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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:
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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.
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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.
4 Analytic Trigonometry
87
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:
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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
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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
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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
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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 .
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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,
4 Analytic Trigonometry
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
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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
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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:
4 Analytic Trigonometry
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:
4 Analytic Trigonometry
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 :
4 Analytic Trigonometry
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