all Elementary Functions Trig Identities

Elementary FunctionsPart 4, Trigonometry

Lecture 4.8a, Trig Identities and Equations

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 21

Trig Identities

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 + y2.

Other examples of identities are:

1 (x+ 3)2 = x2 + 6x+ 9and

2 (a very important one!) A2 −B2 = (A−B)(A+B).


Trig identities vs. trig equations

What is a trig identity? A trig identity is an equation which is true for allinputs (such as angles, θ.)

For example, from the Pythagorean theorem on the unit circle, we knowthat the equation for the unit circle is x2 + y2 = 1 and so this turns intoan identity for trig functions:

(cos θ)2 + (sin θ)2 = 1

cos2 θ + sin2 θ = 1

This is true regardless of the choice of θ.

Other examples of trig identities are:

1 tan θ = sin θcos θ

2 sin(−x) = − sinx3 cos(z) = sin(z + π/2).


Central Angles and Arcs

Some of our trig identities come from our definitions. For example, fromthe definition of the tangent function we know that

tan θ =sin θ

cos θ

We also have some identities given by symmetry. For example, since thesine function is odd then

sin(−x)− sin(x);

Since cosine is even then

cos(−x) = cosx.

By looking at the graphs of sine and cosine we observed that

cosx = sin(x+ π/2).

A trig equation, unlike an identity, may not necessarily be true for allangles θ. In general, with a trig equation, we wish to solve for θ.


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

then it is not sufficient to just list x = 1 as a solution. To find all solutionswe might factor the quadratic x2 − 3x+ 2 = (x− 2)(x− 1) and then setthis equal to zero:

(x− 2)(x− 1) = 0

This equation implies that either x− 2 = 0 (so x = 2) or x− 1 = 0 (sox = 1.) We have found two solutions.

From our understanding about zeroes of a polynomial, we now know thatwe have found all the solutions.Once reason for the concept of factoring is that it aids us in finding allsolutions!



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 findboth the solutions x = 2 and x = −2.

We often remind ourselves of the possibilities of two solutions by writing aplus-or-minus symbol (±) as in the computationx2 = 4 =⇒ x = ±

√4 = ±2.



For many trig problems, we will find solutions within one revolution of theunit circle and then use the period of the trig functions to find an infinitenumber of solutions.

For example, let’s solve the equation

sin θ =

√3

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) =√32 . 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) =√32 . So θ = 2π/3 is a nice solution in the second quadrant.



sin θ =√32

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 sinefunction is periodic with period 2π we can add 2π to any angle withoutchanging 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 = π3 + 2πk or θ = 2π

3 + 2πk

{π3 + 2πk : k ∈ Z} ∪ {2π3 + 2πk : k ∈ Z} .



Sometimes inverse trig functions come in handy. For example, suppose weare solving the equation

tanx = 2.

One solution is x = arctan(2) which is approximately 1.10715. (There isno simple way to write this angle; we need the arctangent function.) Sincetangent has period π and there is only one solution within an interval oflength π, we know that the full set of solutions is

{arctan(2) + πk : k ∈ Z}.


More worked problems.

Solve the equation sin 4θ =√32 .

Solution. If sin 4θ =√32 then from the notes above, we know that

4θ =π

3+ 2πk

or

4θ =2π

3+ 2πk.

So divide both sides by 4 to get

θ = π12 + π

2k

orθ = π

6 + π2k

(where k ∈ Z.)Note that since sin θ has period 2π then sin 4θ must have period 2π

4 = π2 .

This is reflected in our solutions by our adding multiples of π2 to our first

solutions, π/12 and π/6.Smith (SHSU) Elementary Functions 2013 10 / 21


Solve sinx = 12 .

Solution. The equation sinx = 12 has two solutions within one revolution

of the unit circle.

They are x = π6 and x = 5π

6 .

Since sinx has period 2π we know that the collection of all solutions mustbe

x = π6 + 2πk and x = 5π

6 + 2πk.

We can write this in set notation as

{π6 + 2πk : k ∈ Z} ∪ {5π6 + 2πk : k ∈ Z}.



Solve 2 sin(x− π4 ) + 1 = 2.

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.

2 sin(x− π

4) + 1 = 2 =⇒ 2 sin(θ − π

4) = 1 =⇒ sin(θ − π

4) =

1

2.

We know, from the previous problem, that sin(π6 ) =12 and sin(5π6 ) = 1

2and so

θ − π

4=π

6+ 2πk or θ − π

4=

5π

6+ 2πk.

Now we can solve for θ by adding π/4 to both sides of these equations,obtaining solutions

θ =π

4+π

6+ 2πk or θ =

π

4+

5π

6+ 2πk.

The best answer is achieved by getting a common denominator forπ/4 + π/6:

θ = 5π12 + 2πk or θ = 13π

12 + 2πk.



Solve cosx =√32 .

Solution. The equation cosx =√32 has two solutions within one

revolution of the unit circle. Since the cosine function is even, we knowthat when we find one positive solution, the negative solution will alsowork.

So our two solutions are x = π6 and x = −π

6 .

Since cosx has period 2π we know that the collection of all solutions mustbe

{π6 + 2πk : k ∈ Z} ∪ {−π6 + 2πk : k ∈ Z}.



Solve the equation tan2 θ = 3.

Solution. tan2 θ = 3 =⇒ tan θ = ±√3. If tan θ =

√3 then

θ = π3 + 2πk (where k ∈ Z.)

If tan θ = −√3 then θ = 2π

3 + 2πk .

So the final answer is

θ = π3 + 2πk or θ = 2π

3 + 2πk .


Trig Equations

In the next presentation, we will look further at trig equations.

(End)


Elementary FunctionsPart 4, Trigonometry

Lecture 4.8b, Trig Identities and Equations

Dr. Ken W. Smith

Sam Houston State University

2013


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 orB = 0.

If, for example, we need to solve the equation

(tanx− 2)(2 sinx− 1) = 0

then we note that this implies that either tanx− 2 = 0 or sinx− 12 = 0.

These then imply tanx = 2 or sinx = 12 .

In an earlier problem we solved the equation tanx = 2 and got thesolution set

{arctan(2) + πk : k ∈ Z}.In a different problem we solved sinx = 1

2 and found the solution set

{π6+ 2πk : k ∈ Z} ∪ {5π

6+ 2πk : k ∈ Z}.


Some worked problems.

In this case, we want all solutions to both equations. So the solution setto the equation

(tanx− 2)(2 sinx− 1) = 0

is

{arctan(2) + πk : k ∈ Z} ∪ {π6 + 2πk : k ∈ Z} ∪ {5π6 + 2πk : k ∈ Z}.



Solve sin2 x− 5 sinx+ 6 = 0.

Solution. A standard algebra problem might have us solvingz2 − 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 sinx+ 6 into(sinx− 3)(sinx− 2).

Sosin2 x− 5 sinx+ 6 = 0 =⇒ (sinx− 3)(sinx− 2) = 0

=⇒ sinx− 3 = 0 or sinx− 2 = 0.

Now sinx− 3 = 0 =⇒ sinx = 3 and since the sine function has range[−1, 1], this equation has no solutions. Similarly the equationsinx− 2 = 0 has no solutions. So the quadratic equationsin2 x− 5 sinx+ 6 = 0 has NO solutions .



Solve 2 sin2 x− 3 sinx+ 1 = 0.

Solution. We factor 2 sin2 x− 3 sinx+ 1 = (2 sinx− 1)(sinx− 1). So

2 sin2 x− 3 sinx+ 1 = 0 =⇒ (2 sinx− 1)(sinx− 1) = 0

=⇒ 2 sinx− 1 = 0 or sinx− 1 = 0 =⇒ sinx =1

2or sinx = 1.

In the first case (sinx = 12) our solution set is

{π6+ 2πk : k ∈ Z} ∪ {5π

6+ 2πk : k ∈ Z}.

In the second case (sinx = 1) our solution set is

{π2+ 2πk : k ∈ Z}.

So our final solution is

{π6 + 2πk : k ∈ Z} ∪ {5π6 + 2πk : k ∈ Z} ∪ {π2 + 2πk : k ∈ Z} .


Some worked problems

In the next presentation, we will look further at trig identities andequations.

(End)


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