (and likewise for cosine) with b>0b>0:

  • Amplitude =∣a∣=|a|   Period =2πb=\dfrac{2\pi}{b}
  • Phase shift =c=c (right if c>0c>0)   Vertical shift =d=d (midline y=dy=d)

Each number acts at a different stage. Inside, x−cx-c delays the cycle so it starts at x=cx=c, and multiplying by bb makes the input sweep 2π2\pi in only 2πb\tfrac{2\pi}{b} units of xx. Outside, aa scales every height and dd lifts every point, so the wave swings between d−∣a∣d-|a| and d+∣a∣d+|a|.

y=2sin⁡xy=2\sin x (solid) and y=2cos⁡xy=2\cos x (dashed): amplitude 22, period 2π2\pi.

Example
Write and read a graph
xy−2−11234π/4π/23π/4π5π/43π/2y = 1
Midline y=1y=1, peak 44 at x=π2x=\tfrac{\pi}{2}, trough −2-2 at x=πx=\pi.

For y=3sin⁡ ⁣(2(x−π4))+1y=3\sin\!\big(2(x-\tfrac{\pi}{4})\big)+1: amplitude =3=3, period =2π2=π=\dfrac{2\pi}{2}=\pi, phase shift =π4=\dfrac{\pi}{4} right, midline y=1y=1. The graph oscillates between y=−2y=-2 and y=4y=4.

Tip

Tip: Always factor out bb before reading the phase shift: y=sin⁡(2x−π)=sin⁡ ⁣(2(x−π2))y=\sin(2x-\pi)=\sin\!\big(2(x-\tfrac{\pi}{2})\big) shifts right π2\tfrac{\pi}{2}, not π\pi.

Try it
Quick check

For y=−2cos⁡ ⁣(π3x)+5y=-2\cos\!\left(\dfrac{\pi}{3}x\right)+5, give the amplitude, period, range, and the value at x=0x=0.

Show the answer

Answer: amplitude 22; period 66; range [3,7][3,7]; y(0)=3y(0)=3. Period =2ππ/3=6=\dfrac{2\pi}{\pi/3}=6. The midline is 55, so the range is 5±25\pm2. At x=0x=0, −2cos⁡0+5=3-2\cos0+5=3: the negative sign starts the wave at its minimum.

Graphs of Tangent, Cotangent, Secant, Cosecant

Concept
Asymptotes and Periods
  • y=tan⁡xy=\tan x: period π\pi; vertical asymptotes where cos⁡x=0\cos x=0, i.e. x=π2+nπx=\tfrac{\pi}{2}+n\pi.
  • y=cot⁡xy=\cot x: period π\pi; asymptotes where sin⁡x=0\sin x=0, i.e. x=nπx=n\pi.
  • y=sec⁡x=1cos⁡xy=\sec x=\dfrac{1}{\cos x}: asymptotes where cos⁡x=0\cos x=0; period 2π2\pi.
  • y=csc⁡x=1sin⁡xy=\csc x=\dfrac{1}{\sin x}: asymptotes where sin⁡x=0\sin x=0; period 2π2\pi.

y=tan⁡xy=\tan x: period π\pi, asymptotes at x=±π2,±3π2x=\pm\tfrac{\pi}{2},\pm\tfrac{3\pi}{2}.

Example
Secant from cosine
xy−3−2−112−π/2π/2π
Each hump of cos⁡x\cos x (dashed) becomes a U of sec⁡x\sec x; each zero becomes an asymptote.

Because sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}, the secant curve has a U-shaped branch opening up wherever cos⁡x>0\cos x>0 (minimum value 11) and opening down wherever cos⁡x<0\cos x<0 (maximum value −1-1), with asymptotes at each zero of cosine.

Tip

Tip: Sketch sec⁡\sec/csc⁡\csc by first lightly drawing the matching cosine/sine wave: each hill becomes a U opening away from the midline, and each zero becomes an asymptote.

Try it
Quick check

Find the period and the vertical asymptotes of y=tan⁡2xy=\tan 2x.

Show the answer

Answer: period π2\dfrac{\pi}{2}; asymptotes x=π4+nπ2x=\dfrac{\pi}{4}+\dfrac{n\pi}{2}. Tangent repeats every π\pi of its input, and 2x2x covers π\pi when xx covers π2\tfrac{\pi}{2}. Asymptotes where 2x=π2+nπ2x=\tfrac{\pi}{2}+n\pi.

Inverse Trigonometric Functions

Concept
Definitions, Domains, and Ranges

To be invertible, each trig function is restricted:

FunctionDomainRangearcsin⁡x[−1,1][−π2,π2]arccos⁡x[−1,1][0,π]arctan⁡x(−∞,∞)(−π2,π2)\begin{array}{c|c|c} \text{Function} & \text{Domain} & \text{Range} \\\hline \arcsin x & [-1,1] & \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] \\[3pt] \arccos x & [-1,1] & [0,\pi] \\[3pt] \arctan x & (-\infty,\infty) & \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right) \end{array}

arcsin⁡x\arcsin x is the angle in the listed range whose sine is xx (similarly for the others).

Example
Evaluating inverse trig

arcsin⁡ ⁣(22)=π4\arcsin\!\left(\dfrac{\sqrt2}{2}\right)=\dfrac{\pi}{4}   (since sin⁡π4=22\sin\tfrac{\pi}{4}=\tfrac{\sqrt2}{2} and π4\tfrac{\pi}{4} is in range). arccos⁡ ⁣(−12)=2π3\arccos\!\left(-\dfrac12\right)=\dfrac{2\pi}{3}   (the angle in [0,π][0,\pi] with cosine −12-\tfrac12). arctan⁡(−1)=−π4\arctan(-1)=-\dfrac{\pi}{4}.

Example
Composition sin⁡(arccos⁡x)\sin(\arccos x)
θ435
For tan⁡(arcsin⁡35)\tan\left(\arcsin\tfrac35\right): opposite 33, hypotenuse 55, so adjacent 44.

Let θ=arccos⁡x\theta=\arccos x, so cos⁡θ=x\cos\theta=x with θ∈[0,π]\theta\in[0,\pi] (where sin⁡θ≥0\sin\theta\ge0). Then

sin⁡(arccos⁡x)=sin⁡θ=1−cos⁡2θ=1−x2.\sin(\arccos x)=\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-x^2}.

For a numeric case: tan⁡ ⁣(arcsin⁡35)\tan\!\left(\arcsin\tfrac35\right): with sin⁡θ=35\sin\theta=\tfrac35 (QI), cos⁡θ=45\cos\theta=\tfrac45, so the value is 34\tfrac34.

Tip

Tip: For compositions, draw a right triangle with the inner function's ratio, find the missing side by the Pythagorean theorem, then read off the outer function. Watch the range: arccos⁡\arccos outputs QI or QII, arcsin⁡\arcsin and arctan⁡\arctan output QI or QIV.

Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2
Try it
Quick check

Evaluate arccos⁡ ⁣(−32)\arccos\!\left(-\dfrac{\sqrt3}{2}\right) and tan⁡ ⁣(arccos⁡23)\tan\!\left(\arccos\dfrac23\right).

Show the answer

Answer: 5π6\dfrac{5\pi}{6} and 52\dfrac{\sqrt5}{2}. The angle in [0,π][0,\pi] with cosine −32-\tfrac{\sqrt3}{2} is 5π6\tfrac{5\pi}{6}. For the second, adjacent 22 and hypotenuse 33 give opposite 9−4=5\sqrt{9-4}=\sqrt5, so the tangent is 52\tfrac{\sqrt5}{2} (positive, since arccos⁡23\arccos\tfrac23 is in Quadrant I).

Going Deeper: Advanced Trig Ideas

Concept
Product-to-Sum and Sum-to-Product

These identities trade a product for a sum (easier to integrate or evaluate) and vice versa:

cos⁡Acos⁡B=12 ⁣[cos⁡(A−B)+cos⁡(A+B)]sin⁡Asin⁡B=12 ⁣[cos⁡(A−B)−cos⁡(A+B)]sin⁡Acos⁡B=12 ⁣[sin⁡(A+B)+sin⁡(A−B)]\begin{array}{l} \cos A\cos B=\tfrac12\!\left[\cos(A-B)+\cos(A+B)\right] \\[3pt] \sin A\sin B=\tfrac12\!\left[\cos(A-B)-\cos(A+B)\right] \\[3pt] \sin A\cos B=\tfrac12\!\left[\sin(A+B)+\sin(A-B)\right] \end{array}

Running them backward gives sum-to-product, e.g. sin⁡A+sin⁡B=2sin⁡ ⁣A+B2cos⁡ ⁣A−B2\sin A+\sin B=2\sin\!\dfrac{A+B}{2}\cos\!\dfrac{A-B}{2}.

Example
An unusual exact product: cos⁡20∘cos⁡40∘cos⁡80∘\cos20^\circ\cos40^\circ\cos80^\circ

Let P=cos⁡20∘cos⁡40∘cos⁡80∘P=\cos20^\circ\cos40^\circ\cos80^\circ. Multiply and divide by sin⁡20∘\sin20^\circ and use sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta repeatedly:

P=sin⁡20∘cos⁡20∘cos⁡40∘cos⁡80∘sin⁡20∘=12sin⁡40∘cos⁡40∘cos⁡80∘sin⁡20∘.P=\frac{\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ}{\sin20^\circ} =\frac{\tfrac12\sin40^\circ\cos40^\circ\cos80^\circ}{\sin20^\circ}.
=14sin⁡80∘cos⁡80∘sin⁡20∘=18sin⁡160∘sin⁡20∘=18sin⁡20∘sin⁡20∘=18,=\frac{\tfrac14\sin80^\circ\cos80^\circ}{\sin20^\circ} =\frac{\tfrac18\sin160^\circ}{\sin20^\circ} =\frac{\tfrac18\sin20^\circ}{\sin20^\circ}=\frac18,

using sin⁡160∘=sin⁡(180∘−20∘)=sin⁡20∘\sin160^\circ=\sin(180^\circ-20^\circ)=\sin20^\circ. So the exact value is 18\dfrac18.

Example
An unusual exact sum: cos⁡π7+cos⁡3π7+cos⁡5π7\cos\dfrac{\pi}{7}+\cos\dfrac{3\pi}{7}+\cos\dfrac{5\pi}{7}

Call the sum SS. Multiply by 2sin⁡π72\sin\dfrac{\pi}{7} and apply 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B=\sin(A+B)-\sin(A-B) to each term; the pieces telescope:

2sin⁡π7 S=sin⁡2π7+(sin⁡4π7−sin⁡2π7)+(sin⁡6π7−sin⁡4π7)=sin⁡6π7.2\sin\tfrac{\pi}{7}\,S=\sin\tfrac{2\pi}{7}+\big(\sin\tfrac{4\pi}{7}-\sin\tfrac{2\pi}{7}\big)+\big(\sin\tfrac{6\pi}{7}-\sin\tfrac{4\pi}{7}\big)=\sin\tfrac{6\pi}{7}.

Since sin⁡6π7=sin⁡π7\sin\tfrac{6\pi}{7}=\sin\tfrac{\pi}{7}, we get 2sin⁡π7 S=sin⁡π72\sin\tfrac{\pi}{7}\,S=\sin\tfrac{\pi}{7}, hence S=12S=\dfrac12.

Concept
Building a Sinusoid from Data

Given a periodic phenomenon with maximum value MM, minimum value mm, and known timing, model it as y=acos⁡ ⁣(b(t−c))+dy=a\cos\!\big(b(t-c)\big)+d (or with sin⁡\sin):

d=M+m2 (midline),a=M−m2 (amplitude),b=2πperiod.d=\frac{M+m}{2}\ (\text{midline}),\qquad a=\frac{M-m}{2}\ (\text{amplitude}),\qquad b=\frac{2\pi}{\text{period}}.

Choose cos⁡\cos with c=c= (time of the maximum) so no reflection is needed; the midline crossings occur a quarter period from each extreme.

Example
Full sinusoid from tide data
xy3691224681012y = 7
High tide 1212 ft at t=3t=3, low tide 22 ft at t=9t=9; the curve starts at t=0t=0 on the midline.

High tide of 1212 ft occurs at t=3t=3 h; the next low tide of 22 ft at t=9t=9 h. Find a model y(t)y(t) and the height at t=0t=0.

  • Midline d=12+22=7d=\dfrac{12+2}{2}=7; amplitude a=12−22=5a=\dfrac{12-2}{2}=5.
  • High-to-low is half a period, so period =2(9−3)=12=2(9-3)=12 h and b=2π12=π6b=\dfrac{2\pi}{12}=\dfrac{\pi}{6}.
  • Maximum at t=3⇒c=3t=3\Rightarrow c=3, so y(t)=5cos⁡ ⁣(π6(t−3))+7y(t)=5\cos\!\Big(\dfrac{\pi}{6}(t-3)\Big)+7.

At t=0t=0: y(0)=5cos⁡ ⁣(−π2)+7=5(0)+7=7y(0)=5\cos\!\big(-\tfrac{\pi}{2}\big)+7=5(0)+7=7 ft (a midline crossing, as expected a quarter period before the peak).

Concept
Inverse Composition and Range Traps

f−1(f(x))f^{-1}(f(x)) returns xx only when xx already lies in the restricted range of the inverse. Otherwise you must replace the inner angle with the coterminal/reference angle that does lie in range but has the same sine (or cosine, tangent):

arcsin⁡(sin⁡x)=x  ⟺  x∈[−π2,π2],arccos⁡(cos⁡x)=x  ⟺  x∈[0,π].\arcsin(\sin x)=x \iff x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\qquad \arccos(\cos x)=x \iff x\in[0,\pi].

Going the other way, sin⁡(arcsin⁡u)=u\sin(\arcsin u)=u and cos⁡(arccos⁡u)=u\cos(\arccos u)=u hold for all u∈[−1,1]u\in[-1,1] with no trap.

Example
Springing the range trap

Evaluate arcsin⁡ ⁣(sin⁡5π6)\arcsin\!\Big(\sin\dfrac{5\pi}{6}\Big). Here 5π6∉[−π2,π2]\tfrac{5\pi}{6}\notin\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right], so the answer is not 5π6\tfrac{5\pi}{6}. Since sin⁡5π6=12\sin\tfrac{5\pi}{6}=\tfrac12 and the in-range angle with sine 12\tfrac12 is π6\tfrac{\pi}{6},

arcsin⁡ ⁣(sin⁡5π6)=π6.\arcsin\!\Big(\sin\tfrac{5\pi}{6}\Big)=\frac{\pi}{6}.

Similarly arccos⁡ ⁣(cos⁡7π4)\arccos\!\Big(\cos\dfrac{7\pi}{4}\Big): cosine =22=\tfrac{\sqrt2}{2}, and the angle in [0,π][0,\pi] with that cosine is π4\tfrac{\pi}{4}, so the value is π4\dfrac{\pi}{4} (not 7π4\tfrac{7\pi}{4}).

Concept
Linked Gears and Angular-Speed Chains

Where two gears (or pulleys/wheels) mesh, the contact point shares one linear speed: v=r1ω1=r2ω2v=r_1\omega_1=r_2\omega_2. Hence

ω1ω2=r2r1=N2N1,\frac{\omega_1}{\omega_2}=\frac{r_2}{r_1}=\frac{N_2}{N_1},

where NiN_i is the tooth count. Wheels rigidly on the same axle share ω\omega, not vv. Chain these relations to pass speed through a train.

Example
Angular-speed chain

Gear A (rA=8r_A=8 cm) turns at 6060 rev/min and meshes with gear B (rB=2r_B=2 cm). Gear B is fixed on the same axle as gear C (rC=5r_C=5 cm). Find the linear speed at the rim of C.

  • ωA=60⋅2π=120π\omega_A=60\cdot2\pi=120\pi rad/min.
  • Mesh A–B: rAωA=rBωB⇒ωB=rArBωA=82(120π)=480πr_A\omega_A=r_B\omega_B\Rightarrow \omega_B=\dfrac{r_A}{r_B}\omega_A=\dfrac{8}{2}(120\pi)=480\pi rad/min.
  • Same axle B–C: ωC=ωB=480π\omega_C=\omega_B=480\pi rad/min.
  • Rim of C: vC=rCωC=5⋅480π=2400π≈7540v_C=r_C\omega_C=5\cdot480\pi=2400\pi\approx7540 cm/min.
Concept
Counting Solutions in [0,2π)[0,2\pi)

To count solutions of a trig equation on one period, reduce it to F(u)=kF(u)=k where u=nθu=n\theta:

  • A single sin⁡θ=k\sin\theta=k or cos⁡θ=k\cos\theta=k has 2 solutions in [0,2π)[0,2\pi) when ∣k∣<1|k|<1, 1 when ∣k∣=1|k|=1, 0 when ∣k∣>1|k|>1.
  • If the argument is nθn\theta, then θ∈[0,2π)\theta\in[0,2\pi) makes nθn\theta sweep [0,2nπ)[0,2n\pi), i.e. nn full periods, so multiply the base count by nn.
  • Factor first: a product =0=0 contributes the (deduplicated) union of each factor's solutions.
Example
How many solutions?
xy−11π/2π3π/2
Three periods of sin⁡3θ\sin 3\theta, two crossings of y=12y=\tfrac12 each.

How many θ∈[0,2π)\theta\in[0,2\pi) satisfy 2sin⁡(3θ)=12\sin(3\theta)=1, i.e. sin⁡(3θ)=12\sin(3\theta)=\tfrac12? Let u=3θu=3\theta. As θ\theta runs over [0,2π)[0,2\pi), uu runs over [0,6π)[0,6\pi) — three full periods. In each period sin⁡u=12\sin u=\tfrac12 has 22 solutions, so there are 3×2=63\times2=\boxed{6} solutions. Contrast: cos⁡2θ−cos⁡θ=0\cos^2\theta-\cos\theta=0 factors as cos⁡θ(cos⁡θ−1)=0\cos\theta(\cos\theta-1)=0. Then cos⁡θ=0\cos\theta=0 gives 22 solutions (π2,3π2)\big(\tfrac{\pi}{2},\tfrac{3\pi}{2}\big) and cos⁡θ=1\cos\theta=1 gives 11 solution (0)(0); no overlap, so 33 solutions total.

Tip

Tip (Pythagorean simplifying): When an expression mixes sin⁡2\sin^2/cos⁡2\cos^2 or sec⁡2\sec^2/tan⁡2\tan^2, look to collapse it with sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 or sec⁡2θ−tan⁡2θ=1\sec^2\theta-\tan^2\theta=1. For instance

1−cos⁡2θsin⁡θ=sin⁡2θsin⁡θ=sin⁡θ,sec⁡4θ−tan⁡4θ=(sec⁡2θ+tan⁡2θ)(sec⁡2θ−tan⁡2θ)=sec⁡2θ+tan⁡2θ.\frac{1-\cos^2\theta}{\sin\theta}=\frac{\sin^2\theta}{\sin\theta}=\sin\theta,\qquad \sec^4\theta-\tan^4\theta=(\sec^2\theta+\tan^2\theta)(\sec^2\theta-\tan^2\theta)=\sec^2\theta+\tan^2\theta.

Spotting the identity turns a messy fraction or difference of squares into a single term.

Why It Works

Radians are the natural unit because with them sin⁡x≈x\sin x \approx x for small xx — the arc and its sine nearly coincide — which makes lim⁡x→0sin⁡xx=1\lim_{x\to0}\tfrac{\sin x}{x} = 1 and hence ddxsin⁡x=cos⁡x\tfrac{d}{dx}\sin x = \cos x with no stray constants. In degrees the same derivative would carry a factor of π180\tfrac{\pi}{180} forever. Every clean formula in calculus involving trig depends on this choice of unit.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Amplitude, period and shifts
y=asin⁡(b(x−c))+d:amplitude=∣a∣,period=2π∣b∣,shift=c,midline=dy=a\sin\big(b(x-c)\big)+d:\quad \text{amplitude}=|a|,\quad \text{period}=\frac{2\pi}{|b|},\quad \text{shift}=c,\quad \text{midline}=d

What it means. aa stretches the wave, bb squeezes it, cc slides it sideways, dd moves the centre line.

Example. y=3sin⁡(2x)y=3\sin(2x) has amplitude 33 and period 2π2=π\tfrac{2\pi}{2}=\pi.

Why it works. sin⁡\sin repeats when its input advances by 2π2\pi. Here the input is b(x−c)b(x-c), so xx only needs to advance by 2πb\tfrac{2\pi}{b} to complete a cycle — the period shrinks as bb grows.

Tip. Amplitude uses ∣a∣|a| — a negative aa flips the wave but does not make the amplitude negative.