sin3
sin5
73.
sin 4
cos2
74.
sin5
cos3
75.
cos5
sin3
76.
cos4
sin 2
(77–84)
Convert the following sums or differences to products.
77.
sin 5
sin3
78.
sin 4
sin 2
80.
sin 5
sin3
80.
sin 4
sin 2
81.
cos5
cos3
82.
cos4
cos2
83.
cos5
cos3
84.
cos4
cos2

Section 6.1 Trigonometric Identities
Section 6.4
Solving Trigonometric Equations
Objectives
Understanding how to solve trigonometric equations
It is now time to solve equations that have trigonometric functions. To solve equations of this
type you will need to get the equations into factors equal to zero with each factor
containing only one trigonometric function.
General Strategy:
1.
Get the equation equal to zero
2.
Convert all trigonometric functions into the same function by using identities or if
that isn’t possible, then factor the equation into factors where each has only one
type of trigonometric function.
3.
Set each factor equal to zero and solve for the trigonometric function.
4.
Lastly identify which angles make each equation true.
SOLVING TRIGONOMETRIC EQUATIONS
Discussion 1: Solving a Simple Equation
Let’s find all the angles that make 2sin
θ
− 1 = 0.
As we think about our strategy for solving this we see that it is equal to zero already and that there is
only one function.
Therefore, we will begin by getting sin
θ
alone.
2sin
θ
− 1 = 0
2sin
θ
= 1
(added one)
sin
θ
=
1
2
(divided by two)
Now, we need to think about what angle
input would cause the output of the sine
function to be
1
2
.
sin
θ
=
y
on the unit circle. An angle that has a
point on the unit circle with
y
-value of
1
2
is
6
.
We know that
6
is one angle. Now we
need to think about every other angle
whose reference angle is
6
that would
cause the output of the sine function to
be positive.
Sine is positive in the 1
st
and 2
nd
quadrants. So,
we have
6
and
5
6
(whose reference angle in the
2
nd
quadrant is
6
). In addition to these two we
also have every 2
π
revolutions around the unit
circle from these two angles, which will also

Chapter 6 Trigonometric Identities and Equations
make sine equal
1
2
, since sine is periodic with a
period of 2
π
.
Final answer.
θ
=
2
6
n
and
θ
=
5
2
6
n
(
n
any integer)
We could have used the calculator to help us with this problem as long as we are
comfortable with non-exact solutions.
We would still need to solve for sine.
2sin
θ
− 1 = 0
2sin
θ
= 1
(added one)
sin
θ
=
1
2
(divided by two)
But now we could use the calculator and
find the sin
-1
1
2
=
θ
.
Notice that 0.52359… is the same as
6
which we found earlier as one of the many answers.