MATH1013 Calculus I - Hong Kong University of Science and ...

MATH1013 Calculus I

Trigonometric ratios1

Edmund Y. M. Chiang

Department of MathematicsHong Kong University of Science & Technology

September 24, 2014

1Stewart, James, “Single Variable Calculus, Early Transcendentals”, 7th edition, Brooks/Coles, 2012

Briggs, Cochran and Gillett: Calculus for Scientists and Engineers: Early Transcendentals, Pearson 2013

Trigonometric functions Similar triangles Some identities Circular Functions Inverse trigonometric functions

Trigonometric functions

Similar triangles

Some identities

Circular Functions

Inverse trigonometric functions

Recalling trigonometric functions

• Trigonometric ratios have been known to us since Babylonians’time because of the need of astronomical observation. TheGreeks turns the subject into an exact science.

• Earliest definitions Given an angle A with 0 < A < π/2.Then one can form a right angle triangle 4ABC with the sidea opposite to the angle A called the opposite side, the sidenext to the angle A called the adjacent side, and the sideopposite to the right angle, the hypotenuse.

• One can certainly include the consideration that A = 0 orA = π/2 by accommodating a straight line section as ageneralized triangle, which will be useful when we discusstrigonometry ratios for arbitrary triangles later.

Six trigonometric ratios

Figure: (Right angle triangle 4ABC : from Maxime Bocher, 1914)

sinA =opposite side

hypotenuse=

c, cosA =

adjacent side

hypotenuse=

tanA =opposite side

adjacent side=

The other three are

cscA =1

sinA, secA =

cosA, cotA =

Maxime Bocher’s book

Figure: (Bocher’s book (1914))

Maxime Bocher’s bookMaxime Bocher (1867-1918) was an American mathematicianstudied in Gottingen, Germany, in late nineteenth century, then theworld centre of mathematics. He later became a professor inmathematics at the Harvard university.

Figure: (M. Bocher) Courtesy of Harvard University Archives

Similar triangles

Figure: (Chapter One)

Trigonometric Tables

Figure: (Chapter One)

Right-angled triangle example I

Figure: Granville p. 3

Right-angled triangle example II

Some exercises

Right-angled triangle π/4

Right-angled triangle π/3 and π/6

Special angle ratios

A special angle application

Symmetry

• The point of the right-angled triangles show that triangleswith same angle share the same ratios.

• So any of the six ratios are invariant with respect to theactual dimensions (sizes) of the right-angled triangles.

• These ratios reflect some hidden secret of the threedimensional space we live in.

• As a matter of fact, these ratios appear from everyday lifeapplications to the most advanced scientific pursues.

• What we saw about the properties of trigonometric ratiosreflects our space has circular symmetry (i.e., symmetry to dowith circles).

• There are many more hidden symmetries in our space-time.

It really is about circles !!The crux of the matter of why any of the trigonometric ratios onlydepends on the angle and not on the size of the triangle reallymeans we are looking at a circle.

Figure: (Source: Wiki)

Radian measures

Figure: 1.57, Briggs et al

Central idea of radian measures

• The main idea is to use the ratio arc length of a circle to itsradius to represent the angle of the arc subtended at thecentre of the circle.

• The main point is that this ratio is independent of the radii.

• We start with full circles with radius r :

Full circle length

Radius=

r= 2π.

• The last expression illustrates that the ratio is independent ofthe r . That is, we normalize to unit circle

• For the general angle θ results from subtended from an arclength `r , we have

Part of full circle length

Radius=

Part of 2πr

r= Part of 2π = radian.

Why radian measures?• Why we prefer radians over degrees?• Why 360◦ if we want a unit? 360 degrees is arbitrary• It is dimensionaless• The real advantage is the trigonometric functions are built in

for radian measure in Calculus:d

dθsin θ = cos θ,

dθcos θ = − sin θ

• But with degrees (or similarly with other non-radian units),

dθsin θ =

180◦

)cos θ,

dθcos θ = −

180◦

)sin θ

• The real reason is Euler formula:

e iθ = cos θ + i sin θ, e iπ = −1 (i2 = −1).

Otherwise, it would become

e iπθ/180◦

= cos θ + i sin θ.

Trigonometric identities• The size of the right-angled triangle (circle) does not matter.

One immediately obtain the famous Pythagoras theorem

sin2 θ + cos2 θ = 1

where 0 ≤ θ ≤ π/2.• Dividing both sides by sin2 θ and cos2 θ yield, respectively

1 + cot2 θ = csc2 θ and tan2 θ + 1 = sec2 θ.

• Double-angle formulae:

sin 2θ = 2 cos θ sin θ, cos 2θ = cos2 θ − sin2 θ

• Half-angle formulae:

cos θ = 2 cos2θ

2− 1, sin θ = 1− 2 sin2 θ

Addition formulae

sin(x ± y) = sin x cos y ± cos x sin y

cos(x ± y) = cos x cos y ∓ sin x sin y .

•tan(x ± y) =

tan± tan y

1∓ tan x tan y

sin x cos y =1

2[sin(x + y) + sin(x − y)]

cos x cos y =1

2[cos(x + y) + cos(x − y)]

sin x sin y =1

2[cos(x − y)− cos(x + y)].

Proof of Sine additional formula• We standardise the hypotenuse to unit length.

Figure: (Source: Wikipedia)

Ratios beyond π2

• The question is now how to extend the angle beyond π2 , as

the unit circle model clearly indicates such possibility.

Figure: (Source: Bocher, page 34)

Ratios as circular functions: 0 ≤ θ ≤ 2π• Due to the property of similar triangles, it is sufficient that we

consider unit circle in extending to 0 ≤ θ ≤ 2π with the ratiosaugmented with appropriate signs from the xy−coordinateaxes.

Figure: (Source: Bocher, page 38)

Illustration of 0 ≤ θ ≤ π

Figure: Briggs, et al: 1.59

Illustration of 0 ≤ θ ≤ 2π

Ratios beyond 2π

• For sine and cosine functions, if θ is an angle beyond 2π, thenθ = φ+ k 2π for some 0 ≤ φ ≤ 2π. Thus one can write downtheir meanings from the definitions

sin(θ) = sin(φ+k 2π) = sinφ, cos(θ) = cos(φ+k 2π) = cosφ

where 0 ≤ φ ≤ 2π, and k is any integer.

• The case for tangent ratio is slightly different, with the firstextension to −π

2 ≤ θ ≤π2 , and then to arbitrary θ. Thus one

can write down

tan(θ) = tan(φ+ k π) = tanφ

where θ = φ+ kπ for some k and −π2 ≤ φ ≤

Illustration of 2π ≤ θ ≤ 4π

Illustration of −4π ≤ θ ≤ −2π

Figure: 1.62 (Publisher)

Periodic sine and cosine functions

Figure: (Source:Upper: sine, page 45; Lower: cosine, page 46)

Periodic sine/cosec functions (publisher)

Figure: Briggs, et al: 1.63a

Periodic cos/sec functions (publisher)

Figure: Briggs, et al: 1.63b

Periodic tangent function

Figure: (Source: tangent, Borcher page 46)

Periodic tangent functions (publisher)

Figure: 1.64a (publisher)

Periodic cotangent functions (publisher)

Figure: Briggs, et al: 1.64b

Graphs of −4π ≤ θ ≤ −2π

Inverse trigonometric functions• The sine and cosine functions map the [k 2π, (k + 1) 2π] onto

the range [−1, 1] for each integer k . So it is

many −→ one

so an inverse would be possible only if we suitably restrict thedomain of either sine and cosine functions. We note that eventhe image of [0, 2π] “covers” the [−1, 1] more than once.

• In fact, for the sine function, only the subset [π2 ,3π2 ] of [0, 2π]

would be mapped onto [−1, 1] exactly once. That is the sinefunction is one-one on [π2 ,

3π2 ].

• However, it’s more convenient to define the inverse sin−1 x on[−π

2 ,π2 ].

sin−1 x : [−1, 1] −→[− π

•cos−1 x : [−1, 1] −→

[0, π

One-one Sine

Figure: Briggs, et al: 1.71a

Inverse sine

Figure: Stewart: page 67

One-one Cosine

Figure: 1.71b (publisher)

Inverse cosine

Inverse trigonometric functions II

• The tangent has

tan−1 x : (−∞, ∞) −→[− π

]or the whole real axis.

• Examples/Exercises• Find (a) sin−1(

√3/2), (b) cos−1(−

√3/2),

(c) cos−1(cos 3π), (d) sin(sin−1 1/2).• Given θ = sin−1(2/5). Find cos θ and tan θ, (p. 48)

cos−1(cos( 7π6 )).

• Find alternative form of cot(cos−1(x/4)) in terms of x .• Find cos(sin−1(x/3)).• Why sin−1 x + cos−1 x = π/2?

Inverse tangent

Inverse cotangent

Inverse secant

Inverse cosec (publisher)

Remaining inverse functions

Further examples

• (a) tan−1(−1/√

• (b) sec−1(−2),

• (c) sin(tan−1 x).

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