16. Review of Trigonometry
Adapted from: Classical single-variable calculus preparation (radians, directed angles, unit-circle sine/cosine). Rewritten for teaching — not a verbatim textbook dump.
Before derivatives of trig functions, fix radian measure and the unit-circle definitions of sine and cosine.
Angle measure
Degrees: a full turn is 360 ∘ 360^\circ 36 0 ∘ radians .
On a circle of radius 1 1 1 C C C C A CA C A C B CB C B A B ⌢ \overset{\frown}{AB} A B ⌢ 1 1 1 ∠ A C B \angle ACB ∠ A C B one radian .
Fig. 16-1 — arc length 1 on the unit circle defines 1 radian Comparing to a full turn (360 ∘ 360^\circ 36 0 ∘ 2 π 2\pi 2 π
1 radian = 180 π degrees ≈ 57.3 ∘ , 1 ∘ = π 180 radians . 1\text{ radian}=\frac{180}{\pi}\text{ degrees}\approx 57.3^\circ,
\qquad
1^\circ=\frac{\pi}{180}\text{ radians}. 1 radian = π 180 degrees ≈ 57. 3 ∘ , 1 ∘ = 180 π radians . Common conversions (corrected table)
Degrees Radians 30 30 30 π / 6 \pi/6 π /6 45 45 45 π / 4 \pi/4 π /4 60 60 60 π / 3 \pi/3 π /3 90 90 90 π / 2 \pi/2 π /2 180 180 180 π \pi π 270 270 270 3 π / 2 3\pi/2 3 π /2 360 360 360 2 π 2\pi 2 π
Some printings list 180 ∘ 180^\circ 18 0 ∘ π / 2 \pi/2 π /2 π \pi π
Arc length
On a circle of radius r r r θ \theta θ radians cuts off arc length
s = r θ . s=r\theta. s = r θ . (Reason: θ / ( 2 π ) = s / ( 2 π r ) \theta/(2\pi)=s/(2\pi r) θ / ( 2 π ) = s / ( 2 π r )
Fig. 16-3 — s = rθ with θ in radians Check: radians
Pick an answer to see if you are right.
Directed angles
Count a rotation positive if counterclockwise and negative if clockwise. Angles may exceed one full turn (e.g. 5 π / 2 = 2 π + π / 2 5\pi/2=2\pi+\pi/2 5 π /2 = 2 π + π /2 3 π 3\pi 3 π
Fig. 16-4 — +π/2 vs −π/2 Fig. 16-5 — multi-turn directed angles Sine and cosine (unit circle)
Place the origin at O O O A = ( 1 , 0 ) A=(1,0) A = ( 1 , 0 ) O A OA O A θ \theta θ B B B
cos θ = \cos\theta= cos θ = x x x B B B sin θ = \sin\theta= sin θ = y y y B B B
Fig. 16-6 — (cos θ, sin θ) on the unit circle Examples
θ \theta θ B B B cos θ \cos\theta cos θ sin θ \sin\theta sin θ 0 0 0 2 π 2\pi 2 π ( 1 , 0 ) (1,0) ( 1 , 0 ) 1 1 1 0 0 0 π / 2 \pi/2 π /2 ( 0 , 1 ) (0,1) ( 0 , 1 ) 0 0 0 1 1 1 π \pi π ( − 1 , 0 ) (-1,0) ( − 1 , 0 ) − 1 -1 − 1 0 0 0 3 π / 2 3\pi/2 3 π /2 ( 0 , − 1 ) (0,-1) ( 0 , − 1 ) 0 0 0 − 1 -1 − 1
Match with right-triangle ratios
For an acute angle, a similar right triangle with hypotenuse 1 1 1
cos θ = adjacent hypotenuse , sin θ = opposite hypotenuse . \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\qquad
\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}. cos θ = hypotenuse adjacent , sin θ = hypotenuse opposite . Fig. 16-7 — unit hypotenuse matches the classical ratios Special values
θ \theta θ degrees cos θ \cos\theta cos θ sin θ \sin\theta sin θ 0 0 0 0 ∘ 0^\circ 0 ∘ 1 1 1 0 0 0 π / 6 \pi/6 π /6 30 ∘ 30^\circ 3 0 ∘ 3 / 2 \sqrt{3}/2 3 /2 1 / 2 1/2 1/2 π / 4 \pi/4 π /4 45 ∘ 45^\circ 4 5 ∘ 2 / 2 \sqrt{2}/2 2 /2 2 / 2 \sqrt{2}/2 2 /2 π / 3 \pi/3 π /3 60 ∘ 60^\circ 6 0 ∘ 1 / 2 1/2 1/2 3 / 2 \sqrt{3}/2 3 /2 π / 2 \pi/2 π /2 90 ∘ 90^\circ 9 0 ∘ 0 0 0 1 1 1 π \pi π 180 ∘ 180^\circ 18 0 ∘ − 1 -1 − 1 0 0 0 3 π / 2 3\pi/2 3 π /2 270 ∘ 270^\circ 27 0 ∘ 0 0 0 − 1 -1 − 1
Hands On: Convert and evaluate
Basic identities
(16.1) Period 2 π 2\pi 2 π cos ( θ + 2 π ) = cos θ \cos(\theta+2\pi)=\cos\theta cos ( θ + 2 π ) = cos θ sin ( θ + 2 π ) = sin θ \sin(\theta+2\pi)=\sin\theta sin ( θ + 2 π ) = sin θ
(16.2) Even / odd: cos ( − θ ) = cos θ \cos(-\theta)=\cos\theta cos ( − θ ) = cos θ sin ( − θ ) = − sin θ \sin(-\theta)=-\sin\theta sin ( − θ ) = − sin θ
Fig. 16-8 — reflection across the x-axis (16.3) Pythagorean: sin 2 θ + cos 2 θ = 1 \sin^{2}\theta+\cos^{2}\theta=1 sin 2 θ + cos 2 θ = 1 sin 2 θ = 1 − cos 2 θ \sin^{2}\theta=1-\cos^{2}\theta sin 2 θ = 1 − cos 2 θ cos 2 θ = 1 − sin 2 θ \cos^{2}\theta=1-\sin^{2}\theta cos 2 θ = 1 − sin 2 θ
(16.4) Quadrant signs
Fig. 16-9 — ASTC / “all students take calculus” pattern Quadrant sin \sin sin cos \cos cos I + + + + + + II + + + − - − III − - − − - − IV − - − + + +
Check: identities
A sin θ B −cos θ C −sin θ D cos θ
Pick an answer to see if you are right.
Polar coordinates
(16.5) For a point A ( x , y ) ≠ O A(x,y)\neq O A ( x , y ) = O r = x 2 + y 2 r=\sqrt{x^{2}+y^{2}} r = x 2 + y 2 θ \theta θ x x x O A OA O A ( r , θ ) (r,\theta) ( r , θ ) polar coordinates of A A A
x = r cos θ , y = r sin θ . x=r\cos\theta,\qquad y=r\sin\theta. x = r cos θ , y = r sin θ . (Same as scaling the unit-circle point ( cos θ , sin θ ) (\cos\theta,\sin\theta) ( cos θ , sin θ ) r r r
Fig. 16-10 — (r, θ) ↔ (x, y) = (r cos θ, r sin θ) We take as a starting point (proved later / Problem 11 in the source notes):
(16.6) cos ( u − v ) = cos u cos v + sin u sin v \displaystyle \cos(u-v)=\cos u\cos v+\sin u\sin v cos ( u − v ) = cos u cos v + sin u sin v
Then quickly:
(16.7) cos ( u + v ) = cos u cos v − sin u sin v \cos(u+v)=\cos u\cos v-\sin u\sin v cos ( u + v ) = cos u cos v − sin u sin v v v v − v -v − v
(16.8) cos ( π / 2 − v ) = sin v \cos(\pi/2-v)=\sin v cos ( π /2 − v ) = sin v sin ( π / 2 − v ) = cos v \sin(\pi/2-v)=\cos v sin ( π /2 − v ) = cos v u = π / 2 u=\pi/2 u = π /2
(16.9) sin ( u + v ) = sin u cos v + cos u sin v \sin(u+v)=\sin u\cos v+\cos u\sin v sin ( u + v ) = sin u cos v + cos u sin v sin ( u + v ) = cos ( π / 2 − ( u + v ) ) \sin(u+v)=\cos\bigl(\pi/2-(u+v)\bigr) sin ( u + v ) = cos ( π /2 − ( u + v ) )
(16.10) sin ( u − v ) = sin u cos v − cos u sin v \sin(u-v)=\sin u\cos v-\cos u\sin v sin ( u − v ) = sin u cos v − cos u sin v v v v − v -v − v
Check: addition
A cos u cos v − sin u sin v B sin u sin v only C sin u cos v + cos u sin v D cos u cos v + sin u sin v
Pick an answer to see if you are right.
Double- and half-angle
(16.11) cos 2 u = cos 2 u − sin 2 u = 2 cos 2 u − 1 = 1 − 2 sin 2 u \cos 2u=\cos^{2}u-\sin^{2}u=2\cos^{2}u-1=1-2\sin^{2}u cos 2 u = cos 2 u − sin 2 u = 2 cos 2 u − 1 = 1 − 2 sin 2 u v = u v=u v = u sin 2 + cos 2 = 1 \sin^{2}+\cos^{2}=1 sin 2 + cos 2 = 1
(16.12) sin 2 u = 2 sin u cos u \sin 2u=2\sin u\cos u sin 2 u = 2 sin u cos u v = u v=u v = u
(16.13) cos 2 ( u 2 ) = 1 + cos u 2 \displaystyle \cos^{2}\!\left(\frac{u}{2}\right)=\frac{1+\cos u}{2} cos 2 ( 2 u ) = 2 1 + cos u cos u = 2 cos 2 ( u / 2 ) − 1 \cos u=2\cos^{2}(u/2)-1 cos u = 2 cos 2 ( u /2 ) − 1
(16.14) sin 2 ( u 2 ) = 1 − cos u 2 \displaystyle \sin^{2}\!\left(\frac{u}{2}\right)=\frac{1-\cos u}{2} sin 2 ( 2 u ) = 2 1 − cos u
Hands On: Double-angle check
Laws of cosines and sines
(16.15a) Law of Cosines. In △ A B C \triangle ABC △ A B C c c c θ \theta θ a a a b b b c c c C C C
c 2 = a 2 + b 2 − 2 a b cos θ . c^{2}=a^{2}+b^{2}-2ab\cos\theta. c 2 = a 2 + b 2 − 2 ab cos θ . When θ = π / 2 \theta=\pi/2 θ = π /2
(16.15b) Law of Sines.
sin A a = sin B b = sin C c . \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}. a sin A = b sin B = c sin C . Fig. 16-11 — c² = a² + b² − 2ab cos θ Check: triangle laws
If θ = π/2 in the law of cosines, then…
A The formula fails B c² = a² + b² C c² = a² − b² D c = a + b
Pick an answer to see if you are right.
Radians for calculus: 1 rad = 180 / π 1\text{ rad}=180/\pi 1 rad = 180/ π s = r θ s=r\theta s = r θ Unit circle: ( cos θ , sin θ ) (\cos\theta,\sin\theta) ( cos θ , sin θ ) ( 1 , 0 ) (1,0) ( 1 , 0 ) θ \theta θ Polar: x = r cos θ x=r\cos\theta x = r cos θ y = r sin θ y=r\sin\theta y = r sin θ Core engine: cos ( u − v ) \cos(u-v) cos ( u − v ) sin ( u ± v ) \sin(u\pm v) sin ( u ± v ) Triangles: law of cosines c 2 = a 2 + b 2 − 2 a b cos θ c^{2}=a^{2}+b^{2}-2ab\cos\theta c 2 = a 2 + b 2 − 2 ab cos θ sin A / a = ⋯ \sin A/a=\cdots sin A / a = ⋯ Further Reading