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Basic Trigonometric Identities
Reciprocal Identities
1
csc x = sin x
1
sec x = cos x
1
cot x = tan x
Quotient Identities
sin x
tan x = cos x
cos x
cot x = sin x
Pythagorean Identities
sin2x + cos2x = 1
tan2x + 1 = sec2x
1 + cot2x = csc2x
Even-Odd Identities
sin(-x) = -sin x
csc(-x -csc x
cos(-x) = cos x
sec(-x) = sec x
tan(-x) =-tan x
cot(-x) =-cot x
Example
Verify the identity: sec x cot x = csc x.
Solution
The left side of the equation contains the more complicated
expression. Thus, we work with the left side. Let us express this side of the
identity in terms of sines and cosines. Perhaps this strategy will enable us to
transform the left side into csc x, the expression on the right.
1
cos x
sec x cot x =

cos x sin x
Apply a reciprocal identity: sec x = 1/cos x
and a quotient identity: cot x = cos x/sin x.
1
=
= csc x
sin x
Divide both the numerator and the
denominator by cos x, the common factor.
Example
Verify the identity: cosx - cosxsin2x = cos3x..
Solution
We start with the more complicated side, the left side. Factor out
the greatest common factor, cos x, from each of the two terms.
cos x - cos x sin2 x = cos x(1 - sin2 x)
= cos x ·
= cos3 x
cos2
x
Factor cos x from the two terms.
Use a variation of sin2 x + cos2 x = 1.
Solving for cos2 x, we obtain cos2 x =
1 – sin2 x.
Multiply.
We worked with the left and arrived at the right side. Thus, the identity is
verified.
Guidelines for Verifying
Trigonometric Identities
1.
Work with each side of the equation independently of the
other side. Start with the more complicated side and
transform it in a step-by-step fashion until it looks
exactly like the other side.
2.
Analyze the identity and look for opportunities to apply
the fundamental identities. Rewriting the more
complicated side of the equation in terms of sines and
cosines is often helpful.
3.
If sums or differences of fractions appear on one side,
use the least common denominator and combine the
fractions.
4.
Don't be afraid to stop and start over again if you are not
getting anywhere. Creative puzzle solvers know that
strategies leading to dead ends often provide good
problem-solving ideas.
Example
• Verify the identity:
csc(x) / cot (x) = sec (x)
Solution:
csc x
= sec x
cot x
1
sin x = 1
cos x cos x
sin x
1
sin x
1

=
sin x cos x cos x
Example
• Verify the identity:
cos x = cos x  cos x sin x
3
2
Solution:
cos x = cos x  cos x sin x
3
2
cos x = cos x(cos x  sin x)
cos x = cos x(1)
2
2
Example
• Verify the following identity:
tan 2 x - cot 2 x
= tan x - cot x
tan x  cot x
Solution:
tan x - cot x (tan x - cot x) * (tan x  cot x)
=
tan x  cot x
tan x  cot x
2
2
(tan x - cot x) * (tan x  cot x)
=
tan x  cot x
= tan x - cot x
7.2 Trigonometric
Equations
Equations Involving a Single
Trigonometric Function
To solve an equation containing a single
trigonometric function:
• Isolate the function on one side of the
equation.
• Solve for the variable.
Trigonometric Equations
y
y = cos x
1
y = 0.5
–4 
–2 
2
4
x
–1
cos x = 0.5 has infinitely many solutions for –  < x < 
y
y = cos x
1
0.5
2
–1
x
cos x = 0.5 has two solutions for 0 < x < 2
Example
Solve the equation: 3 sin x - 2 = 5 sin x - 1,
Solution
0 ≤ x < 360°
The equation contains a single trigonometric function, sin x.
Step 1 Isolate the function on one side of the equation. We can solve for
sin x by collecting all terms with sin x on the left side, and all the constant
terms on the right side.
3 sin x - 2 = 5 sin x - 1
3 sin x - 5 sin x - 2 = 5 sin x - 5 sin x – 1
-2 sin x - 2 = -1
-2 sin x = 1
sin x = -1/2
x = 210° or x = 330°
This is the given equation.
Subtract 5 sin x from both sides.
Simplify.
Add 2 to both sides.
Divide both sides by -2 and solve for sin x.
Example
Solve the equation:
2 cos2 x  cos x - 1 = 0,
0  x < 2.
Solution
The given equation is in quadratic form 2t2  t - 1 = 0 with t =
cos x. Let us attempt to solve the equation using factoring.
2 cos2 x  cos x - 1 = 0
This is the given equation.
(2 cos x - 1)(cos x  1) = 0
2 cos x - 1= 0
or
Factor. Notice that 2t2 + t – 1 factors as (2t – 1)(2t + 1).
cos x  1 = 0
2 cos x = 1 cos x = -1 cos x = 1/2
Set each factor equal to 0.
Solve for cos x.
x =  x = 2 -=  x = 
The solutions in the interval [0, 2) are /3, , and 5/3.
Example
• Solve the following equation for θ is any
real number. 7 cos  9 = -2 cos
Solution:
7 cos  9 = -2 cos
9 cos = -9
cos = -1
 =  ,3 ,5
 =   2n
Example
• Solve the equation on the interval [0,2)

3
tan =
2
3
Solution:

3
tan =
2
3


7
= and
2 6
6

7
 = and
3
3
Example
• Solve the equation on the interval [0,2)
cos x  2 cos x - 3 = 0
2
Solution:
cos 2 x  2 cos x - 3 = 0
(cos x  3)(cos x - 1) = 0
cos x  3 = 0 cos x - 1 = 0
cos x = -3 cos x = 1
no solution x = 0
x=0
7.3 Sum and
Difference Formulas
The Cosine of the Difference of
Two Angles
cos( -  ) = cos  cos   sin  sin 
The cosine of the difference of two angles
equals the cosine of the first angle times
the cosine of the second angle plus the
sine of the first angle times the sine of
the second angle.
Example
• Find the exact value of cos 15°
Solution
We know exact values for trigonometric functions of 60° and 45°.
Thus, we write 15° as 60° - 45° and use the difference formula for cosines.
cos l5° = cos(60° - 45°)
= cos 60° cos 45°  sin 60° sin 45°
1
2
3
2



2 2
2
2
=
2
6

4
4
=
2 6
4
cos( -) = cos  cos   sin  sin 
Substitute exact values from
memory or use special triangles.
Multiply.
Add.
Example
Find the exact value of cos 80° cos 20°  sin 80° sin 20°.
Solution
The given expression is the right side of the formula for cos( - )
with  = 80° and  = 20°.
cos( -) = cos  cos   sin  sin 
cos 80° cos 20°  sin 80° sin 20° = cos (80° - 20°) = cos 60° = 1/2
Example
• Find the exact value of cos(180º-30º)
Solution
cos(180 - 30)
= cos180 cos 30  sin 180 sin 30
3
1
= -1*
 0*
2
2
3
=2
Example
• Verify the following identity:
5 
2

cos x -  = (cos x  sin x)
4 
2

Solution
5 

cos x 
4 

 5
= cos x cos
 4

 5 
  sin x sin  

 4 
2
2
cos x  sin x
2
2
2
=(cos x  sin x)
2
=-
Sum and Difference Formulas
for Cosines and Sines
cos(   ) = cos  cos  - sin  sin 
cos( -  ) = cos  cos   sin  sin 
sin(    ) = sin  cos   cos  sin 
sin(  -  ) = sin  cos  - cos  sin 
Example
• Find the exact value of sin(30º+45º)
Solution
sin(    ) = sin  cos   cos  sin 
sin( 30  45) = sin 30 cos 45  cos 30 sin 45
1
2
3
2
= 


2 2
2
2
2 6
=
4
Sum and Difference Formulas
for Tangents
The tangent of the sum of two angles equals the tangent of the first
angle plus the tangent of the second angle divided by 1 minus
their product.
tan   tan 
tan(   ) =
1 - tan  tan 
tan  - tan 
tan( -  ) =
1  tan  tan 
The tangent of the difference of two angles equals the tangent of
the first angle minus the tangent of the second angle divided by
1 plus their product.
Example
• Find the exact value of tan(105º)
Solution
•tan(105º)=tan(60º+45º)
tan   tan 
tan(   ) =
1 - tan  tan 
tan 60  tan 45
=
1 - tan 60 tan 45
3 1 1 3
=
=
1- 3 1- 3
Example
• Write the following expression as the sine, cosine, or
tangent of an angle. Then find the exact value of the
expression. 7

7

sin
cos - cos
sin
12
12
12
12
Solution
7

7

sin
cos - cos
sin
12
12
12
12
6
 7  
= sin 
-  = sin
12
 12 12 
= sin

2
=1
Review Quiz Answers
7.4 Double-Angle
and Half-Angle
Formulas
Double-Angle
Identities
sin 2 = 2 sin cos
cos 2 = cos2 – sin2 = 1 – 2 sin2 = 2 cos2 – 1
2tan
tan 2 =
1 - tan 2
Three Forms of the DoubleAngle Formula for cos2
cos 2 = cos  - sin 
2
2
cos 2 = 2 cos  - 1
2
cos 2 = 1 - 2 sin 
2
Power-Reducing Formulas
1 - cos 2
sin  =
2
1  cos 2
2
cos  =
2
1 - cos 2
2
tan  =
1  cos 2
2
Example
• If sin α = 4/5 and α is an acute angle,
find the exact values of sin 2 α and cos
2 α.
Solution
If we regard α as an acute angle of a right triangle, we obtain
cos α = 3/5. We next substitute in double-angle formulas:
Sin 2 α = 2 sin α cos α = 2 (4/5)(3/5) = 24/25.
Cos 2 α = cos2 α – sin2 α = (3/5)2 – (4/5)2 = 9/25 – 16/25 = -7/25.
Half-Angle Identities
x
sin 2 = ±
1 – cos x
2
x
cos 2 = ±
1 + cos x
2
x
tan 2 = ±
1 – cos x
sin x
1 – cos x
=
=
1 + cos x 1 + cos x
sin x
x
where the sign is determined by the quadrant in which 2 lies.
Example
Find the exact value of cos 112.5°.
Solution
Because 112.5° = 225°/2, we use the half-angle formula for cos /2
with  = 225°. What sign should we use when we apply the formula? Because
112.5° lies in quadrant II, where only the sine and cosecant are positive, cos
112.5° < 0. Thus, we use the - sign in the half-angle formula.
225
cos112.5 = cos
2
- 2

1 
1  cos 225
 2 
==
2
2
=-
2- 2
2- 2
=4
2
Half-Angle Formulas for:

1 - cos 
tan =
2
sin 

sin 
tan =
2 1  cos 
Example
• Verify the following identity:
(sin  - cos ) 2 = 1 - sin 2
Solution
(sin  - cos  ) 2
= sin 2  - 2 sin  cos   cos 2 
= sin 2   cos 2  - 2 sin  cos 
= 1 - 2 sin  cos  = 1 - sin 2
7.5 Product-to-Sum
and
Sum-to-Product
Formulas
Product-to-Sum Formulas
1
sin  sin  = cos( -  ) - cos(   )
2
1
cos  cos  = cos( -  )  cos(   )
2
1
sin  cos  = sin(    )  sin(  -  )
2
1
cos  sin  = sin(    ) - sin(  -  )
2
Example
• Express the following product as a sum or difference:
cos 3x cos 2x
Solution
1
cos  cos  = cos( -  )  cos(   )
2
cos 3x cos 2 x
1
= cos(3x - 2 x)  cos(3x  2 x)
2
1
= cos( x)  cos(5 x)
2
Text Example
Express each of the following products as a sum or difference.
a. sin 8x sin 3x
b. sin 4x cos x
Solution
The product-to-sum formula that we are using is shown in each
of the voice balloons.
a.
sin  sin  = 1/2 [cos( - ) - cos( + )]
sin 8x sin 3x = 1/2[cos (8x - 3x) - cos(8x  3x)] = 1/2(cos 5x - cos 11x)
b.
sin  cos  = 1/2[sin( + ) + sin( - )]
sin 4x cos x = 1/2[sin (4x  x)  sin(4x - x)] = 1/2(sin 5x  sin 3x)
Sum-to-Product Formulas
sin   sin  = 2 sin

cos
-
2
2
-

sin  - sin  = 2 sin
cos
2
2

 -
cos   cos  = 2 cos
cos
2
2

 -
cos  - cos  = -2 sin
sin
2
2
Example
• Express the difference as a product:
sin 4x - sin 2x
Solution
sin  - sin  = 2 cos
 
sin
 -
2
2
4x  2x
4x - 2x
sin 4 x - sin 2 x = 2 cos
sin
2
2
6x
2x
= 2 cos sin
= 2 cos 3x sin x
2
2
Example
• Express the sum as a product:
sin x  sin 4x
Solution
sin   sin  = 2 sin

cos
-
2
2
x  4x
x - 4x
sin x  sin 4 x = 2 sin
cos
2
2
5x
- 3x
= 2 sin
cos
2
2
Example
• Verify the following identity:
sin x  sin y
x y
x- y
= tan
cot
sin x - sin y
2
2
Solution
x y
x- y
2 sin
cos
sin x  sin y
2
2
=
sin x - sin y 2 sin x - y cos x  y
2
2
x y
x- y
sin
cos
x y
x- y
2
2
=
= tan
cot
x y
x- y
2
2
cos
sin
2
2
7.6
Inverse Trig Functions
Objective: In this section, we will look at the definitions and
properties of the inverse trigonometric functions. We will recall
that to define an inverse function, it is essential that the function
be one-to-one.
7.6 Inverse Trigonometric Functions
and Trig Equations
y = sin -1 ( x) = arcsin( x)
Domain: [–1, 1]
  
Range: - , 
 2 2
-1
y = cos ( x) = arccos( x)
Domain: [–1, 1]
Range: [0, π]
y = tan -1 ( x) = arctan( x)
Domain: 
 
Range:  - , 

2 2
Let us begin with a simple question:
What is the first pair of inverse functions that pop
into YOUR mind?
f ( x) = x 2
f -1 ( x) = x
This may not be your pair but
this is a famous pair. But
something is not quite right
with this pair. Do you know
what is wrong?
Congratulations if you guessed that the top function
does not really have an inverse because it is not 1-1
and therefore, the graph will not pass the horizontal
line test.
2
y
=
x
.
Consider the graph of
y
Note the two points
on the graph and
also on the line y=4.


f(2) = 4 and f(-2) = 4
so what is an inverse
function supposed
to do with 4?



f -1 (4) = 2 or f -1 (4) = -2 ?
-

By definition, a function cannot generate two different
outputs for the same input, so the sad truth is that this
function, as is, does not have an inverse.
x
So how is it that we arrange for this function to have an
inverse?
2
y=x
y=x
We consider only one half
of the graph: x > 0.
4
The graph now passes
the horizontal line test
and we do have an
inverse:
f ( x) = x for x  0
2
y= x
2




-1
f ( x) = x
Note how each graph reflects across the line y = x onto
its inverse.
x
A similar restriction on the domain is necessary to
create an inverse function for each trig function.
Consider the sine function.
You can see right
away that the sine
function does not
pass the horizontal
line test.
But we can come up
with a valid inverse
function if we restrict
the domain as we did
with the previous
function.
y
y = sin(x)

y = 1/2
-
-
-

-
How would YOU restrict the domain?




x

Take a look at the piece of the graph in the red frame.
We are going to build
the inverse function
from this section of
the sine curve
because:
This section picks up
all the outputs of the
sine from –1 to 1.
This section includes
the origin. Quadrant I
angles generate the
positive ratios and
negative angles in
Quadrant IV generate
the negative ratios.
y

-
-
-





x

-
Lets zoom in and look at some
key points in this section.
I have plotted the special angles on the curve and the
table.
y
y = sin(x)
x
-

2

3

4

6
0

6

4

3

2
f ( x)

-1
3
2
2
2
1
2
0
1
2
2
2
3
2
-
1
-
-
-

-

x

The new table generates the graph of the inverse.
The domain
-1
x
sin ( x )
x sin( x)
of the chosen


-1
To get a good
-1
section of
2
2
 
look at the
the sine-is
3

, 

3
 2 2
graph
of
the
2
3
3
2
So the range
2

inverse

2
of the arcsin
2
4
function, we
4
2
  
1

is

1
will “turn the
- 2 , 2 
2
6
6
2
tables”
on
0
0
0
0
The range of
1


1
the sine
2
6
the chosen
6
2
function.
2


2
section of the
2
4
4
2
sine is [-1 ,1]
3


3
so the domain
2
3
3
2
of the arcsin is


1
1
[-1, 1].
2
2
Note how each point on the original graph gets “reflected”
onto the graph of the inverse.
    
 ,1 to 1, 
2   2
y = arcsin(x)
y
y = sin(x)
 3   3  
 ,
 

 3 2  to  2 , 3 

 


 2   2  
 ,
 

 4 2  to  2 , 4 

 

-

etc.
You will see the
inverse listed
as both:
arcsin( x) and sin -1 ( x)
-
x
In the tradition of inverse functions then we have:


 
sin   = 1  arcsin( 1) = or sin -1 (1) =
2
2
2
 3 
 3 
3
 
-1
 = or sin 
=
sin   =
 arcsin 



3 2
 2  3
 2  3
Unless you are
instructed to
use degrees,
you should
assume that
inverse trig
functions will
generate
outputs of real
numbers (in
radians).
The thing to remember is that for the trig function the
input is the angle and the output is the ratio, but for the
inverse trig function the input is the ratio and the output
is the angle.
The other inverse trig functions are generated by using
similar restrictions on the domain of the trig function.
Consider the cosine function:
What do you
think would be
a good domain
restriction for
the cosine?
Congratulations if
you realized that
the restriction we
used on the sine
is not going to
work on the
cosine.
y = cos(x)
-
-
y

-

-




x

The chosen section for the cosine is in the red frame. This
section includes all outputs from –1 to 1 and all inputs in
the first and second quadrants.
Since the domain and range for the cosine are 0,  and -1,1,
the domain and range for the inverse cosine are -1,1 and 0 ,  .
y = cos(x)
y
y
y = arccos(x)





-
-
-





x



-
-

x
The other trig functions require similar restrictions on
their domains in order to generate an inverse.
Like the sine function, the domain of the section of the
  
tangent that generates the arctan is - ,  .
 2 2
y
y
y=arctan(x)


y=tan(x)




-
-


x
-
-

-
-
-
-
-
-
  
D =  - ,  and R = - ,  
 2 2
  
D = - ,   and R =  - , 
 2 2

x

The table below will summarize the parameters we have
so far. Remember, the angle is the input for a trig function
and the ratio is the output. For the inverse trig functions
the ratio is the input and the angle is the output.
arcsin(x) arccos(x) arctan(x)
Domain -1  x  1 -1  x  1 -   x  
Range
-

2
x

2
0 x  -

2
x
When x<0, y=arcsin(x) will be in which quadrant?

2
y<0 in IV
When x<0, y=arccos(x) will be in which quadrant? y>0 in II
When x<0, y=arctan(x) will be in which quadrant? y<0 in IV
The graphs give you the big picture concerning the
behavior of the inverse trig functions. Calculators are
helpful with calculations (later for that). But special
triangles can be very helpful with respect to the basics.
60
45
2
2
2
1
30
45

3
2
Use the special triangles above to answer the following.
Try to figure it out yourself before you click.
 3
=
arccos 

2


csc -1 ( 2) =
3
 

30  or  because cos30  =
2
 6


30  or  because csc30  = 2 / 1 = 2
 6

OK, lets try a few more. Try them before you peek.
60

45
2
1
2
2
30
45
3
2
 1 

1


arcsin 
 = 45 (or ) because sin 45 =
4
2
 2
tan -1 ( 3 ) =

arcsin 


3
60 (or ) because tan 60 =
= 3
3
1
-1 
 = - 45 (or -  ) because sin - 45 = - 1
2
4
2




Negative inputs for the arccos can be a little tricky.
y
60

2
2
3
1
30

-1
3
 = 180 - 60 = 120

 -1 
arccos   = 
 2 

60

x

x -1
to check : cos 120 = =
r
2



From the triangle you can see that arccos(1/2) = 60 degrees.
But negative inputs for the arccos generate angles in
Quadrant II so we have to use 60 degrees as a reference
angle in the second quadrant.
You should be able to do inverse trig calculations without
a calculator when special angles from the special
triangles are involved. You should also be able to do
inverse trig calculations without a calculator for
quadrantal angles.
Its not that bad. Quadrantal
angles are the angles between
the quadrants—angles like
-

2
or - 90 , 0 or 0 ,



2
y = cos(x)
y

or 90 ,  or 180
To solve arccos(-1) for example,
you could draw a quick sketch of
the cosine section:
And observe that arccos(-1) =

-
-
-

-




 ,-1
x

But a lot of people feel comfortable using the following
sketch and the definitions of the trig ratios.
For arccos(-1) for example,
you can observe that, since
cos  =
y
(0, 1)
r=1
x
r the point (-1, 0) is
the one we want. That point
is on the terminal side of
 = .
x -1
So, since cos( ) = =
= -1,
r 1
arccos( -1) =  .
x
(1, 0)
(-1, 0)
(-1, 0)
Finally, we encounter the composition of trig functions
with inverse trig functions. The following are pretty
straightforward compositions. Try them yourself before
you click to the answer.
 -1  - 3  
 = ?
sin  sin 


2



so
 -1  - 3  
- 3


 = sin  =
sin sin 


2
2




First, what do we know about
 ?
We know that  is an angle whose sine is
- 3
.
2
Did you suspect from the beginning that this was the
answer because that is the way inverse functions are
SUPPOSED to behave? If so, good instincts but….
Consider a slightly different setup:
 

arcsin sin 120 =
 3
 = 60.
arcsin 

2


This is also the
composition of two
inverse functions but…
Did you suspect the answer was going to be 120
degrees? This problem behaved differently because
the first angle, 120 degrees, was outside the range of
the arcsin. So use some caution when evaluating the
composition of inverse trig functions.
The remainder of this presentation consists of
practice problems, their answers and a few complete
solutions.
Find the
exact value
of each
expression
without using
a calculator.
When your
answer is an
angle,
express it in
radians.
Work out the
answers
yourself
before you
click.
 -1 
1. sin  
 2 
2. arccos - 1
10. sec -1 2 
-1
3. tan -1 - 1
 1 
4. arctan 

 3
5. arcsin 0 
 1 
6. cos 

 2
-1

7. arctan - 3
8. sin -1 - 1

- 3

9. cos 

2


-1
 -1 
11. arccos

 2
   
12. arcsin  sin  -  
  2 
13. arcsin sin 270
 


 -1  
14. tan  arccos  
 2 

  - 
15. arccos cos
 
  3 
 -1  1  
16. sin  cos  -  
 2 

Answers for problems 1 – 9.
1.
2.
3.
4.
 -1  - 
sin   =
6
 2 
arccos - 1 = 
-
-1
tan - 1 =
4
 1  
arctan 
=
 3 6
-1
5. arcsin 0  = 0
 1  
6. cos 
=
 2 4
-
7. arctan - 3 =
3
-
-1
8. sin - 1 =
2
 - 3  5
=
9. cos 
 6
2


-1
Negative ratios for arccos
generate angles in Quadrant II.
y
1
2
- 3

x
-1


The reference angle is


6
6  5
- =
so the answer is  =  - =
6
6 6
6
10. sec -1 2  = cos -1 1 / 2  = 
y
3
 - 1  3
11. arccos
=
 2 4
   

12. arcsin  sin  -   = 2
  2 
2
3
60
 

  -
15. arccos cos
  3

1 
  = arccos  =

2 3
 -1  1  
3
 2 
16. sin  cos  -   = sin   =
 2 
 3  2


-1
-
13. arcsin sin 270 = arcsin - 1 = -90 =
2

 -1  
 2 
14. tan  arccos   = tan   = - 3
y
2
3







14.
x

15.
1
x

2
- 3
Review Answers for Test