4.8
Trig Identities and Equations
An identity is an equation that is true for all values of the variables. Examples of identities might be
“obvious” results like
2x + 2x = 4x
or
(x + y)2 = x2 + 2xy + y 2 .
Other examples of identities are:
1. (x + 3)2 = x2 + 6x + 9
and
2. (a very important one!) A2 − B 2 = (A − B)(A + B).
4.8.1
Trig identities vs. trig equations
What is a trig identity? A trig identity is an equation which is true for all inputs (such as angles, θ.) For
example, from the Pythagorean theorem on the unit circle, we know that the equation for the unit circle
is x2 + y 2 = 1 and so this turns into an identity for trig functions:
cos2 θ + sin2 θ = 1
This is true regardless of the choice of θ.
Other examples of trig identities are:
1. tan θ =
sin θ
cos θ
2. sin(−x) = − sin x
3. cos(z) = sin(z + π/2).
Some of our trig identities come from our definitions. For example, from the definition of the tangent
function we know that
sin θ
tan θ =
cos θ
We also have some identities given by symmetry. For example, since the sine function is odd then
sin(−x) − sin(x);
Since cosine is even then
cos(−x) = cos x.
By looking at the graphs of sine and cosine we observed that
cos x = sin(x + π/2).
A trig equation, unlike an identity, may not necessarily be true for all angles θ. In general, with a
trig equation, we wish to solve for θ.
189
4.8.2
Find all solutions to a trig equation
When we “solve” an equation, it is important that we find all solutions. For example, if we are solving
the equation
x2 − 3x + 2 = 0
then it is not sufficient to just list x = 1 as a solution. To find all solutions we might factor the quadratic
x2 − 3x + 2 = (x − 2)(x − 1) and then set this equal to zero:
(x − 2)(x − 1) = 0
This equation implies that either x − 2 = 0 (so x = 2) or x − 2 = 0 (so x = 1.) We have found two
solutions. From our understanding about zeroes of an equation, we now know that we have found all the
solutions.
Once reason for the concept of factoring is that it aids us in finding all solutions!
Another example: suppose we want to solve the equation x2 = 4. It is not sufficient to notice that
x = 2 is a solution, but we want to find both the solutions x = 2 and x = −2. (We often remind ourselves
of the possibilities √
of two solutions by writing a plus-or-minus symbol (±) such as in the computation
x2 = 4 =⇒ x = ± 4 = ±2.
For many trig problems, we will find solutions within one revolution of the unit circle and then use the
period of the trig functions to find an infinite number of solutions. For example, let’s solve the equation
√
3
sin θ =
.
2
Since the sine here is positive then the angle θ must be in quadrants
I or II. Recall our discussion of the
√
unit circle and 30-60-90 triangles and recognize that sin(π/3) = 23 . So θ = π/3 is a nice solution in the
first quadrant for our equation.
√
In the second quadrant, we note that 2π/3 has reference angle π/3 and so sin(2π/3) = 23 . So θ = 2π/3
is a nice solution in the second quadrant.
However, these two solutions are not all the solutions! Since the sine function is periodic with period
2π we can add 2π to any angle without changing the value of sine. So if θ = π/3 is a solution then so are
θ = π/3 − 2π, θ = π/3 + 2π, θ = π/3 + 4π, θ = π/3 + 6π, ... etc.
We can write this infinite collection of solutions in the form
θ = π/3 + 2πk
where we understand that k is a whole number, that is, k ∈ Z.
Similarly, if θ = 2π/3 is a solution then so are
2π/3 + 2πk (k ∈ Z)
We collect these together as our solution:
θ=
π
2π
+ 2πk or θ =
+ 2πk
3
3
√
(where k is an integer.) Or we might write all our solutions to the equation sin θ =
{
π
2π
+ 2πk : k ∈ Z} ∪ {
+ 2πk : k ∈ Z} .
3
3
190
3
using set notation:
2
Sometimes inverse trig functions come in handy. For example, suppose we are solving the equation
tan x = 2.
One solution is x = arctan(2) which is approximately 1.10715. (There is no simple way to write this
angle; we need the arctangent function.) Since tangent has period π and there is only one solution within
an interval of length π, we know that the full set of solutions is
{arctan(2) + πk : k ∈ Z}.
More worked problems.
√
3
1. Solve the equation sin 4θ =
.
2
√
3
then from the notes above, we know that
Solution. If sin 4θ =
2
π
4θ = + 2πk
3
or
2π
4θ =
+ 2πk.
3
So divide both sides by 4 to get
π
π
θ=
+ k
12
2
or
π π
θ= + k
6
2
(where k ∈ Z.)
π
Note that since sin θ has period 2π then sin 4θ must have period 2π
4 = 2 . This is reflected in our
π
solutions by our adding multiples of 2 to our first solutions, π/12 and π/6.
2. Solve sin x =
1
.
2
1
has two solutions within one revolution of the unit circle. They
2
5π
and x = 6 . Since sin x has period 2π we know that the collection of all solutions must
Solution. The equation sin x =
are x =
be
π
6
x=
π
6
+ 2πk and x =
5π
6
+ 2πk.
We can write this in set notation as
{ π6 + 2πk : k ∈ Z} ∪ { 5π
6 + 2πk : k ∈ Z}.
3. Solve 2 sin(x − π4 ) + 1 = 2.
Solution. Let’s not worry about the expression x −
x − π4 as an angle in its own right.
π
4
at the beginning of this problem. Think of
First we subtract 1 from both sides and then divide both sides by 2.
2 sin(x −
π
π
π
1
) + 1 = 2 =⇒ 2 sin(θ − ) = 1 =⇒ sin(θ − ) = .
4
4
4
2
191
π
1
5π
1
We know, from the previous problem, that sin( ) = and sin( ) = and so
6
2
6
2
θ−
π
π
π
5π
= + 2πk or θ − =
+ 2πk.
4
6
4
6
Now we can solve for θ by adding π/4 to both sides of these equations, obtaining solutions
θ=
π π
π 5π
+ + 2πk or θ = +
+ 2πk.
4
6
4
6
The best answer is achieved by getting a common denominator for π/4 + π/6:
θ=
5π
13π
+ 2πk or θ =
+ 2πk.
12
12
√
4. Solve cos x =
3
.
2
√
3
Solution. The equation cos x =
has two solutions within one revolution of the unit circle.
2
Since the cosine function is even, we know that when we find one positive solution, the negative
solution will also work. So our two solutions are x = π3 and x = − π3 . Since cos x has period 2π we
know that the collection of all solutions must be
{ π3 + 2πk : k ∈ Z} ∪ {− π3 + 2πk : k ∈ Z}.
5. Solve the equation tan2 θ = 3.
√
√
π
Solution. tan2 θ = 3 =⇒ tan θ = ± 3. If tan θ = 3 then θ = + 2πk (where k ∈ Z.)
3
√
2π
If tan θ = − 3 then θ =
+ 2πk . So the final answer is
3
θ=
4.8.3
π
2π
+ 2πk or θ =
+ 2πk .
3
3
Use algebra in solving trig identities
The standard rules of algebra are still important in solving trig equations. We will often use the algebra
fact that if AB = 0 then either A = 0 or B = 0.
If, for example, we need to solve the equation
(tan x − 2)(2 sin x − 1) = 0
then we note that this implies that either tan x − 2 = 0 or sin x − 21 = 0. These then imply tan x = 2 or
sin x = 12 . In an earlier problem we solved the equation tan x = 2 and got the solution set
{arctan(2) + πk : k ∈ Z}.
In a different problem we solved sin x =
{
1
2
and found the solution set
π
5π
+ 2πk : k ∈ Z} ∪ {
+ 2πk : k ∈ Z}.
6
6
192
In this case, we want all solutions to both equations. So the solution set to the equation
(tan x − 2)(2 sin x − 1) = 0
is
{arctan(2) + πk : k ∈ Z} ∪ { π6 + 2πk : k ∈ Z} ∪ { 5π
6 + 2πk : k ∈ Z}.
Some worked problems.
1. Solve sin2 x − 5 sin x + 6 = 0.
Solution. A standard algebra problem might have us solving z 2 − 5z + 6 = 0 by factoring the
quadratic polynomial into (z − 3)(z − 2) and solving (z − 3)(z − 2) = 0. Here we can do something
very similar; we factor sin2 x − 5 sin x + 6 into (sin x − 3)(sin x − 2). So
sin2 x − 5 sin x + 6 = 0 =⇒ (sin x − 3)(sin x − 2) = 0 =⇒ sin x − 3 = 0 or sin x − 2 = 0.
Now sin x − 3 = 0 =⇒ sin x = 3 and since the sine function has range [−1, 1], this equation has
no solutions. Similarly the equation sin x − 2 = 0 has no solutions. So the quadratic equation
sin2 x − 5 sin x + 6 = 0 has NO solutions .
2. Solve 2 sin2 x − 3 sin x + 1 = 0.
Solution. We factor 2 sin2 x − 3 sin x + 1 = (2 sin x − 1)(sin x − 1). So
2 sin2 x − 3 sin x + 1 = 0 =⇒ (2 sin x − 1)(sin x − 1) = 0
=⇒ 2 sin x − 1 = 0 or sin x − 1 = 0 =⇒ sin x =
1
or sin x = 1.
2
In the first case (sin x = 12 ) our solution set is { π6 + 2πk : k ∈ Z} ∪ { 5π
6 + 2πk : k ∈ Z}. In the second
case (sin x = 1) our solution set is { π2 + 2πk : k ∈ Z}. So our final solution is
π
{ π6 + 2πk : k ∈ Z} ∪ { 5π
6 + 2πk : k ∈ Z} ∪ { 2 + 2πk : k ∈ Z} .
Homework.
As class homework, please complete Worksheet 4.8, Trig Identities and Equations available
through the class webpage.
193