Sign In
| 13 Theory slides |
| 9 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Consider the two given equalities. First, analyze if any of them can be simplified. Then, focus on how many values of x make each of them true. What is the main difference between the equalities?
The first equality is an equation that can be solved to find the few, if any, values of x that make the equation true. However, the second equality can be simplified to x=x and, therefore, is an equation that is true for all values of x for which the expressions in the equation are defined. Given that characteristic, the second equality is called an identity.
A trigonometric identity is an equation involving trigonometric functions that is true for all values for which every expression in the equation is defined.
Two of the most basic trigonometric identities are tangent and cotangent Identities. These identities relate tangent and cotangent to sine and cosine.
The tangent of an angle θ can be expressed as the ratio of the sine of θ to the cosine of θ.
tanθ=cosθsinθ
Similarly, the cotangent of θ can be expressed as the ratio of the cosine of θ to the sine of θ.
cotθ=sinθcosθ
Two proofs will be written for this identity, one using a right triangle and the other using a unit circle.
In a right triangle, the tangent of an angle θ is defined as the ratio of the length of the opposite side k to the length of the adjacent side ℓ.
Consider a unit circle and an angle θ in standard position.
It is known that the point of intersection P of the terminal side of the angle and the unit circle has coordinates (cosθ,sinθ).
Draw a right triangle using the origin and P(cosθ,sinθ) as two of its vertices. The length of the hypotenuse is 1 and the lengths of the legs are sinθ and cosθ.
Two more proofs will be written for this identity, one of them using just a right triangle and the other using a unit circle.
In a right triangle, the cotangent of an angle θ is defined as the ratio of the length of the adjacent side ℓ to the length of the opposite side k.
LHS⋅sinθcosθ=RHS⋅sinθcosθ
LHS/tanθ=RHS/tanθ
cot(θ)=tan(θ)1
Rearrange equation
There are also trigonometric identities which show that some trigonometric functions are reciprocals of others.
The trigonometric ratios cosecant, secant, and cotangent are reciprocals of sine, cosine, and tangent, respectively.
cscθ=sinθ1
secθ=cosθ1
cotθ=tanθ1
Consider a right triangle with the three sides labeled with respect to an acute angle θ.
LHS⋅sinθ1=RHS⋅sinθ1
LHS⋅opphyp=RHS⋅opphyp
Rearrange equation
Thinking of different ways to solve the firework challenge, Paulina found herself thinking about her time learning trigonometric identities earlier in the school year. She really enjoyed those lessons.
A few of her favorite exercises included the following where she was asked to simplify these expressions.
tanθ=cosθsinθ
(ba)m=bmam
b/ca=ba⋅c
Cross out common factors
Cancel out common factors
1⋅a=a
1a=a
cotθ=sinθcosθ, secθ=cosθ1
Multiply fractions
ba=b/cosθa/cosθ
a⋅1=a
cscθ=sinθ1, tanθ=cosθsinθ
Multiply fractions
Cross out common factors
Cancel out common factors
Some techniques, like the following, are helpful when verifying if trigonometric identities are true.
One of the most known trigonometric identities relates the square of sine and cosine of any angle θ. This identity can be manipulated to obtain two more identities involving other trigonometric functions.
For any angle θ, the following trigonometric identities hold true.
Definition | Substitute | Simplify | |
---|---|---|---|
sinθ | HypotenuseLength of opposite side to ∠θ | 1opp | opp |
cosθ | HypotenuseLength of adjacent side to ∠θ | 1adj | adj |
It can be seen that if the hypotenuse of a right triangle is 1, the sine of an acute angle is equal to the length of its opposite side. Similarly, the cosine of the angle is equal to the length of its adjacent side.
By the Pythagorean Theorem, the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Therefore, for the above triangle, the sum of the squares of sinθ and cosθ is equal to the square of 1.
LHS/cos2θ=RHS/cos2θ
Write as a sum of fractions
aa=1
Write as a power
bmam=(ba)m
cosθsinθ=tanθ
cosθ1=secθ
Commutative Property of Addition
LHS/sin2θ=RHS/sin2θ
Write as a sum of fractions
aa=1
Write as a power
bmam=(ba)m
sinθcosθ=cotθ
sinθ1=cscθ
1+cot2θ=csc2θ
The first identity can be shown using the unit circle and the Pythagorean Theorem. Consider a point (x,y) on the unit circle in the first quadrant, corresponding to the angle θ. A right triangle can be constructed with θ.
The angle is in the first quadrant, where cosine is positive. Therefore, it can be concluded that the cosine of θ is 53.
sinθ=54, cosθ=53
ba/dc=ba⋅cd
Multiply fractions
Cross out common factors
Cancel out common factors
The angle is in the second quadrant, where sine is positive. Therefore, the sine of β is 815.
cosβ=-87, sinβ=815
ba/dc=ba⋅cd
Multiply fractions
Cross out common factors
Cancel out common factors
Given the value of a trigonometric function and the quadrant of the angle, use one of the Pythagorean Identities to find the value of another trigonometric function. Round the answer to two decimal places.
Maya told Jordan, who was late for the graduation party, all about the contest and asked her to recall the trigonometric identities they learned on her way over. While thinking about the identities, Jordan remembered that, when they studied the topic, Maya sent her one interesting identity and decided to find that text.
Use the known trigonometric identities to rewrite the left- and right-hand sides of the equation until they match.
secθ=cosθ1, cscθ=sinθ1
Multiply fractions
tanθ=cosθsinθ, cotθ=sinθcosθ
ba=b⋅sinθa⋅sinθ
ba=b⋅cosθa⋅cosθ
Add fractions
Studying angles and their trigonometric values in a right triangle more closely reveals further relationships between sine and cosine, and between tangent and cotangent.
For any angle θ, the following trigonometric identities hold true.
sin(2π−θ)=cosθ
cos(2π−θ)=sinθ
tan(2π−θ)=cotθ
Consider a right triangle. The measure of its right angle is 90∘ or 2π radians. Let θ be the radian measure of one of the acute angles. Since the sum of two acute angles in a right triangle is 2π, the measure of the third acute angle must be 2π−θ.
sin(2π−θ)=cosθ
This identity is true for all angles, not just those that make it possible to construct a right triangle. Using similar reasoning, the corresponding identities for cosine and tangent can be proven.
When dealing with negative angles, identities that relate trigonometric values of negative and positive angles become very useful.
The function y=sin(x) has odd symmetry, and y=cos(x) has even symmetry, which can be seen from their graphs. As a result, the corresponding identities hold true.
sin(-θ)=-sinθcos(-θ)=cosθtan(-θ)=-tanθ
The identities will be proven using a unit circle. Consider an arbitrary ∠θ on a unit circle. Let A be the point that the angle forms on the circle.
Recall that the values on the x-axis are represented by cosine and the value on the y-axis are represented by sine. Therefore, the coordinates of A are (cosθ,sinθ).
Additionally, after the reflection, the angle between A′, the origin, and the x-axis is -θ. This means that the lengths of the newly created horizontal and vertical segments are cos(-θ) and sin(-θ). Therefore, the coordinates of A′ can also be written as (cos(-θ),sin(-θ)).
tan(θ)=cos(θ)sin(θ)
sin(-θ)=-sin(θ)
cos(-θ)=cos(θ)
Put minus sign in front of fraction
cos(θ)sin(θ)=tan(θ)
Jordan, still on the bus ride to the graduation party, feels excited about getting to solve a trig problem to win tickets to see Ali Styles live. She still has a bit of time before arriving so she used her tablet to look up a 10-minute online quiz as practice.
Here are the expressions that Jordan is asked to simplify. What answers should be submitted to score a perfect 100%?
cos(θ)=sin(90∘−θ)
sin(θ)=cos(90∘−θ)
Subtract term
A−A=0
sin(-θ)=-sin(θ)
cos(-θ)=cos(θ)
Put minus sign in front of fraction
tan(θ)=cos(θ)sin(θ)
cos(2π−θ)=sin(θ)
csc(θ)=sin(θ)1
b/ca=ba⋅c
a⋅a=a2
1a=a
cot(θ)=tan(θ)1
Multiply fractions
tan(-θ)=-tan(θ)
tan(2π−θ)=cotθ
Write as a product of fractions
Put minus sign in front of fraction
Start by applying the Cofunction and Negative Angle Identities for cosine. Then find by which factor the fraction should be expanded so that the Pythagorean Identity could be used.
cos(90∘−θ)=sin(θ)
cos(-θ)=cos(θ)
ba=b⋅(1+cosθ)a⋅(1+cosθ)
(a−b)(a+b)=a2−b2
1a=1
1−cos2(θ)=sin2(θ)
ba=b/sinθa/sinθ
Distribute 3
Write as a sum of fractions
ba=a⋅b1
csc(θ)=sin(θ)1
ca⋅b=a⋅cb
cot(θ)=sin(θ)cos(θ)
Factor out 3
Use two of the Reciprocal Identities for tangent and secant. Then apply one of the Pythagorean Identities.
ca⋅b=a⋅cb
cosθsinθ=tanθ
Write as a product of fractions
cos2θ1=sec2θ
Write the exact value of each expression.
We want to find the exact value of cot θ given that tan θ = 3. To do so, we will use one of the Reciprocal Identities. cot θ = 1/tan θ, tan θ ≠ 0 Let's do it!
Now we want to find the exact value of sin θ given that cos θ = 14. To do so, we will use one of the Pythagorean Identities. sin^2 θ +cos^2 θ =1 Let's do it!
We found the value of sin θ, but now we need to determine the correct sign of the value. We are told that θ lies between 0^(∘) and 90^(∘). Therefore, θ is in Quadrant I.
In this quadrant, the sine of θ is positive. Therefore, we will only keep the positive solution. sin θ =sqrt(15)/4
Similar to Part B, we can find the exact value of cos θ given that sin θ = 35. Let's use the Pythagorean Identity again. sin^2 θ +cos^2 θ =1 Let's substitute the known value of sine and calculate the corresponding value of cosine.
Now we need to determine the sign of cos θ. We are told that θ lies between 90^(∘) and 180^(∘). Therefore, θ is in Quadrant II.
In this quadrant, the cosine of θ is negative. Therefore, we will only keep the negative solution. cos θ =- 4/5
Finally, we are given that cosθ=- 27 and are asked to find the value of csc θ. Let's start by recalling one of the Reciprocal Identities that involves cosecant. cscθ=1/sinθ As we can see, there is a relationship between cscθ and sinθ. To use this relationship, we first need to find the value of sinθ. Let's use the same Pythagorean Identity as before that relates sine and cosine. sin^2 θ +cos^2 θ =1 Let's substitute the known value of cosθ and solve the equation for sinθ.
Now let's determine the sign of sin θ. We are told that θ lies between 270^(∘) and 360^(∘). Therefore, θ is in Quadrant IV.
In this quadrant, the sine of θ is negative, so we will only keep the negative solution. sin θ =- 3sqrt(5)/7 Now that we know the value of sin θ, we can find the value of cscθ using the Reciprocal Identity that we wrote earlier.
Finally, we can rationalize the denominator by multiplying both the numerator and the denominator by sqrt(5). Let's do it!
Simplify the given expressions.
We need to simplify the given trigonometric expression. sec^2 θ - tan^2 θ We can do it using one of the Pythagorean Identities. Let's start by recalling it! 1+tan^2 θ = sec^2 θ With this identity, we can substitute 1+tan^2 θ for sec^2 θ in our expression and simplify it.
Therefore, the given expression simplifies to 1. sec ^2 θ - tan ^2 θ = 1
We are asked to simplify the following trigonometric expression. sin θ sec θ/cot θ We can do it by using the Reciprocal Identities and the Tangent Identity. Let's start by recalling two of the Reciprocal Identities. sec θ = 1/cos θ, cos θ ≠ 0 [1.1em] tan θ = 1/cot θ, cot θ ≠ 0 Since we have cotθ in our expression, let's rewrite the second identity to express cotθ in terms of tanθ. tan θ = 1/cot θ ⇕ cot θ = 1/tan θ Now, we can substitute 1cos θ for sec θ and 1tan θ for cot θ in our expression. Then we will simplify it.
Finally, let's use the Tangent Identity. tan θ = sin θ/cos θ, cos θ ≠ 0 With this identity, we can substitute tan θ for sin θcos θ in our expression and simplify it.
We have simplified the given expression. sin θ sec θ/cot θ = tan ^2 θ
Determine whether each equation is an identity.
We are asked to check if the given equation is a trigonometric identity. sinθtanθ=1/cosθ-cosθ Let's start on the left-hand side and use Trigonometric Identities to rewrite it. If we obtain the expression on the right-hand side, then the equation is an identity. Let's recall the Tangent Identity. tanθ = sin θ/cos θ With this identity, we can substitute sin θcos θ for tan θ into the expression on the left-hand side.
Now we will use one of the Pythagorean Identities. Note that we need to rearrange the terms so that we can use it for our expression. sin^2 θ + cos^2 θ = 1 ⇕ sin^2 θ = 1 - cos^2 θ Let's substitute 1 - cos^2 θ for sin^2 θ in our expression.
We started on the left-hand side of the equation and used Trigonometric Identities to arrive at the right-hand side. Therefore, the given equation is an identity.
We want to determine whether the given equation is a trigonometric identity. sec θ - cos θ = tan θ sin θ Similarly to Part A, we will start on the left-hand side and use Trigonometric Identities to try to arrive at the right-hand side. First, let's recall one of the Reciprocal Identities. sec θ = 1/cos θ With this identity, we can substitute 1cos θ for sec θ in the left-hand side.
Now we will use one of the Pythagorean Identities. Note that we need to rearrange the terms so that we can use it for our expression. sin^2 θ + cos^2 θ = 1 ⇕ sin^2 θ = 1 - cos^2 θ Let's substitute sin^2 θ for 1 - cos^2 θ in our expression.
Finally, let's apply the Tangent Identity. tan θ = sinθ/cos θ With this identity, we can substitute tan θ for sin θcosθ in our expression.
We started on the left-hand side of the equation and by using Trigonometric Identities obtained the expression on the right-hand side. Therefore, the given equation is always true and, by definition, is an identity. sec θ - cos θ = tan θ sin θ ✓
Write the first trigonometric expression in terms of the second.
We can start by recalling the Pythagorean Identity that relates cotangent and cosecant. 1 + cot^2 θ = csc^2 θ Let's rewrite this identity until cotθ is isolated on the left-hand side.
We have expressed cotθ in terms of csc θ.
We will use some of the Trigonometric Identities to rewrite tangent in terms of cosine. First, let's recall the Tangent Identity. tan θ = sin θ/cos θ Now we will use one of the Pythagorean Identities. Note that we will need to rearrange the terms so that we can use it in the expression we have in the right-hand side of the above identity. sin^2 θ + cos^2 θ = 1 ⇕ sin θ =± sqrt(1 - cos^2 θ) Now we can substitute ± sqrt(1 - cos^2 θ) for sin θ in tan θ= sin θcos θ.
We have written tan θ in terms of cos θ. tan θ = ± sqrt(1 - cos^2 θ)/cos θ