Beyond the right tri­an­gle

In ear­lier classes, trigonom­e­try lived inside right tri­an­gles, so every angle was acute. That is too small a world. A wheel keeps turn­ing past ninety degrees, and a clock hand goes round and round. So here an angle becomes a rota­tion of any size, pos­i­tive or neg­a­tive, mea­sured in degrees or in radi­ans, and sine and cosine become func­tions of every real num­ber. You will con­vert between degrees and radi­ans, use l=rθl = r\theta, define the six trigono­met­ric func­tions on the unit cir­cle, and work out their signs, peri­ods and val­ues for large or neg­a­tive angles.

Angles and their mea­sure

Think of an angle as the amount a ray rotates about its start­ing point. Anti­clock­wise turns count as pos­i­tive and clock­wise turns as neg­a­tive. One full turn is 360∘360^\circ, and each degree is split fur­ther: 1∘=60′1^\circ = 60', 1′=60′′1' = 60''.

Degrees are a human choice. The radian comes from the cir­cle itself: it is the angle sub­tended at the cen­tre by an arc whose length equals the radius. A full turn is 2π2\pi radi­ans, so

Circle with centre O and two radii of length r enclosing an arc whose length is also r; the angle between the radii is marked 1 rad, about 57 degrees 16 minutes.
One radian: the arc is as long as the radius

π rad=180∘,1 rad≈57∘16′,1∘=π180 rad.\displaystyle \pi \text{ rad} = 180^\circ, \qquad 1 \text{ rad} \approx 57^\circ 16', \qquad 1^\circ = \frac{\pi}{180} \text{ rad}.

This is why radi­ans are so con­ve­nient. An arc of length ll on a cir­cle of radius rr sub­tends θ=lr\displaystyle \theta = \tfrac{l}{r} radi­ans, which gives the tidy for­mula l=rθl = r\theta. There is no stray con­ver­sion fac­tor to remem­ber.

Exam­ple 1. 75∘=75×π180=5π12\displaystyle 75^\circ = 75 \times \tfrac{\pi}{180} = \tfrac{5\pi}{12} rad. 52∘30′=52.5∘=7π24\displaystyle 52^\circ 30' = 52.5^\circ = \tfrac{7\pi}{24} rad.

Exam­ple 2. 44 rad =4×180π≈229∘5′\displaystyle = 4 \times \tfrac{180}{\pi} \approx 229^\circ 5' (using π≈227\displaystyle \pi \approx \tfrac{22}{7}: 4×180×722=229111∘\displaystyle \tfrac{4 \times 180 \times 7}{22} = 229\tfrac{1}{11}^\circ).

Exam­ple 3. A pen­du­lum 6060 cm long swings through an arc of 2222 cm. The angle it turns through is 2260=1130\displaystyle \tfrac{22}{60} = \tfrac{11}{30} rad =21∘= 21^\circ (with π=227\displaystyle \pi = \tfrac{22}{7}).

Pendulum of length 60 cm shown at both ends of its swing; the bob travels along a dashed arc of 22 cm, and the angle at the pivot is 22/60 = 11/30 radian.
Exam­ple 3: a 60 cm pen­du­lum swing­ing through a 22 cm arc

Exam­ple 4. A clock's minute hand is 2.12.1 cm long. In 2020 min­utes it turns through 13\displaystyle \tfrac{1}{3} of 2π2\pi, so its tip trav­els 2.1×2π3=1.4π≈4.4\displaystyle 2.1 \times \tfrac{2\pi}{3} = 1.4\pi \approx 4.4 cm.

The trigono­met­ric func­tions

Here is the key idea. Draw the unit cir­cle x2+y2=1x^2 + y^2 = 1. Turn a ray through angle xx (in radi­ans) from the pos­i­tive xx-axis; it meets the cir­cle at a point, and we sim­ply define that point to be P=(cos⁡x,sin⁡x)P = (\cos x, \sin x). For an acute angle this agrees with the tri­an­gle def­i­n­i­tions you already know, but now it works for any angle at all. The other four func­tions fol­low:

Unit circle x squared + y squared = 1 with a radius at angle x from the positive x-axis meeting the circle at P = (cos x, sin x); the horizontal leg is cos x and the vertical leg sin x.
On the unit cir­cle the point at angle x is (cos x, sin x)

tan⁡x=sin⁡xcos⁡x,cot⁡x=cos⁡xsin⁡x,sec⁡x=1cos⁡x,csc⁡x=1sin⁡x.\displaystyle \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x}, \quad \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x}.

Because PP lies on the unit cir­cle, sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 comes for free. Divide it through by the square of cosine, and then of sine, and you get 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x and 1+cot⁡2x=csc⁡2x1 + \cot^2 x = \csc^2 x.

Zeros. sin⁡x=0\sin x = 0 at x=nπx = n\pi, and cos⁡x=0\cos x = 0 at x=(2n+1)π2\displaystyle x = (2n + 1)\tfrac{\pi}{2}.

Signs depend on the quad­rant. Many stu­dents remem­ber the order as "all, sine, tan­gent, cosine" for the pos­i­tive ones in I, II, III and IV:

I II III IV
sin⁡\sin, csc⁡\csc ++ ++ −- −-
cos⁡\cos, sec⁡\sec ++ −- −- ++
tan⁡\tan, cot⁡\cot ++ −- ++ −-
Circle divided into four quadrants: I all positive, II sin and csc positive, III tan and cot positive, IV cos and sec positive.
Which func­tions are pos­i­tive in each quad­rant

Ranges. Since the point never leaves the unit cir­cle, −1≤sin⁡x,cos⁡x≤1-1 \le \sin x, \cos x \le 1. On the other hand tan⁡x\tan x and cot⁡x\cot x take every real value, and ∣sec⁡x∣,∣csc⁡x∣≥1\lvert \sec x \rvert, \lvert \csc x \rvert \ge 1.

Peri­ods. One full turn brings the point back to where it started, so sin⁡\sin, cos⁡\cos, sec⁡\sec, csc⁡\csc repeat every 2π2\pi. The func­tions tan⁡\tan and cot⁡\cot repeat sooner, every π\pi. Reflect­ing in the hor­i­zon­tal axis also gives sin⁡(−x)=−sin⁡x\sin(-x) = -\sin x and cos⁡(−x)=cos⁡x\cos(-x) = \cos x.

In prac­tice, to find a value you set­tle the sign from the quad­rant, and find the size from a right tri­an­gle or a known angle.

Worked val­ues

Exam­ple 5. Sup­pose sin⁡x=−817\displaystyle \sin x = -\tfrac{8}{17} with xx in quad­rant III. Only tan­gent and cotan­gent are pos­i­tive there, so cos⁡x=−1517\displaystyle \cos x = -\tfrac{15}{17}, tan⁡x=815\displaystyle \tan x = \tfrac{8}{15}, cot⁡x=158\displaystyle \cot x = \tfrac{15}{8}, sec⁡x=−1715\displaystyle \sec x = -\tfrac{17}{15} and csc⁡x=−178\displaystyle \csc x = -\tfrac{17}{8}.

Exam­ple 6. Sup­pose tan⁡x=−724\displaystyle \tan x = -\tfrac{7}{24} with xx in quad­rant IV. Then sec⁡2x=1+49576=625576\displaystyle \sec^2 x = 1 + \tfrac{49}{576} = \tfrac{625}{576}. Secant is pos­i­tive in IV, so sec⁡x=2524\displaystyle \sec x = \tfrac{25}{24}, giv­ing cos⁡x=2425\displaystyle \cos x = \tfrac{24}{25} and sin⁡x=−725\displaystyle \sin x = -\tfrac{7}{25}.

Exam­ple 7. For big angles, strip off whole peri­ods first. cos⁡25π6=cos⁡(4π+π6)=32\displaystyle \cos \tfrac{25\pi}{6} = \cos\left(4\pi + \tfrac{\pi}{6}\right) = \tfrac{\sqrt{3}}{2}. Next, tan⁡(−13π4)=−tan⁡(3π+π4)=−1\displaystyle \tan\left(-\tfrac{13\pi}{4}\right) = -\tan\left(3\pi + \tfrac{\pi}{4}\right) = -1. And sin⁡(−1470∘)=−sin⁡(1470∘−4×360∘)=−sin⁡30∘=−12\displaystyle \sin(-1470^\circ) = -\sin(1470^\circ - 4 \times 360^\circ) = -\sin 30^\circ = -\tfrac{1}{2}.

Prac­tice ques­tions

  1. Con­vert to radi­ans: 30∘30^\circ; 135∘135^\circ; −210∘-210^\circ; 40∘30′40^\circ 30'.
  2. Con­vert to degrees: 2π5\displaystyle \tfrac{2\pi}{5}; −7π6\displaystyle \tfrac{-7\pi}{6}; 33 rad (use π=227\displaystyle \pi = \tfrac{22}{7}).
  3. An arc of 3333 cm sub­tends 3π4\displaystyle \tfrac{3\pi}{4} at the cen­tre. Radius? (Use π=227\displaystyle \pi = \tfrac{22}{7}.)
  4. The wheels of a cart turn 180180 times a minute. Through how many radi­ans does a wheel turn in one sec­ond?
  5. Two cir­cles have arcs of equal length sub­tend­ing 72∘72^\circ and 54∘54^\circ. Ratio of radii?
  6. Find the other five func­tions: cos⁡x=35\displaystyle \cos x = \tfrac{3}{5}, xx in IV; csc⁡x=135\displaystyle \csc x = \tfrac{13}{5}, xx in II; cot⁡x=43\displaystyle \cot x = \tfrac{4}{3}, xx in III.
  7. Eval­u­ate: sin⁡17π3\displaystyle \sin \tfrac{17\pi}{3}; cos⁡1125∘\cos 1125^\circ; tan⁡29π6\displaystyle \tan \tfrac{29\pi}{6}; csc⁡(−750∘)\csc(-750^\circ).
  8. In which quad­rant is xx if sin⁡x<0\sin x < 0 and tan⁡x>0\tan x > 0?
  9. Prove sin⁡x1+cos⁡x+1+cos⁡xsin⁡x=2csc⁡x\displaystyle \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2\csc x.
  10. What is the least pos­i­tive period of sin⁡2x\sin 2x? Of tan⁡3x\tan 3x?

Answers

Show answers
  1. π6\displaystyle \tfrac{\pi}{6}; 3π4\displaystyle \tfrac{3\pi}{4}; −7π6\displaystyle -\tfrac{7\pi}{6}; 9π40\displaystyle \tfrac{9\pi}{40}.
  2. 72∘72^\circ; −210∘-210^\circ; 171911∘≈171∘49′\displaystyle 171\tfrac{9}{11}^\circ \approx 171^\circ 49'.
  3. r=333π/4=14\displaystyle r = \tfrac{33}{3\pi/4} = 14 cm.
  4. 33 turns per sec­ond: 6π6\pi rad.
  5. l=r12π5=r23π10\displaystyle l = r_1 \tfrac{2\pi}{5} = r_2 \tfrac{3\pi}{10}: r1:r2=3:4r_1 : r_2 = 3 : 4.
  6. sin⁡=−45\displaystyle \sin = -\tfrac{4}{5}, tan⁡=−43\displaystyle \tan = -\tfrac{4}{3}, cot⁡=−34\displaystyle \cot = -\tfrac{3}{4}, sec⁡=53\displaystyle \sec = \tfrac{5}{3}, csc⁡=−54\displaystyle \csc = -\tfrac{5}{4}. sin⁡=513\displaystyle \sin = \tfrac{5}{13}, cos⁡=−1213\displaystyle \cos = -\tfrac{12}{13}, tan⁡=−512\displaystyle \tan = -\tfrac{5}{12}, cot⁡=−125\displaystyle \cot = -\tfrac{12}{5}, sec⁡=−1312\displaystyle \sec = -\tfrac{13}{12}. tan⁡=34\displaystyle \tan = \tfrac{3}{4}, sin⁡=−35\displaystyle \sin = -\tfrac{3}{5}, cos⁡=−45\displaystyle \cos = -\tfrac{4}{5}, csc⁡=−53\displaystyle \csc = -\tfrac{5}{3}, sec⁡=−54\displaystyle \sec = -\tfrac{5}{4}.
  7. −32\displaystyle -\tfrac{\sqrt{3}}{2}; 12\displaystyle \tfrac{1}{\sqrt{2}}; −13\displaystyle -\tfrac{1}{\sqrt{3}}; −2-2.
  8. III.
  9. The left side is sin⁡2x+(1+cos⁡x)2sin⁡x(1+cos⁡x)=2+2cos⁡xsin⁡x(1+cos⁡x)=2sin⁡x\displaystyle \frac{\sin^2 x + (1 + \cos x)^2}{\sin x(1 + \cos x)} = \frac{2 + 2\cos x}{\sin x(1 + \cos x)} = \frac{2}{\sin x}.
  10. π\pi; π3\displaystyle \tfrac{\pi}{3}.