Trig Identities An identity is an equation that is true for all - - PowerPoint PPT Presentation

Elementary Functions

Part 4, Trigonometry Lecture 4.8a, Trig Identities and Equations

• Dr. Ken W. Smith

Sam Houston State University

2013

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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

• r

(x + y)2 = x2 + 2xy + y2. Other examples of identities are:

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

and

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

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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 + y2 = 1 and so this turns into an 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 θ cos θ 2 sin(−x) = − sin x 3 cos(z) = sin(z + π/2).

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Central Angles and Arcs

Some of our trig identities come from our definitions. For example, from the 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 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.

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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 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. 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

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 (±) as in the computation x2 = 4 = ⇒ x = ± √ 4 = ±2.

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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 . 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) =

√ 3 2 . 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) =

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

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Finding all solutions to a trig equation

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 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:

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Finding all solutions to a trig equation

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}.

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More worked problems.

Solve the equation sin 4θ =

√ 3 2 .

• Solution. If sin 4θ =

√ 3 2 then from the notes above, we know that

4θ = π 3 + 2πk

• r

4θ = 2π 3 + 2πk. So divide both sides by 4 to get θ = π

12 + π 2 k

• r

θ = π

6 + π 2 k

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

4 = π 2 .

This is reflected in our solutions by our adding multiples of π

2 to our first

solutions, π/12 and π/6.

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More worked problems.

Solve sin x = 1

2.

• Solution. The equation sin x = 1

2 has two solutions within one revolution

• f the unit circle.

They are x = π

6 and x = 5π 6 .

Since sin x has period 2π we know that the collection of all solutions must be x = π

6 + 2πk and x = 5π 6 + 2πk.

We can write this in set notation as {π

6 + 2πk : k ∈ Z} ∪ { 5π 6 + 2πk : k ∈ Z}.

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More worked problems.

Solve 2 sin(x − π

4 ) + 1 = 2.

• Solution. Let’s not worry about the expression x − π

4 at the beginning of

this problem. Think of x − π

4 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( π

6 ) = 1 2 and sin( 5π 6 ) = 1 2

and 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,

• btaining solutions

θ = π 4 + π 6 + 2πk or θ = π 4 + 5π 6 + 2πk. The best answer is achieved by getting a common denominator for π/4 + π/6:

5π 13π

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More worked problems.

Solve cos x =

√ 3 2 .

• Solution. The equation cos x =

√ 3 2 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 = − π 6 .

Since cos x has period 2π we know that the collection of all solutions must be {π

6 + 2πk : k ∈ Z} ∪ {− π 6 + 2πk : k ∈ Z}.

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More worked problems.

Solve the equation tan2 θ = 3.

• Solution. tan2 θ = 3 =

⇒ tan θ = ± √

• 3. If tan θ =

√ 3 then θ = π

3 + 2πk (where k ∈ Z.)

If tan θ = − √ 3 then θ = 2π

3 + 2πk .

So the final answer is θ = π

3 + 2πk or θ = 2π 3 + 2πk .

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Trig Equations

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

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Elementary Functions

Part 4, Trigonometry Lecture 4.8b, Trig Identities and Equations

• Dr. Ken W. Smith

Sam Houston State University

2013

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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 − 1

2 = 0.

These then imply tan x = 2 or sin x = 1

2.

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 6 + 2πk : k ∈ Z} ∪ {5π 6 + 2πk : k ∈ Z}.

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Some worked problems.

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} ∪ { π

6 + 2πk : k ∈ Z} ∪ { 5π 6 + 2πk : k ∈ Z}.

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Some worked problems.

Solve sin2 x − 5 sin x + 6 = 0.

• Solution. A standard algebra problem might have us solving

z2 − 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 .

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Some worked problems.

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 2 or sin x = 1. In the first case (sin x = 1

2) our solution set is

{π 6 + 2πk : k ∈ Z} ∪ {5π 6 + 2πk : k ∈ Z}. In the second case (sin x = 1) our solution set is {π 2 + 2πk : k ∈ Z}. So our final solution is { π

6 + 2πk : k ∈ Z} ∪ { 5π 6 + 2πk : k ∈ Z} ∪ { π 2 + 2πk : k ∈ Z} .

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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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