list of trigonometric identities - usna · 2014-01-28 · 1/28/2014 physics handout series.tank:...

Trigonometry – Basic Proofs and Methods

Contact: [email protected]

Concepts of primary interest:

Geometric representations

Proofs of the cos(-) and sin(-) identities

Laws of Sines and of Cosines

Derivative of sin() w.r.t.

Radian measure

Derivatives of trigonometric functions

Hyperbolic sinh(u), cosh(u), …..; sinh-1(z), cosh-1(z), …..

Tools of the Trade

The unit circle

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

O: side opposite

A: side adjacent to

H: hypotenusesin

cos

tan

OH

AH

OA

Table of Values to be Memorized

(rad.) sin cos tan

0 0 0 1 0

300 ½ 3½/2 (.866) 3-½ (.5774)

370 36.87o .6435 3/5

4/5 3/4

450 2-½ (.707) 2-½ (.707) 1

530 +53.130 .9273 4/5

3/5 4/3

600 3½/2 (.866) ½ 3½ (1.732)

900 1 0

The ‘370’ and ‘530’ entries are to represent a 3-4-5 triangle.

1/28/2014 Physics Handout Series.Tank: Basic Trig Trig-2

The trigonometric functions are geometric in nature so geometric arguments are to be used to

develop the fundamental identities and to prove that:

0

sin( )1Limit

This limit plus a few trigonometric identities are required to the prove that: sin( )

cos( )d

d

. Given

this anchor, the derivatives of the remaining trigonometric functions can be computed.

The trigonometric functions are ubiquitous in the description of physical problems as are their

derivatives. The goal is to base these derivatives on geometric arguments. Afterward, the results are

to be reinforced by alternative developments based on the known derivatives of other functions.

The familiar results sin( ) cos( )

cos( ) sin( )andd d

d d

are valid if and only if the angle

is measured in radians. Radian measure of angles is assumed throughout these math method guides.

Radian measure is defined in terms of an angle with is vertex at the center of a circle. The angle is

the ratio of the arc length that subtends the angle to the radius of the circle sR where s is the

length of the arc segment between A and B. For the remainder of this section, we will set R = 1.

A

B

x

y Circle of Radius R

O

s

Exercise: The derivative is defined as a limit.


0 0

sin( ) sin( ) sin( ) sin( )0

das

d

Put your calculator in the degree mode and compute the limit above for the sequence of angles =

100, 10, 0.10, 0.010, 0.0010, ... and compare the result with that for using the radian mode and the

sequence = 0.1 rad, 0.01 rad, 0.001 rad, 0.0001 rad, ... . Compare the limit in degree mode with

the value π /180.

Serious errors can result if angles are expressed in units other than radians as all

relations between linear and angular quantities assume radian measure. Always

convert angular values to radians before any calculations. Convert back to the units

used in the statement of the problem only after the calculations are complete.

EXAMPLE: The Unit Circle and Trig Identities: Many trigonometric identities can be proven

with a few geometric steps if a unit circle (a circle of radius one) is used. We shall show that:

cos( - ) = cos cos + sin sin. [Trig.1]

Note that the points C, D, E and F are located by dropping perpendiculars.

OC = cos( - )

OD = cos(AD = sin()

OE = OD cos() = cos() cos()

AF = AD sin() = sin() sin()

EC = AF OC = OE + AF

The final geometric equation for the length of the line from O to C is equivalent to the identity that

was to be proved. [You should justify the identification of the angle ADF as .]


Exercise: Use the figures above to establish the identities for OE = cos(-) and AE = sin(-).

Exercise: Use the identity above and the definition cosA B A B

to show x x y yA B A B A B

.

Describe the angle represented by .

Exercise: Prove that: sin( = sin cos + cos sin. [Trig.2]

The extended unit circle:


Table TR1: Trigonometric and Hyperbolic Functions

2 2sin cos 1x x sin costan x xx

sin( ) sin cos cos sinx y x y x y cos( ) cos cos sin sinx y x y x y


tan tantan( )

1 tan tan

x yx y

x y

2

2 tantan(2 )

1 tan

xx

x

sin(x) sin(y) = ½[cos(x - y) – cos(x + y)] cos(x) cos(y) = ½[cos(x - y) + cos(x + y)]

sin(x) cos(y) = ½[sin(x - y) + sin(x + y)]

sin(2 ) 2sin cosx x x 2 2cos(2 ) cos sinx x x

2 1 cos(2 )sin

2

xx

2 1 cos(2 )

cos2

xx

2 21 tan sec ;1 1cos sinsec csc

3sin(3 ) 3sin 4 sinx x x 3cos(3 ) 4cos 3 cosx x x

cos( ) sin( )ie i cosh( ) sinh( )xe x x

sin2

i ie e

i

cos2

i ie e

sinh( )2

x xe ex

cosh( )

2

x xe ex

2

2

1tan

1

i i i

i i i

e e e

i e e i e

2

2

1tanh

1

x x x

x x x

e e ex

e e e

2 2cosh sinh 1x x 2 21 tanh sechx x

Exercise: Begin with sin2 + cos2 = 1, and show that 1 + tan2 = sec2.

Exercise: Use these identities to develop expression for sin[x] + sin[y] and cos[x] + cos[y]. Continue

to find a similar expression for sin[x] + cos[y].

Answers: sin(x) + sin(y) = 2 sin[½(x + y)] cos[½(x - y)],

cos(x) + cos(y) = 2 cos[½(x + y)] cos[½(x - y)]

sin(x) + cos(y) = sin(x) + sin(y + ½ ) = 2 sin[½(x + y + ½ )] cos[½(x - y - ½ )]

Expansions for small arguments

function

Small argument limit x << 1


cos(x)

1 - 1/2 x2

+ 1/24 x4 + O[x6]

0

2

(2 )!1m

m

mxm

sin(x)

x - 1/6 x3

+ 1/120 x5 + O[x7]

0

2 1

(2 1)!1m

m

mxm

tan(x)

x + 1/3 x3

+ 2/15 x5 + O[x7] 2

0

2 1

(2 )!1 ( 4) (1 4 )m m m

mm

mxmB

The B2m are Bernoulli numbers. That is: We do not have s simple explicit expansion for tangent. It is

therefore a curiosity that arctan(x) = 0

2 1

2 11 1m

m

mxm for x

.

Law of Sines:

Consider a general triangle with sides of length

A, B and C. The angles, and are opposite the

sides A, B and C respectively. A perpendicular is

dropped from one vertex to the side opposite. It

has length D where D = B sin = A sin. It

follows that the ratio of the sine of each vertex

angle to the length of the side opposite is a

constant for the triangle. A/sin = B/sin C/sin

Exercise: Show that C sin = A sin where is the angle opposite to the side of length C or between

the sides of length A and B.

Law of Cosines: C2 = A2 + B2 – 2 A B cos

Use C = x + y or C2 = x2 + y2 + 2 x y. The side D

divides the triangle into two right triangles.

x2 = B2 – D2 and y2 = A2 – D2, cos2 = 1 – sin2.

Note that: x = B cos and y = A cos. Combining

with and the law of sines relation for the ratio of

sides A and B:

C2 = A2 + B2 + 2 A B (cos cos - sin sin)

A

B

C

D


The last equation transforms into the standard from of the law of cosines. C2 = A2 + B2 – 2 A B cos

Exercise: Show that C2 = A2 + B2 + 2 A B (cos cos - sin sin) is equivalent to the more standard

form of the law of cosines, C2 = A2 + B2 – 2 A B cos. Use = - ( +). Identify the angle .

Exercise: Given that C A B

, compute C2 as C C

. Relate the result to the law of cosines. How is

the angle between the directions of A

and B

related to ?

The Great Inequalities:

There are two inequalities associated with the results above. One is the Schwarz inequality which

states that the magnitude of the inner product of two vectors is less or equal to the product of the

magnitudes of the two vectors.

A B A B

The second is the triangle inequality which states that the length of any side of a triangle is

bounded by the absolute value of the difference in the lengths of the other two sides and the sum of

the lengths of the other two sides.

C A B

A B

2 2 2 2 cosA B ABC

A

B

C

D x

y C = x + y

= - -


A B A B A B

For now, both results follow from the definition of the inner product, cos ABA B A B

and the

standard bounds on the range of cos (-1 cos 1). The angle AB is the angle between the

directions of A

and B

.

Derivatives of trigonometric functions

The next task is to prove that sin( )

cos( )d

d

using the trig identity above and the limit of sin

/.

0 0

00

0 0

sin( ) sin( ) sin( ) cos( ) cos( )sin( ) sin( )

sin( ) cos( ) 1 cos( )sin( )

cos( )sin( ) sin( )cos( ) cos

where several theorems on limits and trig identities have been invoked. See the exercise below.

Exercise: Use cos(+)= cos cos - sin sin = 1 – 2 sin2 to show that

1 - cos(2 ) = 2 sin2 so that:

00

0

2sin( )sin( )cos( ) 1 2 2 sin 022 2

Once

0

sin( ) cos( ) 1

has been shown to vanish in the exercise above, the derivative

reduces to: sin( )d

d

= 0

sin( )cos( )

. The evaluation of the limiting value of sin

/ remains

as the last hurdle between us and the well-known result: sin( )

cos( )d

d

.

The standard limits are to be assumed.


0

sin 0Limit

and 0

cos 1Limit

The area of a circle is 212 2 r so the area of a

pie wedge with angle is 212 r . Working

with the unit circle below, we conclude that the

area of the pie wedge Ocd of radius cos()

is less than the area of the triangle Ocb which is

less than the area of the unit radius pie wedge

Oab.

Area(Ocd) < Area(cb) < Area(Oab)

21 1 12 2 2cos ( ) cos( )sin( )

a

b

x

+y Circle of Radius 1

O c

d

cos()

Divide by 12 cos( ) :

sin( ) 1cos( )

cos( )

0 0

sin( ) 1 sin( )cos( ) 1

cos( )1Limit Limit

0

sin( )1Limit

[Trig.3]

Alternative Derivations (The developments below are circular. Why?)

The sine function can be represented as a power series

3 2 11 ( 1)sin( ) ( )

3 ! (2 1)!

nnt t t t

n

which can be differentiated term by tern to yield

2 2sin( ) 1 ( 1)1 cos( )

2 ! (2 )!

nnd t

dtt t t

n

Euler's identity can be used in reverse:

sin( ) ( )sin( )

2 2 2; cos( )

it it it it it ite e d t ie i e e et

i dt it

[Trig.4]


Exercise: Use the third order Taylor’s series expansions for sin() and cos() to show that :

cos() < sin() < for small angles .

The Philosophy of these Handouts: You can use principles and results more effectively if you

believe them. You can use them even more effectively when you fought and bled to prove them.

Elegant proofs are of little value. Whenever possible, the proof or solution should be in the same

context and 'language' as the problems to which it is to be applied. It should grapple with the issues

that arise when the result is used.

The Pythagorean Theorem: To show that c2 = a2 + b2 for a right triangle.

Consider the right triangle illustrated. As + = 90o, four congruent copies of this triangle may be

formed into a square of side c as shown. The area of the square is the area of the four triangles plus

the area of the internal square that has a side length of (b - a). Thus

c2 = 4 (1/2 ab) + (b - a)2 = a2 + b2 [QED]

a

b

c

c

c

(b - a)

Ex: A second proof follows by dropping a perpendicular to the hypotenuse to form several similar

triangles. The length c can now be represented as the sum of two pieces. Complete the proof.

Hyperbolic Functions: Definitions:


sinh( )2

x xe ex

cosh( )

2

x xe ex

1 2sech( ) cosh( ) x xx x

e e

2

2

sinh( ) 1tanh( )

cosh( ) 1

x x x

x x x

x e e ex

x e e e

2 2cosh ( ) sinh ( ) 1x x 2 21 tanh ( ) sech ( )x x

sinh(x + y) = sinh(x) cosh(y) + sinh(y) cosh(x)

tanh( ) tanh( )tanh( )

1 tanh( ) tanh( )

x yx y

x y

cosh(x + y) = cosh(x) cosh(y) + sinh(y) sinh(x)

sinh(2x) =2 sinh(x) cosh(x)

cosh(2x) = sinh2(x) + cosh2(x)

= 2 sinh2(x) +1=2 cosh2(x) -1

2

2 tanh( )tanh(2 )

1 tanh ( )

xx

x

Plots:

Plot[{Cosh[x],Sinh[x]},{x,-3,3},PlotRange {-10,10}] Plot[{Tanh[x],Sech[x]},{x,-3,3},PlotRange {-1,1}]

Limiting Forms for large and small arguments

function

Large argument limit x >> 1

Small argument limit x << 1

cosh(x)

½ e x [ 1 + e -2x ]

1 + 1/2 x2

+ O[x4]

sinh(x)

½ e x [ 1 - e -2x ]

x + 1/6 x3

+ O[x5]


tanh(x)

1 – 2 e -2x

x – 1/3 x3

+ O[x5]

The symbol collection ‘+ O[x4]’ is read as plus terms of order x4 and higher.

The symbol ' ' indicates that the representation is exact.

**** See Problem 11 ****

Finding inverse functions for the hyperbolic functions:

Deduce inverses such as sinh-1(z) = u by inverting sinh(u) = z to find u(z). It follows that 2 z = eu - e-

u. Multiplying by eu and rearranging, 2 2 1 0u ue z e which is quadratic in eu 2 1ue z z .

As eu must be positive, the plus sign is chosen. It follows that u = 2ln 1z z = sinh-1(z).

Exercise: Verify that 2ln 1z z might represent sinh-1(z). Compute sinh(2). Compute

2ln sinh(2)sinh(2) 1

. Discuss.

Tools of the Trade

Trig Function Conversion using triangles

1

x

21 x

2

2sin cos 1 ; tan

1

xx x

x

x

12

1 x

22 1

cos sin 1 ; tanx

x xx

x

1

21 x

2 2

1tan cos ; sin

1 1

xx

x x

The Unit Circle for Trig Functions

The unit circle is a circle of radius ONE centered on the origin:


x

y

1

(cos,sin)

UNIT CIRCLE for SIN & COS x = cos and y = sin

x

y

1

(cos,sin)

UNIT CIRCLE for SIN & COS x = cos and y = sin

(cos(-,sin(-))

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

NOTE: Choosing to be about 250 makes it easy to distinguish between the angle

and its complement, /2 - .

The x and y coordinates of a point on the circle (x, y) are (cos, sin). Using a circle with the points for and for - marked, one observes that the x coordinates are the same while the y coordinates are negatives of each other.

cos(-) = cos() sin(-) = - sin() [Trig.5]

Cosine is an even function while sine is an odd function.


Exercise: Prepare a unit circle with points for the angle and for -. Find the relations between

cos() and cos(-) and between sin() and sin(-).

Exercise: Prepare a unit circle with points for the angle and for +. Find the relations between

cos() and cos(+) and between sin() and sin(+).

The relations for the tangent and other trig functions are deduced by basing them on the rules for

sine and cosine.

x

y

1

(cos,sin)

UNIT CIRCLE for SIN & COS

m and p are integers:

0 ; 2sin[ ]

0 ; 2 1

for m even m pm

for m odd m p

1 ; 2

cos[ ] 11 ; 2 1

m for m even m pm

for m odd m p

0 4 4,8,12,...

1 4 1 1,5,9,...sin[ ]

0 4 2 2,6,10,...2

1 4 3 3,7,11,...

for m p

for m pm

for m p

for m p

1 4 4,8,12,...

0 4 1 1,5,9,...cos[ ]

1 4 2 2,6,10,...2

0 4 3 3,7,11,...

for m p

for m pm

for m p

for m p


1 2

1 2

1 2

1 2

0 8

8 1

1 8 2

8 3sin[ ]

0 8 44

8 5

1 8 6

8 7

for m p

for m p

for m p

for m pm

for m p

for m p

for m p

for m p

1 2

1 2

1 2

1 2

1 8

8 1

0 8 2

8 3cos[ ]

1 8 44

8 5

0 8 6

8 7

for m p

for m p

for m p

for m pm

for m p

for m p

for m p

for m p

Solving a particular equation: a cos + b sin = c

This equation can be cast in the form of a trig identity by dividing by 2 2a b .

2 2 2 2 2 2 2 2

cos sin or sin cos cos sina b c c

a b a b a b a b

leading to: 2 2

sinc

a b

. Solutions for exists if and only if 2 2c a b as

|sin

Given that must be 2 2 2 2

arcsin and arccosa b

a b a b

, a single solution value is

determined, and the values of = 2 2

arcsinc

a b

- . Note that arcsin(x) returns two possible

values for |x| < 1 and hence two solution values are expected for .

Alternative Solution: a cos + b sin = c

a cos - c = - b sin square

a2 cos2 - 2a c cos + c2 = b2 [ 1 – cos2

[a2 + b2] cos2 - 2a c cos + [c2 - b2] = 0

Solve the quadratic for cos and pick the allowed root for cos. Use inverse cosine remembering that it returns two values. Note that squaring the equation introduces an extraneous root. Beware of the evil root. Table TR2: Hyperbolic Functions with Identities

sinh( )2

x xe ex

cosh( )

2

x xe ex


2 2cosh sinh 1x x 2

2

sinh( ) 1tanh( )

cosh( ) 1

x x x

x x x

x e e ex

x e e e

csch(x) = [sinh(x)]-1 1 - tanh2x = sech2x 1 - coth2x = csch2x

sinh(x + y) = sinh(x) cosh(y) + cosh(x) sinh(y)

sinh(x - y) = sinh(x) cosh(y) - cosh(x) sinh(y) cosh(x + y) = cosh(x) cosh(y) + sinh(x) sinh(y)

cosh(x - y) = cosh(x) cosh(y) - sinh(x) sinh(y) tanh( ) tanh( )

tanh( )1 tanh( ) tanh( )

x yx y

x y

tanh( ) tanh( )tanh( ) tanh( ) 1

tanh( )

x yx y

x y

sinh(x) sinh(y) =½[ cosh(x + y) - cosh(x - y)] cosh(x) cosh(y) =½[ cosh(x + y) + cosh(x - y)] sinh(x) cosh(y) =½[ sinh(x + y) + sinh(x - y)]

Hyperbolic Function Conversion Table

sinh(u) = x; cosh(u) = 21 x ; tanh(u) =

21xx

cosh(u) = x; sinh(u) = 2 1x ; tanh(u) = 12xx

tanh(u) = x; sinh(u) =

21xx

; cosh(u) = 2

11 x

DeMoivre’s Theorem and Trig Identities

The Euler relation ei = cos + i sin anchors the development of the theorem. It follows that

ei = cos + i sincos(n) + i sin(n) (cos + i sinn

coupling this identity with the Binomial theorem:

(cos + i sinn = 1 1 2 2! !(cos ) (cos ) ( sin ) (cos ) ( sin ) ... ( sin )

( 1)!1! ( 2)!2!n n n nn n

i i in n

Consider the case n = 2: 2 2(cos ) 2 cos sin (sin ) cos(2 ) sin(2 )i i . Equating the real parts and separately the imaginary parts,

2 2(cos ) (sin ) cos(2 ); 2cos sin sin(2 ) [Trig.6]

Exercise: Complete the proofs that 2 21 cos(2 ) 1 cos(2 )sin and cos

2 2

.

APPENDIX: A second approach to the limit [ sin/] as goes to zero


Do not read this section unless you have time to spare. It is a second proof of the identity, and only

marginal gain results from its study.

Consider the figure below inscribed in a pie-wedge of the unit circle with small angle The plan

is to show that:

cos = Arc(c,d) < a-d = sin = c-b < Arc(a,b) =

For each small horizontal step x on the right-hand figure, the distance moved along the arc is

1/ 22

1dy

xdx

. Moving along the line a-d, the slope is greater than that of the Arc(c,d) so

Arc(c,d) < a-d. Similarly, the slope of the Arc(a,b) is greater than that of the line segment c-b = a-d

so c-b < Arc(a,b). It follows that cos < sin < for <

/2.

Using this result, 0 0

cos( ) sin( ) sin( )cos( ) 1or

As approaches zero, cos( ) approaches one so the limit of interest is sandwiched between 1

and 1. 0

sin( )lim 1

a

b

-y

+x Circle of Radius 1

O c

d

c

d

sin()

cos(

)

cos() +x

+y

a

b

sin() cos()

Going way back to the expression for the derivative of sin():

0 0

sin( ) sin( ) cos( ) cos( )sin( ) sin( ) sin( )lim cos( ) cos( )

d

d


Granted, this proof was not very efficient, but it was geometric, and, as trigonometry is a geometric

field, its proofs should be geometric.

Exercise: Show that the straight line segments a-d and c-b have the length sin().

Problems

b.) Use cos(+)= cos cos - sin sin = 1 – 2 sin2 to show that

1 – cos(2 ) = 2 sin2. Use this identity to show that:

00

cos( ) 1sin 02

2.) Prove that cos(-) = cos cos + sin sin using a geometric argument.

3.) Compute the derivative tan( )d

d

given that sin( )d

d

and cos( )d

d

are known.

4.) Use the figure to the right as the basis for

establishing that:

a.) cos(+) = cos cos - sin sin

b.) sin(+) = cos sin + cos sin

Hint: What are the lengths CB and OB?

5.) Prove that cos( )

sind

d

. Use the identity (1 – cos[) = 2 sin2[/2] and the standard result

that if all the limits are defined, the limit of a product is the product of the limits.

6.) Prepare a unit circle drawing with a line x = - 0.6 and a line y = 0.8. Place dots at the points

where the x = - 0.6 line intersects the circle and small circles at the points where the line y = 0.8

intersects the circle. What are the possible values of given that = cos-1(- 0.6)? What are the

possible values of given that =sin-1(+0.8)? What are the possible values of given that

= cos-1(- 0.6) and that = sin-1(+0.8)?


7.) Given the equation x(t) = A cos(t) + B sin(t) = C cos(t + ), find expressions for C and as

functions of A and B. Find the inverse transformations that give A and B in terms of C and . Be

aware that each inverse trig functions has two solution angles. Give a procedure to remove this

ambiguity to ensure that the correct value of is found. Note that: C is the amplitude of the motion

and by convention C 0. All the values are real. It is said that sine trails cosine in phase by ½ so

that it is common to adopt the notation A cos(t) + B sin(t) = C cos(t - ) as added a sine to a

cosine retards the phase. Give the expressions for C and in terms of A and B.

8.) Two vectors lie in the x-y plane. The vector A

has a magnitude A and makes an angle of with

respect to the x axis while B

has a magnitude of B and makes an angle of with respect to the x

axis. Find the x and y components of the vectors. The inner product cos( )A B A B

. Use the

trig identity for cos(-) to express the inner product in terms of the components of the vectors. The

inner product gauges the degree to which two vectors are parallel. How is the dot product related to

the identity for cos( - )?

9.) Two vectors lie in the x-y plane. The vector A

has a magnitude A and makes an angle of with

respect to the x axis while B

has a magnitude of B and makes an angle of with respect to the x

axis. Find the x and y components of the vectors. Compute their cross product using the determinant

method. Collect the trig factors into a single trig function using the trig identities. The cross product

A B

gauges the degree to which two vectors are perpendicular. It is proportional to what function

of what angle? Express the result for the z component of the cross product in terms of the

components Ax, Ay, Bx, and By. How is the cross product related to the identity for sin( - )?

10.) Use the ‘Finding inverse functions for the ppear lic’ method to find cosh-1(x) and tanh-1(x).

11.) Hyperbolic functions: (a.) Given the values of the powers of e below, complete the

table for the hyperbolic functions by using the values of xe and the definitions of the

hyperbolic functions. DO NOT just use the cosh key on your calculator.


function\arg:x -2 -1 0 1 2

e x 0.135 0.368 1.000 2.718 7.389

cosh(x)

sinh(x)

tanh(x)

(b.) Prepare a sketch of cosh, sinh and tanh vs. x.

(c.) Generate large argument approximations for cosh, sinh and tanh by factoring xe out of

the numerators of cosh and sinh and out of the numerator and denominator of tanh. Reduce

the results using 2( 1)2!1 1 ...

n n nx nx x .

(d.) Use: 2 32! 3!1 ...x x xe x to generate small argument expansions for cosh, sinh and tanh.

(e.) Find the large argument expansion of [cosh(x) – sinh(x)].

12.) By considering the real and imaginary parts of ei ei

prove the standard identities for

sin( + ) and cos( + ). Hint: What is ei

13.) By considering /12 as /3 - /4, evaluate cot(/12).

14.) Study the graphs of cosh(x), sinh(x) and tanh(x). Prepare graphs of cosh-1(x), sinh-1(x) and

tanh-1(x).

15.) Begin with the definitions for cosh(z), sinh(z) and tanh(z). Solve equations analogous to:

cosh( )2

z ze ez u

to find z(u) and hence find explicit representations for cosh-1(u), sinh-1(u) and tanh-1(u).


16.) Prepare sketches of sin(), cos(), tan(), sec(), csc() and ctn(). Note ctn() cot().

Prepare plots for – 2 < < 2.

17.) Prepare sketches of inverse functions for sin(), cos(), tan(), sec(), csc() and ctn().

18.) Assume that cos[-] = cos cos + sin sin. Use the unit circle technique to confirm that

sin[] = cos[/2 - ]. Hence sin[-] = cos[/2 – (-] = cos[(/

2 + ) – ]. Expand and apply the

unit circle technique repeatedly to show that sin[-] = cos sin - sin cos

19.) It has been established that 0

sin( )1Limit

. What is the

0

tan( )Limit

?

20.) Use the results of the previous problem to essentially complete a geometric development of the

relation d(tan) = [sec]2 d. Begin by preparing an excellent drawing. Use 0

tan( )Limit

= 1.

The length EC is approximately sec d which follows from the limit of tan/ as 0.

21.) Use the expressions for cos() and for cos(+) to find expressions for sin sin and

for cos cos that are linear combinations of trig functions (to the first power).


22.) Study the previous problem. Find an expression for sin cos that is a linear combination of trig

functions (to the first power).

23.) Prepare sketches of sin, cos and tan for – 2 2. Not that each of these functions is

either even or odd about every multiple of ½ . Conclude that any product of integer powers of trig

functions is even or odd about every multiple of ½ .

24.) Develop the expression for tan( ) given in the identities table TR1. The tangent of /3 is 3½.

What are the tangents of /6 and /4? Use these values to find the tangents of /12 and 5/12.

25.) Use the expression for tangent of a sum of angles and the binomial expansion to show directly

that d(tan)/d = sec2(). tan tan

tan( )1 tan tan

and [1 + x ]n 1 + n x for |x | << 1.

26.) Show that: a.) the product sinh(x) cosh(y) can be written as a sum of hyperbolic sines, b.) the

product sinh(x) sinh(y) can be written as a sum of hyperbolic cosines, c.) the product cosh(x) cosh(y)

can be written as a sum of hyperbolic cosines and d.) comment about the character of the first

product that leads to its being represented in terms of hyperbolic sines while the last two products

are represented in terms of hyperbolic cosines.

27.) Use the results of the previous problem to find expressions for sinh(x + y) and cosh(x + y) in

terms of hyperbolic functions of x and of y.

28.) Show that cosh2(x) – sinh2(x) = 1 and that tanh2(x) + sech2(x) = 1. Begin with the definitions of

the hyperbolic functions in terms of exponentials.

29.)

1sin( )sin( ) cos( ) cos( )21cos( )cos( ) cos( ) cos( )2

1sin( )cos( ) sin( ) sin( )2

x y x y x y

x y x y x y

x y x y x y


Given the Maclaurin series sin(x) x and cos(x) 1 - ½ x2 which are correct to order x2. Use these

expansions to verify the three trig identities for products of sines and cosines to second order for

small arguments.

30.) Derive the relation 3sin(3 ) 3sin 4 sinx x x using only the relations that are found above it in

Table TR1. Prepare a sketch of sin3x for 0 < x < . Sketch ¾ sinx and ¼ sin(3x) on the same plot.

Does the identity appear reasonable?

31.)

1sin( )sin( ) cos( ) cos( )21cos( )cos( ) cos( ) cos( )2

1sin( )cos( ) sin( ) sin( )2

x y x y x y

x y x y x y

x y x y x y

[Trig.7]

Use these identities to develop expression for sin[x] + sin[y] and cos[x] + cos[y]. Continue to find a

similar expression for sin[x] + cos[y]. Refer to Table TR1 for additional identities.

32.) Show that (1 + tanh(x)) ln[1 + tanh(x)] + (1 – tanh(x)) ln[1 – tanh(x)] is equivalent to

2 x tanh(x) -2 ln[cosh(x)].

Check:

f[x_] =(1 + Tanh[x]) Log[(1 + Tanh[x])] +(1 – Tanh[x]) Log[(1 – Tanh[x])]+ 2 Log[Cosh[x]] – 2 x Tanh[x]

N[f[1]] = 1.11022 1016 ; N[f[3]] = 2.22045 1016 ; close enough!!

33.) a.) Expand both sides of these equations to second order in x and y to show that they are valid to

that order. Sin(x) + sin(y) = 2 sin[½(x + y)] cos[½(x – y)],

cos(x) + cos(y) = 2 cos[½(x + y)] cos[½(x – y)]

c.) Use these relation to show that: cos(x) cos(y) = ½ [ cos(x – y) + cos(x + y)].

34.) Show that: cos sin(2) – 2 cos(2) sin = 2 (sin)3.

35.) Express sin[3x] as a third degree polynomial in sin[x]. The identities for sine and cosine of sums

of angles are recommended along with the relations:


sin[2nx] = 2 sin[nx] cos[nx]; cos[2nx] = 1 – 2 sin2[nx]; cos2[nx] = 1 – sin2[nx]

answer: sin[3x] = 3 sin[x] – 4 sin3[x]

Why do only odd powers ppear in the result?

36.) Express sin[5x] as a fifth degree polynomial in sin[x]. The identities for sine and cosine of sums

of angles are recommended along with the relations:

sin[2nx] = 2 sin[nx] cos[nx]; cos[2nx] = 1 – 2 sin2[nx]; cos2[nx] = 1 – sin2[nx]

answer: sin[5x] = 5 sin[x] – 20 sin3[x] + 16 sin5[x]

Why do only odd powers appear in the result?

37.) a.) Motivate the result that, for odd integers k, sin[kx] = k sin[x] + c3 sin3[x] + … + ck sink[x]

where the coefficients satisfy the equations:

3 53

3 5;3! 3! 5! 5! 2

ck k k kc c

b.) Find the analogous relation for 7

7!

k .

38.) Express cos[2x] and cos[4x] as second and fourth degree polynomial in cos[x

answer: cos[2x] = 2 cos2[x] – 1

cos[4x] = 8 cos4[x] – 8 cos2[x] + 1

Why do only even powers appear in the results?

39.) Find the angles in the range 0 for which cos[4] = cos[6].

Hint: Is cosine and even or odd function?

40.) Sketch a pentagon of equal sides inscribed in a circle of radius 1. What is the area of the

pentagon divided by 5? Find the area between one side of the pentagon and the boundary of the

circle. One approach divides the region into strips of length cos - cos[/5], width dy = d(sin) and

integrates over the range -/5 to /5. Describe your method if you believe that it is easier.


41.) Express cos(1t) + cos(2t + ) as twice the product of two cosines. Discuss the relation in

terms of two beam interference

42.) Two straight rods are configured to swing in the x-y plane on pivots. The first rod is pivoted to

swing about the origin, and the second rod swings about the other end of the first rod. The other end

of the second rod is at the position P = (xP, yP). The system represents the vector addition

A B C

.

P

x

y

C

A

B

A B C

a.) Each of the rods has a length of 5 meters, and the point P is ( 4 m, 4 m). Use the laws of cosines

and sines to find the angles and .

b.) Represent each vector in terms of its x and y components using the angles, = + , and data

given. Solve the equations to find and and hence .

c.) Show that the length of the resultant vector is 4 2 . Solve the problem a third time by finding the

angles in the triangle illustrated and then rotating the figure to represent the problem in part a.


4 2

43.) Apply de Moivre’s theorem to prove that:

a.) 3 33 31 14 4 4 4sin sin sin 3 and cos cos cos3 .

b.) 4 43 31 1 1 18 2 8 8 2 8sin cos 2 cos 4 and cos cos 2 cos 4 .

c.) Explain why each identity above contain all sines or all cosines on the right.

44.) Prove that: 1 tan( )

sec(2 ) tan(2 )1 tan( )

xx x

x

. Use the identities in Table TR1.

45.) Relation between the inner product and cosine of the angle:

A

B

C

The law of cosines identifies C2 as A2 + B2

– 2A B cos. Represent C2 as:

(Bx – Ax)2 + (By – Ay)

2 +(Bz – Az)2 .

Continue to conclude that:

| || | cos x x y y z zA B A B A B A B A B

We will consider the representation of the

inner product using cosine as its definition.

The representation of the inner product in

terms of the components of the vectors will

be dubbed a Parseval Relation.


46.) Kite Theorem

The two black lines represent the crossed frame members of

the kite. They cross at a point cutting the members into parts

of arbitrary ratio (but not at an end point) and with crossing

angles of and - . Draw four lines that join the mid points

of the four red sides. Show that the resulting quadrilateral is

a parallelogram with the same interior angles as the frame

member crossing angles and that encloses an area equal to

one half the area of the kite. Warm up using = /2.

The proof that the area enclosed in the quadrilateral in the general case is one half the area of the

kite is relatively straight forward and general.

45.) Find expressions for sin[x]cos[y] in terms of sines of one half the sum and difference of x and y.

46.) Begin with the Law of Sines and show that: tan[½( )]

tan[½( )]

a b

a b

. Note that is the angle

opposite side a. This relation is sometimes called the law of tangents. It is valid for any pair of sides.


c

a

b

47.) Continuing with the figure above, prove Mollweide’s Formula: cos[½( )]

sin[½ ]

a b

c

.

Again, begin with the Laws of Sines; Use the half angle relations; Use sin[] = sin[ - ].

References:

1. The Wolfram web site: mathworld.wolfram.com/

2. Noether's Theorem may be discussed in your intermediate mechanics textbook.

2. K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical Methods for Physics and Engineering,

2nd Ed., Cambridge, Cambridge UK (2002).

3.) Essential Physics by Frank W. K. Firk.

APPENDIX: Changing the Magnitude and Direction of a Vector

Concepts of primary interest:

Vector attributes – magnitude and direction

Infinitesimal additions parallel – change magnitude

Infinitesimal additions perpendicular – change direction

A vector quantity has two attributes magnitude and direction. For small additions to a vector,

the claim is that a small piece added parallel to the original vector changes the magnitude of

the vector (not the direction) and that a small piece added perpendicular to the original vector

changes the direction (not the magnitude).


In the drawing to the right, the original

vector is v

, and the small addition is v .

The original vector: ˆv vv , the magnitude

of the vector times its direction. The small

addition is represented in terms of its

components parallel and perpendicular to v

.

ˆ ˆv ev v v

v

v

v

v

v

v ˆ v

v

ˆ e

v v ˆ v

The direction parallel to v

is its direction v while the direction e is perpendicular to v

.

The drawing suggests that the magnitude of the vector increases by v

and that the

direction of v v makes an angle of v

v

with respect to the original v

. Recall that

the tangent is the side opposite over the adjacent and that v

is small so that tan .

These speculations are to be confirmed using more analytic techniques. A vector is its magnitude times its direction ( ˆv vv ) and the magnitude is the square root of the inner (dot)

product of the vector with itself (v v v v ). Rearranging: ˆ v

v vvv v

.

The derivative process includes taking the limit that the changes become infinitesimal so its machinery is to be invoked to handle the small part. First, the time rate of change of the magnitude of v

is computed. (Note that v without the arrow is the magnitude always!)

1

2

dvdv dv dv dv vv v v vdv d dtdt dt dt dt dvv v dtdt dt v vv v v v

Multiplying by t, it follows that:

dv dvdv dvt t v vdt dtdt dt

The change in the magnitude of the vector is just equal to the component of the small

addition that is parallel to the original vector. If the additional piece is anti-parallel then

v

is negative and the magnitude is decreased.

Exercise: Identify the rules for differentiation used in the line:


1

2

dvdv dv dv dv vv v v vdv d dtdt dt dt dt dvv v dtdt dt v vv v v v

.

Attacking the direction with a time derivative yields:

3/ 2 3

ˆ21 1ˆ

2

ˆdvdv v v vdv dvv dtd d dtdt dt dv dv vdt dtdt dt v v v

vv vv v v v v v

In the last pair of parentheses, the component of dvdt

parallel to v

is subtracted from dv

dt

leaving only the perpendicular part of dvdt

.

ˆ ˆdv dv dvv edt dt dt

1 1ˆ ˆ

ˆd dv dv dvv edt dt dtdt v v

v

The last equation demonstrates that the component added perpendicular to the original vector causes the change in the direction for the resultant as compared to the original. Multiplying by t as before:

1 ˆˆ ˆˆ vdv dv v edt dt vv

v

.

Exercise: The equation above seems to suggest that it is more difficult to change the

direction of a vector if it has a larger magnitude. Prepare a sketch that supports this

interpretation.

Problems 1.) The position vector for a particle in a plane can be represented as:

ˆˆ( ) ( ) cos( sin( )Pr t R t j

where the angle is a function of time. Compute

PP

drdr dtdt

. Resolve Pdr

into components parallel to and perpendicular to Pr

. Compute

P

dr

dt

and compare it to Pdr

dt

. Compare 1 P

Pdrr dt

and d/dt.

2.) Given the initial vector ˆ ˆ400 300v i j

and the small addition ˆ ˆ ˆ ˆ4 3 3 4v i j i j

compute the magnitudes of v

and of v v

and the angles that each makes with respect to the x axis. Use the equations developed in this handout to estimate v and v . What does the quantity

ˆˆ

vv

represent? Why was v

divided into two parts? What are the magnitudes of the pieces

enclosed in parentheses? Why is ˆ ˆ ˆ ˆ4 3 3 4v i j i j

resolved into the two terms in

parentheses?

list of trigonometric identities - usna · 2014-01-28 · 1/28/2014 physics handout series.tank:...

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