Trig Identities
An identity is an equation that is true for all values of the variables.
Examples of identities might be “obvious” results like
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Part 4, Trigonometry
Lecture 4.8a, Trig Identities and Equations
2x + 2x = 4x
or
(x + y)2 = x2 + 2xy + y 2 .
Dr. Ken W. Smith
Other examples of identities are:
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2013
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2013
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1
(x + 3)2 = x2 + 6x + 9
and
2
(a very important one!) A2 − B 2 = (A − B)(A + B).
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Elementary Functions
2013
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Trig identities vs. trig equations
Central Angles and Arcs
What is a trig identity? A trig identity is an equation which is true for all
inputs (such as angles, θ.)
Some of our trig identities come from our definitions. For example, from
the definition of the tangent function we know that
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:
sin θ
cos θ
We also have some identities given by symmetry. For example, since the
sine function is odd then
tan θ =
sin(−x) − sin(x);
(cos θ)2 + (sin θ)2 = 1
Since cosine is even then
cos2 θ + sin2 θ = 1
cos(−x) = cos x.
This is true regardless of the choice of θ.
By looking at the graphs of sine and cosine we observed that
Other examples of trig identities are:
sin θ
1 tan θ = cos θ
2 sin(−x) = − sin x
3 cos(z) = sin(z + π/2).
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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.
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2013
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Finding all solutions to a trig equation
Finding 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
Another example: suppose we want to solve the equation x2 = 4.
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 − 1 = 0 (so
x = 1.) We have found two solutions.
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
√ (±) as in the computation
2
x = 4 =⇒ x = ± 4 = ±2.
From our understanding about zeroes of a polynomial, 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!
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Finding all solutions to a trig equation
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Finding all solutions to a trig equation
√
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
√
sin θ =
3
.
2
sin θ =
3
2
We found two solutions to this equation: θ = π/3 and θ = 2π/3.
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
Since the sine here is positive then the angle θ must be in quadrants I or II.
θ = π/3 + 2πk
where we understand that k is a whole number, that is, k ∈ Z.
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.
Similarly, if θ = 2π/3 is a solution then so are
2π/3 + 2πk (k ∈ Z)
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.
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We collect these together as our solution:
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Finding all solutions to a trig equation
More worked problems.
√
Solve the equation sin 4θ =
3
2 .
√
Solution. If sin 4θ =
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
3
2
then from the notes above, we know that
π
4θ = + 2πk
3
or
4θ =
2π
+ 2πk.
3
So divide both sides by 4 to get
θ=
π
12
+ π2 k
θ=
π
6
+ π2 k
or
{arctan(2) + πk : k ∈ Z}.
(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 solutions by our adding multiples of π2 to our first
solutions, π/12 and π/6.
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Elementary Functions
2013
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More worked problems.
and x =
1
2
has two solutions within one revolution
5π
6 .
Since sin x has period 2π we know that the collection of all solutions must
be
x = π6 + 2πk and x = 5π
6 + 2πk.
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{ π6 + 2πk : k ∈ Z} ∪ { 5π
6 + 2πk : k ∈ Z}.
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Solution. Let’s not worry about the expression x − π4 at the beginning of
this problem. Think of x − π4 as an angle in its own right.
First we subtract 1 from both sides and then divide both sides by 2.
π
π
π
1
2 sin(x − ) + 1 = 2 =⇒ 2 sin(θ − ) = 1 =⇒ sin(θ − ) = .
4
4
4
2
1
π
5π
We know, from the previous problem, that sin( 6 ) = 2 and sin( 6 ) = 12
and so
π
π
5π
π
θ − = + 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
π π
π 5π
θ = + + 2πk or θ = +
+ 2πk.
4
6
4
6
We can write this in set notation as
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Solve 2 sin(x − π4 ) + 1 = 2.
Solution. The equation sin x =
of the unit circle.
π
6
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More worked problems.
Solve sin x = 12 .
They are x =
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The best answer is achieved by getting a common denominator for
π/4 +Smith
π/6:(SHSU)
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2013
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More worked problems.
More worked problems.
√
Solve cos x =
3
2 .
Solve the equation tan2 θ = 3.
√
Solution. The equation cos x = 23 has two solutions within one
revolution of the unit circle. 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 =
π
6
and x =
− π6 .
√
√
Solution. tan2 θ = 3 =⇒ tan θ = ± 3. If tan θ = 3 then
θ = π3 + 2πk (where k ∈ Z.)
√
If tan θ = − 3 then θ =
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+ 2πk .
So the final answer is
Since cos x has period 2π we know that the collection of all solutions must
be
{ π6 + 2πk : k ∈ Z} ∪ {− π6 + 2πk : k ∈ Z}.
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2π
3
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θ=
Smith (SHSU)
π
3
+ 2πk or θ =
2π
3
+ 2πk .
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Trig Equations
Elementary Functions
Part 4, Trigonometry
Lecture 4.8b, Trig Identities and Equations
In the next presentation, we will look further at trig equations.
(End)
Dr. Ken W. Smith
Sam Houston State University
2013
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Elementary Functions
2013
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Some worked problems.
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 −
These then imply tan x = 2 or sin x =
1
2
= 0.
is
1
2.
{arctan(2) + πk : k ∈ Z} ∪ { π6 + 2πk : k ∈ Z} ∪ { 5π
6 + 2πk : k ∈ Z}.
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
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
and found the solution set
π
5π
+ 2πk : k ∈ Z} ∪ {
+ 2πk : k ∈ Z}.
6
6
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Some worked problems.
Some worked problems.
Solve sin2 x − 5 sin x + 6 = 0.
Solve 2 sin2 x − 3 sin x + 1 = 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.
Solution. We factor 2 sin2 x − 3 sin x + 1 = (2 sin x − 1)(sin x − 1). So
Here we can do something very similar; we factor sin2 x − 5 sin x + 6 into
(sin x − 3)(sin x − 2).
So
sin 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 .
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2013
=⇒ 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
π
5π
+ 2πk : k ∈ Z} ∪ {
+ 2πk : k ∈ Z}.
6
6
In the second case (sin x = 1) our solution set is
π
{ + 2πk : k ∈ Z}.
2
So our final solution is
{
2
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2 sin2 x − 3 sin x + 1 = 0 =⇒ (2 sin x − 1)(sin x − 1) = 0
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π
{ π6 + 2πk : k ∈ Z} ∪ { 5π
6 + 2πk : k ∈ Z} ∪ { 2 + 2πk : k ∈ Z} .
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2013
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Some worked problems
In the next presentation, we will look further at trig identities and
equations.
(End)
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