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4 Analytic Trigonometry

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Chapter 4 Analytic Trigonometry

4.1 Inverse Trigonometric Functions

The trigonometric functions act as an operator on the variable (angle) x, resulting in an

output value y. Suppose this process is reversed: given a y-value, is it possible to work

backward to determine the angle x that produced y?

In simplest terms, we wish to solve equations such as

.,1sec,tan,sin52

21 etcxxx

In some cases the value x can be determined “by inspection”. For example, the simple

algebraic equation 21sin x implies that

6x is one possible solution. However, this

method does not work this easy in general: for example, by inspection what is the

solution to 52tan x ?

Review of Inverse Graphs and Inverse Functions

A function )(xfy generates a set of points )}(|),{( xfyyx that are plotted on a

Cartesian (x-y) coordinate axis system. The result is the graph of the function.

The inverse graph of a function )(xfy is a graph of the set of points

)}(|),{( xfyxy . In simplest terms, the inverse graph of )(xfy is a new graph in

which each ordered pair ),( yx of )(xfy has its coordinates reversed to ),( xy . The

points ),( xy and ),( yx are symmetrical across the line xy ; this allows a simple way

to sketch the inverse graph of a function )(xfy . However, in most cases the inverse

graph will not be a function since it fails the vertical line test.

To ensure that the inverse graph of a function )(xfy is itself a function, the given

function must be one-to-one. A function is one-to-one if )()( bfaf implies that

ba . Visually, a function that is one-to-one passes the horizontal line test.

To summarize, a function that passes the horizontal line test is said to be one-to-one, and

if this condition is met, its inverse graph will be a function as well. If this is true, then the

inverse function is denoted as )(1 xfy .

The Inverse Sine Function (Arcsine)

Consider the function xxfy sin)( . Its graph is given in the next page:


70

The sine function is not one-to-one since it does not pass the horizontal line test.

However, if the domain of xxf sin)( is restricted to the interval 22

x , then it is

one-to-one and its inverse graph is therefore a function:

Therefore, the inverse sine function (also called the arcsine function, written arcsin(x)),

is defined to be the inverse graph of the function xxf sin)( on the closed interval

22x . The domain of the inverse sine function is 11 x and the range is

22)(xf . The inverse sine function returns values in the 1

st quadrant (if 0x ) or

the 4th

quadrant (if 0x ).

Example 1: Determine values for (a) 21arcsin , (b)

2

3arcsin , (c) 2

2arcsin ,

(d) )1arcsin( and (e) )arcsin( .

Solutions: The results for (a), (b) and (c) are based on the normal measures of the sine

function in the first and fourth quadrants. Thus, 62

1arcsin , 32

3arcsin and

42

2arcsin . For (d), 2

)1arcsin( and for (e), the solution is not defined since is

outside the domain.

Example 2: Using a calculator set to degree mode, what is )4.0arcsin( ?

Solution: The solution to two decimal places is the angle 23.58 degrees.


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Example 3: Using a calculator in radian mode, what is )arcsin(61 ?

Solution: To two decimal places, the result is 0.17 radian.

The Inverse Cosine Function (Arccosine)

The cosine function )cos()( xxf is not one-to-one over its usual domain of the Real

numbers. However, when restricted to the closed interval x0 , the cosine function

is one-to-one, and hence its inverse graph is also a function, which will be defined at the

inverse cosine function (also written )arccos(x ).

The inverse cosine graph )arccos()( xxf is defined on the domain 11 x with a

range of )(0 xf . It returns values in the 1st quadrant (if 0x ) or the 2

nd quadrant

(if 0x ).

Example 4: Determine values for (a) 21arccos , (b)

2

3arccos , (c) 2

2arccos , (d)

)1arccos( and (e) )arccos( .

Solutions: The results for (a), (b) and (c) are based on the normal measures of the cosine

function in the first and second quadrants. Thus, 32

1arccos , 62

3arccos and

43

2

2arccos . For (d), 0)1arccos( and for (e), the solution is not defined since

is outside the domain.

Example 5: What radian measure solves )215.0arccos( ?

Solution: The answer is 1.354 radians.

Example 6: If 32)sin(x , find )cos(x .

Solution: This is identical to the question ))(cos(arcsin32 . On a right triangle, the

opposite measure of one of the non-right angles is 2 and the hypotenuse is 3. With these


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facts, the Pythagorean Theorem allows for the adjacent leg to be solved: 523 22.

Therefore, the cosine of this angle is 3

5 .

The Inverse Tangent Function (Arctangent)

Like the sine and cosine function, the tangent function )tan()( xxf is not one-to-one

unless restricted to a smaller domain, which is usually chosen to be 22

x . Thus

restricted, the inverse tangent function (written )arctan(x ) is defined as the inverse

graph of the tangent function.

The domain of the arctangent function is x and its range is 22

)(xf . It

returns values in the 1st quadrant (if 0x ) or the 4

th quadrant (if 0x ). Notably, it has

two horizontal asymptotes at 2

y as x , and 2

y as x .

Example 7: Determine the values to (a) )3arctan( , (b) )1arctan( and (c) )arctan(2

.

Solutions: For (a), 3

)3arctan( , for (b), 4

)1arctan( and for (c), the result is

undefined, although it trends to positive infinity as 2

x .

Example 8: Evaluate ))(sin(arctan41 .


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Solution: The arctan of 41 is an angle whose opposite leg is 1 and adjacent leg is 4.

Therefore, the hypotenuse is 1741 22. Thus, the sine of this angle is the opposite

divided by the hypotenuse: 17

17

17

141 ))(sin(arctan .

Important Results

1 1 1 1 1 1sin ( ) sin ( ), tan ( ) tan ( ), cos ( ) cos ( )t t t t t t

Example 9: Evaluate the following expressions

a) 1sin (1) b) 1sin ( 1) c) 1sin ( 1/ 2) d) 1cos (1) e) 1cos ( 1)

f) 1 2cos

2 g)

1tan (1) h) 1tan ( 3) i) 1 3tan

3

Solution:

a) 1 1sin (1) sin (sin( / 2)) / 2

b) 1 1sin ( 1) sin (1) / 2

c) 1 1sin ( 1/ 2) sin (1/ 2) sin(sin( / 6)) / 6

d) 1 1cos (1) cos (cos0) 0

e) 1 1cos ( 1) cos (1) 0

f) 1 12 2cos cos / 4 3 / 4

2 2

g) 1 1tan (1) tan (tan( / 4)) / 4

h) 1 1 1tan ( 3) tan ( 3) tan (tan( /3)) / 3

i) 1 1 13 3

tan tan tan tan( / 6) / 63 3

Common Errors to Avoid

Students sometimes make the erroneous assumption that the “arcsin” cancels the “sin”,

for example, xx))(arcsin(sin , which is not necessarily true! Pay special attention to

the domain appropriate to the particular function being evaluated. Consider these

following examples:


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Example 10: Is it true that 1010

))(arcsin(sin ?

Solution: Yes, since 2102

.

Example 11: Is it true that 4

54

5 ))(arcsin(sin ?

Solution: No. The argument 4

5 is outside the domain of arcsine. The arcsine function

returns the smallest such angle corresponding to the input. Since 4

5 lies in the 3rd

quadrant, the arcsine function will return an angle in the 4th

quadrant equivalent to 4

5 .

The correct result is 44

5 ))(arcsin(sin .

Example 12a: Is it true that 4

34

3 ))(arccos(cos ?

Solution: Yes. The argument 4

3 is within the domain of the arccosine function

430 .

Example 12b: Is it true that 4

34

3 ))(cos(arccos ?

Solution: No. The value 4

3 is outside the domain 11 x of the arccosine function.

This expression is undefined.

Generalized Expressions

The argument can be left as an independent variable and expressions combined to derive

relationships between the sine, cosine and tangent operations with the arcsine, arccosine

and arctangent inverse operations.

Example 13: Simplify the expression ))(cos(arcsin x .

Solution: The )arcsin(x suggests a result y such that xy)sin( . Sketching a right

triangle with x at the opposite leg relative to angle y, and 1 at the adjacent leg to y, the

Pythagorean Theorem gives the hypotenuse as having length 12x . Therefore, the

cosine of this angle (y) is the adjacent leg divided by the hypotenuse:

1

1))(cos(arcsin)cos(

2xxy .

Example 14: Simplify the expression ))(arccos(sin3

Solution: Since 63

cossin , rewrite ))(arccos(sin3

as ))(arccos(cos6

. Since 6

is on

the domain of the arccosine function, the expression reduces to 66

))(arccos(cos .


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Example 15: Solve the equation 01sinsin2 2 xx in the interval 20 x .

Solution: Factor the equation:

0)1)(sin1sin2(

01sinsin2 2

xx

xx

Set each factor equal to zero and solve for x:

621

21 )arcsin(sin01sin2 xxxx

2)1arcsin(1sin01sin xxxx

Note that the arcsine function returns values in quadrants 1 and 4 only. Furthermore,

other solutions may need to be inferred from the ones provided by the arcsine operation:

Since 6

x is a solution, there also exists a solution in quadrant 2 by symmetry.

Therefore, 6

5x is also a solution to the equation.

The solution 2

x provided by the arcsine operation is correct but outside the desired

bounds 20 x . This is easily remedied by selecting 2

3x as the equivalent

solution.

Therefore, the solution set to this equation is },,{2

36

56

.

Example 16: Solve the equation 03tan4tan2 xx in the interval 22

x .

Solution: Factor the equation:

0)1)(tan3(tan

03tan4tan 2

xx

xx

Each factor is set equal to zero and solved for the variable:

radiansxxxx ...249.1)3arctan(3tan03tan

radiansxxxx4

)1arctan(1tan01tan

The solution set (in radians) is },249.1{4

.


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4.2 Trigonometric Identities

An identity is an equation that is always true for all values in its domain. For example,

the equation ))((22 bababa is an identity since it is true for all a and b. In a

similar manner, many trigonometric expressions can be expressed in different forms, thus

forming trigonometric identities.

Reciprocal Identities

From the definitions of the six trigonometric functions, the tangent, cotangent, secant and

cosecant functions can all be expressed as identities involving the sine and/or cosine

functions:

xx

xx

x

xx

x

xx

sin

1csc

cos

1sec

sin

coscot

cos

sintan

Simple algebraic manipulations can generate more identities. As an example, since

x

xx

cos

sintan , it is possible to generate a new identity by multiplying the cos x and

deriving xxx costansin . However, restrictions placed on the domain x are

maintained throughout any further manipulations. Thus, the expression

xxx costansin is true as long as Nnxn

,2

)12(, since these restrictions are in

place because of the presence of the tangent function.

The Pythagorean Identities

The sine and cosine functions are defined on the unit circle and are related by the

Pythagorean identity:

1cossin 22 xx

With simple algebra, new corollary identities can be formed:

xx

xx

22

22

sin1cos

cos1sin

These identities are true for all x .

Furthermore, the Pythagorean Identity can be divided through by x2sin or x2cos to

generate more identities:

Divide by x2sin :

xxxx

x

x

xxx 22

22

2

2

222 csccot1

sin

1

sin

cos

sin

sin1cossin

Divide by x2cos :

xxxx

x

x

xxx 22

22

2

2

222 sec1tan

cos

1

cos

cos

cos

sin1cossin

These identities are true for all x for which the expressions are defined.


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General Identities and Demonstrations

The purpose of employing identities is to reduce a large trigonometric expression into

something smaller and easier to manipulate. The next few examples illustrate the method

to demonstrate the truthfulness of each identity. Until the proof is demonstrated, the

(potential) identity is called a theorem.

The general method is to use whatever identity is appropriate at any given step – this

often takes practice to “recognize” the right identity. Furthermore, there may be more

than one way to demonstrate the truthfulness of the identity. The only requirement that

must not be violated is that each side of the identity must be handled separately. Any

algebraic maneuver that assumes the equality is true cannot be used since the truthfulness

of the equation has yet to be shown! In other words, you cannot use what you are trying

to prove. In most cases, this removes the method of cross-multiplication from

consideration. Otherwise, normal algebraic techniques are used as needed. Always work

from the most complicated form first.

Example 1: Prove that xx

x

x

x

x

x

cossin

cos1

cos

sin

sin

cos1

Solution: The left side of the theorem shows two rational expressions being summed.

Rewrite the left side as a sum by using a common denominator. The common

denominator of xsin and xcos is their product, xxcossin . The numerators must be

adjusted accordingly:

xx

xxx

x

x

x

x

x

x

x

x

cossin

sincos)cos1(

sin

sin

cos

sin

cos

cos

sin

cos1 2

The expression on the right side can be simplified by distributing the xcos :

xx

xxx

xx

xxx

cossin

sincoscos

cossin

sincos)cos1( 222

.

Since 1cossin 22 xx , the numerator reduces to

xx

x

xx

xxx

cossin

1cos

cossin

sincoscos 22

Thus, we have shown by transitivity the following:

xx

xxx

x

x

x

x

x

x

x

x

x

x

x

x

cossin

sincos)cos1(

sin

sin

cos

sin

cos

cos

sin

cos1

cos

sin

sin

cos1 2

xx

x

xx

xxx

cossin

1cos

cossin

sincoscos 22

Therefore, the left side of the theorem is equal to the right side. The theorem is proven; it

is an identity.


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Example 2: Prove that x

x

xx

xx

sec

sin

cscsec

cossin

Solution: The left side of the theorem contains “more” expressions with which to

manipulate. Convert all functions into equivalent forms in terms of sine and cosine:

xx

xx

xx

xx

sin1

cos1

cossin

cscsec

cossin

The denominators of the two small expressions in the denominator of the main

expression can be combined by summing over a common denominator:

xxxx

xxx

xxx

xx

xxxxxx

cossincossin

cossincos

cossinsin

sin1

cos1

cossincossincossin

Reciprocating the combined expression in the main denominator, the “ xx cossin ”

expressions cancel:

xxxx

xxxx

xx

xxxx

cossincossin

cossin)cos(sin

cossin

cossincossin

Since x

xsec

1cos , the expression xxcossin now becomes

x

x

xxxx

sec

sin

sec

1sincossin

Thus, the theorem is proven.

Notice that we never once manipulated the expression xx

secsin directly in this proof. We

were able to show the relationship entirely by manipulating the expressions from the left

side of the original theorem.

Shift and Reflection Identities

The sine and cosine functions are identical in shape and differ only by a horizontal shift.

Specifically, the sine function is the cosine function shifted 2

units to the right.

Conversely, the cosine function is the sine function shifted 2

units to the left. Therefore,

two useful identities can be stated:

)sin(cos

)cos(sin

2

2

xx

xx

Furthermore, the sine function shifted units to the left or right results in an inverted

sine function. The same is true for the cosine function. These two identities are

summarized below:

xx

xx

cos)cos(

sin)sin(


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Shifting the sine and cosine functions 2

units in opposite directions result in inverted

cosine and sine functions:

xx sin)cos(2

xx cos)sin(2

Of course, the sine and cosine functions have a period of 2 . Therefore, two more

identities can be stated:

xx

xx

cos)2cos(

sin)2sin(

The tangent function has a period of so its shift identity is

xx tan)tan(

Shift identities for the secant, cosecant and cotangent functions can be derived by

converting into sine and cosine functions.

A vertical reflection of any function can be determined by replacing the argument x with

x . The result is a graph that is reflected across the y-axis. The cosine function is

already symmetrical to the y-axis so it will remain unchanged. The sine and tangent

functions, both symmetrical to the origin, will invert across the x-axis. Thus, the

following reflection identities can be stated:

xx

xx

xx

tan)tan(

sin)sin(

cos)cos(

For now, the “proofs” of these identities are done visually. In the next section, a method

will be developed to analytically prove all such shift and reflection identities.

Example 3: Prove that xx cos)sin(2

.

Solution: Factor the negative within the argument of the sine function:

))(sin()sin(22

xx

The leading negative within the argument can be “moved” to in front of the sine function:

)sin())(sin(22

xx

Since xx cos)sin(2

, then xxx cos)cos()sin(2

.


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4.3 Sum and Difference Formulas

The sum and difference forms of the sine and cosine functions are the following four

forms:

)sin(),cos(),sin( yxyxyx and )cos( yx .

The trigonometric function can be viewed as operators on the variable(s) x and y, but they

are not linear operators. The function/operators cannot be distributed across addition and

subtraction. Hence, statements like yxyx sinsin)sin( are false. Unfortunately

these are common mistakes students make when considering the sum and difference

forms of the trigonometric functions.

In this section we will present a geometric proof for all these four forms.

We prove that

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

Suppose that OX be the initial side and the

terminal side OY makes an angle x with initial

side and the terminal side OZ makes angles y

with the side OY. Take a point P on the side

OZ and make perpendicular lines PQ and PS

on OX and OY respectively. Further we make

perpendicular line ST and SR according to the

adjacent diagram. Now observe that the angle

TPS is equal to the angle x. The angle POR is

equal to x + y.

Z

P

T x Y

S

y

x

O Q R X

sin( )

cos sin sin cos

PQ PT TQx y

OP OP

PT TQ PT SR

OP OP OP OP

PT PS SR OS

PS OP OS OP

x y x y

And

sin( ) sin cos( ) cos sin( )

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

x y x y x y

x y x y x x x x

Now we have to prove the result for cosine:

Remember that


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

cos cos sin sin , sin( ) sin , sin cos , cos( ) cos2

x y x y x y x y x y

x y x y x x x x x x

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

We now present the algebraic proof.

The sum and difference forms can be derived from an analysis of the angles drawn on a

unit circle, and the familiar distance formula. We consider the case )cos( yx first, from

which the other three forms can easily be derived as corollaries.

Proof of

cos(x – y) = cos x cos y + sin x sin y

Consider the unit circle in the first quadrant (without loss of generality). Let ray xr be

drawn with angle x, and yr be drawn with angle y, and let yx . Therefore, ray xr

intersects the circle at point )sin,(cos xx and ray yr intersects the circle at point

)sin,(cos yy .

Rigidly rotate the circle clockwise through an angle of y, so that the original ray yr sits

along the positive x-axis while original ray xr now has an angle of elevation yx and

therefore intersects the unit circle at the point ))sin(),(cos( yxyx .

The core of the proof is to show that the distance between points )sin,(cos xx and

)sin,(cos yy is the same as the distance from point ))sin(),(cos( yxyx to the point

)0,1( . See the following diagram:

Figure. Diagram showing rays xr and yr (left), and the rotation through an angle y (right).

The distance between )sin,(cos xx and )sin,(cos yy is found by the distance formula:


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yxyx

yyxxyyxx

yxyxyxyxD

sinsin2coscos22

sinsinsin2sincoscoscos2cos

)sin(sin)cos(cos))sin,(sin),cos,((cos2222

22

The Pythagorean identity 1cossin 22 xx was used twice in the second-to-third step.

The distance between ))sin(),(cos( yxyx and (1,0) is

)cos(22

)(sin1)cos(2)(cos

)0)(sin()1)(cos())0,1()),sin(),((cos(22

22

yx

yxyxyx

yxyxyxyxD

The Pythagorean identity 1)(cos)(sin 22 yxyx was used in the second-to-third

step.

Since the two distances are equal, relate them by equality:

yxyxyx sinsin2coscos22)cos(22

Squaring both sides removes the radicals. The constants cancel, and the above equation

algebraically reduces to:

yxyxyx sinsincoscos)cos( (A)

Therefore, the difference formula for the cosine function is proved.

Proofs of the Remaining Forms

The other three forms can be derived using the difference form of the cosine function (A)

above. For example, using the shift identities xx sin)cos(2

and xx cos)sin(2

,

we can make the substitution 2

x for x into the above form (A) to get

yxyxyx sin)sin(cos)cos()cos(222

(B)

The left side of (B) can be re-arranged as )cos()cos(22

yxyx , which equals

)sin( yx . On the right side of (B), we use the substitutions xx sin)cos(2

and

xx cos)sin(2

. Therefore, equation (B) can be written as

yxyxyx sincoscossin)sin(

This proves the difference formula for the sine function.


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The sum forms can be proven using the symmetry identities xx sin)sin( and

xx cos)cos( . Therefore, the term y is substituted with y in equation (A) to get

yxyxyx

yxyxyx

sinsincoscos)cos(

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

This proves the sum form for the cosine function.

The same substitution is made in equation (B):

yxyxyx

yxyxyx

sincoscossin)sin(

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

This proves the sum form for the sine function.

The Sum and Difference Identities for the Sine and Cosine Functions

yxyxyx

yxyxyx

yxyxyx

yxyxyx

sinsincoscos)cos(

sincoscossin)sin(

sinsincoscos)cos(

sincoscossin)sin(

The following examples illustrate some of the ways these formulas can be used:

Example 1: Calculate )sin(12

exactly.

Solution: The angle 12

is the difference of angles 3

and 4

: 4312

. Therefore,

4

26

2

2

21

2

2

2

3

43434312))(())(()sin()cos()cos()sin()sin()sin(

Example 2: Calculate )105cos( exactly.

Solution: 105 degrees can be written as the sum of 60 and 45 degrees. Therefore,

4

62

2

2

2

3

2

2

21 ))(())(()45sin()60sin()45cos()60cos()4560cos()105cos(

Example 3: Prove that xx cos)cos( .

Solution: Use the sum form of the cosine function:

xxxxxx cos)0(sin)1(cossinsincoscos)cos(


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Example 4: Evaluate 1012

tan( ) exactly.

Solution: The angle 5

6

10

12 is the sum of angles

3 and

2. Therefore,

5 1sin sin sin cos cos sin

6 2 3 2 3 2 3 2

Alternate proof: 5 1

sin sin cos6 2 3 3 2

Sum and Difference Angle Formulas For Tangent

The sum and difference formulas for the tangent function can be derived using the sum

and difference forms for the sine and cosine function.

For )tan( yx , we rewrite this in terms of sine and cosine:

)cos(

)sin()tan(

yx

yxyx

The quotient is rewritten using the sum forms:

yxyx

yxyx

yx

yx

sinsincoscos

sincoscossin

)cos(

)sin(

Multiply the numerator and denominator by the expression yx coscos

1 and simplify:

yx

yx

yxyx

yxyx

yxyx

yxyx

yx

yx

tantan1

tantan

)sinsincos)(cos(

)sincoscos)(sin(

sinsincoscos

sincoscossin

coscos1

coscos1

Therefore, the sum formula for the tangent function is:

yx

yxyx

tantan1

tantan)tan(

The difference formula is found by replacing y with y :


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yx

yx

yx

yxyx

tantan1

tantan

)tan(tan1

)tan(tan)tan(

The two forms can be summarized together as

yx

yxyx

tantan1

tantan)tan(

Example 5: Evaluate )tan(127 exactly.

Solution: The angle 127 is the sum of angles

3 and

4. Therefore,

2

)13(

131

13

tantan1

tantan)tan()tan(

2

43

43

43127

In the final step, the denominator was rationalized by multiplication of the conjugate.


86

4.4 Double and Half-Angle Formulas

A natural extension of the sum and difference formulas (section 4.3) are the double-angle

formulas. For example, the expression )2sin( x is a double-angle expression, since the

argument “ x2 ” is interpreted as “double the angle x”.

Using the sum formula for sin x, the expression )2sin( x can be rewritten as:

xxxxxxxxx cossin2sincoscossin)sin()2sin(

This is an extremely useful identity known as the double-angle identity for sine.

The double-angle identity for the cosine function is derived similarly:

xxxxxxxxx 22 sincossinsincoscos)cos()2cos(

Using the Pythagorean identities (section 4.2) for x2sin or x2cos , the double-angle

identity for cosine can be rewritten in two corollary forms:

xxxx 222 sin21sin)sin1()2cos(

1cos2)cos1(cos)2cos( 222 xxxx

For the tangent function, its double-angle identity is

x

x

xx

xxxxx

2tan1

tan2

tantan1

tantan)tan()2tan(

All forms can be summarized as the double angle identities: These formulas are very

useful and can be incorporated into many of the common identity proofs that are

encountered.

The Double-Angle Identities

2 2 2 2

2

sin(2 ) 2sin cos

cos(2 ) cos sin 2cos 1 1 sin

2 tantan(2 )

1 tan

x x x

x x x x x

xx

x

Example 1: Prove that )2sin(1)cos(sin 2 xxx

Solution: Expand the left side by multiplication:

xxxxxx 222 coscossin2sin)cos(sin

The first and third terms sum to 1 by the Pythagorean Identity. The middle term is

sin(2x). Therefore, the identity is proven.


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Example 2: Let angle x be given in the following diagram. Calculate )2sin( x and

)2cos( x .

Solution: The length of the hypotenuse is found by the Pythagorean formula:

6574 22hyp

Therefore, for )2sin( x , we use its double angle formula:

6556

65

7

65

4 ))((2cossin2)2sin( xxx .

For )2cos( x , any one of the three double-angle formulas may be used:

6533

6516492

65

42

65

722 )()(sincos)2cos( xxx

The Pythagorean identity could also have been used to determine )2cos( x :

6533

42251089

42251089

422531362

655622

)2cos(

1)(1)2(sin1)2(cos

x

xx

It is interesting to note that the numbers 33, 56 and 65 form an integer solution to the

Pythagorean formula: 222 655633 . In fact, the double-angle formulas for sine and

cosine always generate a Pythagorean “triple” of integer solutions, as the next example

illustrates:

Example 3: Suppose a right triangle has angle acute angle x as one of its interior angles.

Let the opposite leg be a and the adjacent leg be b. Determine )2sin( x and )2cos( x and

show that the numerators and the denominator form an integer Pythagorean triple.

Solution: The hypotenuse is 22 bahyp . For )2sin( x , we get

222222

2))((2cossin2)2sin(ba

ab

ba

b

ba

axxx .

Similarly,

22

22

2222

2222 )()(sincos)2cos(ba

ab

ba

a

ba

bxxx .

The expressions ab2 , 22 ab and 22 ab are integers and form an integer solution to

the Pythagorean formula:


88

True

ababab

abababba

ababab

222?

4224

222?

422422

222?

2222

)(2

)(24

)()()2(

The equality is true. This is an interesting way to generate integer solutions to the

Pythagorean formula, and shows also that the number of possible integer solutions to the

Pythagorean formula is infinite.

Example 4: If 52sin x and x is in the first quadrant, find )2sec( x .

Solution: We will need the cosine of x. Since the opposite leg is 2 and the hypotenuse 5,

the adjacent leg is 2125 22adj . Therefore, 5

21cos x . To determine

)2sec( x , rewrite it as

)2cos(

1)2sec(

xx

The double-angle formula for cosine gives

25172

522

5

2122 )()(sincos)2cos( xxx .

Therefore, )2sec( x is 1725

The Half-Angle Formulas

Using the double-angle formulas for sine and cosine, we can derive similar formulas for

the half-angle of sine and cosine. The usual approach is to start with the double-angle

formula for cosine, entirely in terms of cosine. Let v be a temporary dummy variable:

1cos2)2cos( 2 vv

Let 2xv , so therefore xv2 . The above formula becomes

1)(cos2cos2

2 xx

Then algebraically solve for the term )cos(2x :

2cos1

22cos1

2

2

2

2 )cos()(cos1)(cos2cos xxxxxx

This establishes the half-angle formula for cosine. The sign is selected depending on the

quadrant in which the angle x is located. To determine the half-angle formula for sine,

use the Pythagorean identity )(cos1)sin(2

2

2xx :

2cos1

2cos1

22

2cos12

2cos1

2)()(1)(1)sin( xxxxx


89

Be careful to note that the two formulas both contain cos x and they look quite similar.

The half-angle formula for tangent is derived by the following steps:

x

x

x

x

x

x

x

cos1

cos1

)cos(

)sin()tan(

2cos1

2cos1

2

2

2

This is one common way to write the half-angle formula for tangent. Another way to

write the formula is to rationalize the denominator by multiplying by the conjugate:

x

x

x

x

x

x

x

x

x

xx

sin

cos1

sin

)cos1(

cos1

)cos1(

)cos1(

)cos1(

cos1

cos1)tan(

2

2

2

2

2

The half-angle formulas are summarized below:

The Half-Angle Formulas

x

x

x

xxxxxx

sin

cos1

cos1

cos1)tan(

2

cos1)cos(

2

cos1)sin(

222

Example 5: Let angle x be given in the following diagram. Calculate )sin(2x and )cos(

2x .

Solution: The hypotenuse is 65 . Therefore,

130

65765

652

765

2

1

2cos1

2

65

7

)sin( xx

In the final step, the internal fraction was rationalized by multiplying top and bottom by

65 .

The value for )cos(2x is found via a similar calculation:

130

65765

652

765

2

1

2cos1

2

65

7

)cos( xx


90

4.5 Product-to-Sum and Sum-to-Product Formulas

Occasionally it is desirable to convert the product of unlike sine and/or cosine terms into

the sum or difference of two separate sine or cosine terms, and occasionally the reverse is

true. In any case, the product-to-sum and sum-to-product identities allow us to make

these conversions. These formulas are especially useful in integral calculus.

The Product-to-Sum Identities

The product-to-sum identities arise from a simple algebraic manipulation of the sum and

difference formulas for the sine and cosine function. For example, we write both the sum

and difference formulas for the cosine function:

yxyxyx

yxyxyx

sinsincoscos)cos(

sinsincoscos)cos(

Subtracted, the yx coscos terms cancel and we get:

yxyxyx sinsin2)cos()cos(

Divide out the 2 and we have one such identity:

)]cos()[cos(sinsin21 yxyxyx

Summed, the yxsinsin terms cancel, and diving by 2, we get another similar identity:

)]cos()[cos(coscos21 yxyxyx

The process is repeated with the sum and difference formulas for the sine function, and

two more identities are derived in a similar manner. The set of product-to-sum identities

are given in the next page.

The Product-to-Sum Identities

)]cos()[cos(sinsin21 yxyxyx

)]cos()[cos(coscos21 yxyxyx

)]sin()[sin(cossin21 yxyxyx

)]sin()[sin(sincos21 yxyxyx

Consider the following example:

Example 1: Write xx 3sinsin as the sum of two individual trigonometric terms.

Solution: The identity )]cos()[cos(sinsin21 yxyxyx is used here. We get the

following:

)]4cos()2[cos()]3cos()3[cos(3sinsin21

21 xxxxxxxx


91

Recall that the cosine absorbs negatives, so that the final result is

)]4cos()2[cos(3sinsin21 xxxx

The Sum-to-Product Identities

These identities are proven by performing the product-to-sum identities in reverse.

Consider the identity )]sin()[sin(cossin21 vuvuvu . Multiply the 2 to remove the

fraction, and let vux and vuy . Solving for u and v in terms of x and y yields

the following: 2

yxu and

2

yxv . Thus, the equation becomes:

yxyxyx

vuvuvu

sinsin2

cos2

sin2

)]sin()[sin(cossin21

The other three identities are solved in a similar manner. The four sum-to-product

identities are:

The Sum-to-Product Identities

2cos

2sin2sinsin

yxyxyx

2sin

2cos2sinsin

yxyxyx

2cos

2cos2coscos

yxyxyx

2sin

2sin2coscos

yxyxxy

Example 2: Rewrite the sum of xx 3cos2cos as the product of two trigonometric

functions.

Solution: The identity produces the following:

)cos()cos(2)cos()cos(22

32cos

2

32cos23cos2cos

21

25

21

25 xxxx

xxxxxx

By applying the appropriate product-to-sum identity, the latter result can easily be

rewritten into its sum form.

These formulas can be used to prove various identities:

Example 3: Prove the identity: )6cos4(coscot6sin4sin xxxxx .


92

Solution: The appropriate sum-to-product identities are used on both sides of the

equation:

)cos()5sin(2)cos()5sin(2

))sin()5sin(2(sin

cos)cos()5sin(2

))sin()5sin(2(cot)cos()5sin(2

2

46sin

2

46sin2cot

2

64cos

2

64sin2

)6cos4(coscot6sin4sin

xxxx

xxx

xxx

xxxxx

xxxxx

xxxx

xxxxx

Carefully follow each step and notice the cancellation taking place as well as the

absorption of negative signs by the cosine function.

Example 4: Prove the identity: qp

qpqp

coscos

sinsin

2tan

Solution: It’s probably easiest to work the right side of this equation first:

2cos

2cos2

2cos

2sin2

coscos

sinsin

qpqp

qpqp

qp

qp

The 2

cosqp

factors cancel as well as the 2s. The identity is thus proven:

2tan

2cos

2sin

2cos

2cos2

2cos

2sin2

coscos

sinsin qp

qp

qp

qpqp

qpqp

qp

qp

In the next section we will see an example how these formulas can be used to solve a

trigonometric equation.

4.6 Solution of Trigonometric Equations

An equation is that involves one or more of the various trigonometric functions is called a

trigonometric equation. The variable is in the argument position of the trigonometric

functions, and normally polynomial and exponential terms are not included. For

example,


93

0)1cos(3,1)2sin( xx and 1sin2cos2 xx

are all considered trigonometric equations, while

0cos22 xxx

is not considered a trigonometric equation since it involves (in this case) polynomial

terms. Equations of this last type are usually impossible to solve by purely algebraic

means. We will not consider these latter forms in this section.

Linear Trigonometric Equations

A linear trigonometric equation involves just one trigonometric function, usually itself a

linear term. In these cases, solutions are easily found by the usual rules of algebra, and

the correct use of the inverse trigonometric function at the appropriate time. Bear in

mind all solutions will be in radians, and that you will be responsible for locating all

solutions of the equation taking into account symmetries.

Example 1: Solve for x: 01sin2 x . State the solution sets as follows: (a) all

solutions, (b) solutions within 20 x .

Solution: Adding the 1 and diving the 2 results in this equivalent equation:

21sin x

Now is the correct time to employ the inverse sine function as an operator:

)(sin

)(sin)(sinsin

211

2111

x

x

From our knowledge of the inverse sine function, the primary solution is 6

x .

However, there is another solution. Recall the sine is positive in quadrants I and II; by

symmetry, the secondary solution is 6

5x .

Thus, for (a), the set of all solutions is found by adding integer multiples of 2 to each

solution:

Qnnn },2,2{6

56

where set Q represents the integers. For (b), the restriction 20 x simply allows us

to avoid listing all (infinitely many) solutions of this equation. The solution set is

therefore just the set },{6

56

.

Example 2: Solve for x: 52)12cos(6 x . State the solution sets as follows: (a) all

solutions, (b) solutions within 20 x .

Solution: Isolate the cosine term:

21)12cos( x


94

Now we use the inverse cosine as an operator:

)(cos12

)(cos))12(cos(cos

211

2111

x

x

The value for )(cos211 is

3 (primary, quadrant I), and by symmetry,

35 (secondary,

quadrant IV). We now have two equations:

nxnx 212&2123

53

Both solve the same way, yielding

nxnx6

356

3 &

Thus, to answer (a), the set of all solutions is

},{6

356

3 nn

To find the solution restricted to 20 x , we need to evaluate for values of n. When

n = 0, we get the two principal solutions 6

356

3 & xx ; these two solutions lie in

quadrants I and II, respectively. When n = 1, we generate two more solutions

6311

637 & xx , which lie in quadrants III and IV, respectively. When n = 2, we

generate two more solutions, but they place us back to the original solutions in quadrants

I and II. Therefore, to answer part (b), the solution set of this equation when restricted to

20 x is

},,,{6

3116

376

356

3

Note that there was no need to evaluate for 1n since doing so generates solutions

redundantly.

Example 3: Solve for x: 0sinsincos2 xxx . Give all solutions in the interval

20 x .

Solution: In this case we can factor xsin :

0)1cos2(sin

0sinsincos2

xx

xxx

Each factor can be solved for 0 separately. Setting 0sin x yields the solutions 0x

and x . Setting 01cos2 x yields 21cos x which in turn yields solutions

3x

and 3

5x . Therefore, the solution set to this function is },,,0{3

53

.

Comment: the restriction 20 x is common in trigonometric equations since it

requests only those solutions in the first four quadrants, and no more (i.e. no


95

redundancies). Note that this restriction includes 0 but excludes 2 since at this latter

value, we are “repeating” the unit circle once again. Thus, in the previous example, there

was no need to state the solution 2x since it is redundant to the solution 0x .

Comment: In the preceding example, do NOT divide by xsin . In doing so, you remove

all solutions that the xsin factor provides, and you potentially divide by zero, which is

never a legal arithmetic maneuver.

Non-linear Trigonometric Equations

If a trigonometric term is raised to a power other than one, it is considered non-linear.

The usual rules of algebra still apply, and we solve these forms using whatever methods

are appropriate.

Example 4: Solve for x in the interval 20 x : 432sin x .

Solution: The sine term is squared, so we take the radical, including both positive and

negative solutions:

2

3

432

432

sin

sin

sin

x

x

x

Thus, we need to solve two equations for x:

2

3sin x and 2

3sin x

In the first case, the inverse sine function as an operator yields 32

31 )(sin xx as

the primary solution, and by symmetry, 3

2x as the secondary solution.

In the second case, the inverse sine operator yields 3

52

31 )(sin xx as the

primary solution and by symmetry, 3

4x as a solution. Thus, there are four solutions

and the solution set is },,,{3

53

43

23

.

Example 5: Solve for x in the interval 20 x : xx 2sin2cos1 .

Solution: This equation involves both sine and cosine terms, as well as a squared term.

In cases like this, it is usually best to get all trigonometric terms into a common form.

Thus, we re-arrange this equation a bit, then replace the x2sin term using the identity

xx 22 cos1sin . This will result in a quadratic equation in terms of cosine only:


96

01coscos2

01cos)cos1(2

0cos1sin2

sin2cos1

2

2

2

2

xx

xx

xx

xx

The last equation is now a quadratic equation with xcos as the unknown. Factor:

0)1)(cos1cos2(

01coscos2 2

xx

xx

Each factor is then solved for zero:

35

321 ,cos01cos2 xxxx

and

xxx 1cos01cos

Therefore, the solution set is },,{3

53

.

Other Forms

There is no limit to the number of possible equations involving trigonometric functions.

Some may be easy to solve, some not so easy, and some impossible. You should always

look for possible identities to simplify the equation, collecting terms, and avoiding the

dividing out of a trigonometric function. The following example is one of many possible

trigonometric equations:

Example 6: Solve for x in the interval 20 x : xxx tancscsec2

Solution: Replacing each of these forms in terms of sine and cosine, we can greatly

simplify this equation and write it in terms of a single trigonometric function:

xx

x

x

xx

xxx

cos

1

cos

1

cos

sin

sin

1

cos

1

tancscsec

2

2

2

Cross-multiplying and collecting the terms to one side yields 0coscos2 xx . We

factor:

0)1(coscos

0coscos2

xx

xx

Each factor is solved for its solutions:


97

23

2,0cos xxx

01cos01cos xxx

However, none of them are acceptable! Why is this? Note that the xsec term is not

defined when 0cosx , so the two solutions must be discarded, and the xcsc term is not

defined when 0x , so it too must be discarded. The solution set is empty: .

Example 7: Solve for x in the interval 20 x : 03sin2sin xx

Solution: These two sine terms are “unlike”; they cannot be combined conveniently by

addition. Therefore, we use an appropriate sum-to-product identity (section 4.5) to assist:

0)cos()sin(2

03sin2sin

21

25 xx

xx

Therefore, each factor can be set to 0 and solved. We explore the solutions generated by

the factor 0)sin(25 x first:

)2(&)20(2&200)sin(52

52

25

25

25 nxnxnxnxx

The solutions are generated by evaluating for .,2,1,0 etcn By direct evaluation we get

these solutions:

510

58

56

54

52

2,1,0

0x

xn

x

xn

x

xn .

The last solution 5

10 is the same as 2 and can be ignored, and there is no need to

evaluate for higher values of n since this will simply repeat our known solutions again.

The solutions generated by the factor 0)cos(21 x , we get the following:

)2(2&)2(22&20)cos(2

322

321

221

21 nxnxnxnxx

The only meaningful solution that results from these two equations is x . All others

are outside the restriction 20 x . Therefore, the solution set to this equation is:

},,,,,0{58

56

54

52

If xxxf 3sin2sin)( , then the solution set can be illustrated as the x-intercepts of this

graph in the interval 20 x :


98

Figure: Graph for example 7

4.7 Law of Sines and Cosines

Law of Cosines

Suppose that a triangle has sides of lengths a, b, and c corresponding to the angles

, , and as shown in the figure.

Then 2 2 2

2 2 2

2 2 2

2 cos

2 cos

2 cos

a b c bc

b c a ca

c a b ab

C

b a

A c B

And Law of Sines given below

sin sin sinA B C

a b c

Examples

1. A triangle has sides of length 8, and 6, and the angle between these sides is 30

degrees. Find the length of the third side.

Solution We use the formula (Use your calculator for simplification) A

2 2 2

2 2 2

2 cos

6 8 2(8)(6)cos30

a b c bc

a 8 6

The third side has the length equals a. B C

2. Use law of Sines to find the indicated side CB = x, where angle CAB = 56

degrees and angle CBA = 65 degrees, AB 27 cm.

C

Solution We have angle ACB = 180 – (56+65) = 59

sin 59 sin 56

27

sin 56(27) 26.11

sin 59

x

x

A 56 65 B

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