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Trigonometric functions are often used to perform calculations in real-life fields, such as architecture, optics, and trajectories. In many cases, multiple trigonometric functions appear in one expression, potentially causing confusion. Fortunately, rules that relate different trigonometric functions exist, providing the ability to simplify calculations. This lesson will present some of them. ### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Challenge

Best friends Paulina and Maya graduated high school last Friday. At the graduation party, they, along with the rest of the graduates and guests, were admiring bright and colorful fireworks.
Their physics teacher came up with a contest to play for a prize — three tickets to an Ali Styles concert! He said that the height of the firework $h$ and horizontal displacement $x$ were related by the following equation.

External credits: @pch.vector

$h=2v_{2}cos_{2}θ-gx_{2} +cosθxsinθ $

Here, $v$ is the initial velocity of the projectile, $θ$ is the angle at which it was fired, and $g$ is the acceleration due to gravity. The first student, who can rewrite the equation so that $tanθ$ is the only trigonometric function, will win the tickets. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["v","THETA","g"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\">h<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["\\dfrac{- gx^2}{2v^2}(1+\\tan^2(\\theta))+x\\tan(\\theta)","-\\dfrac{gx^2}{2v^2}(1+\\tan^2(\\theta))+x\\tan(\\theta)","\\dfrac{-gx^2}{2v^2}+\\dfrac{-gx^2}{2v^2}\\tan^2(\\theta)+x\\tan(\\theta)","-\\dfrac{gx^2}{2v^2}-\\dfrac{gx^2}{2v^2}\\tan^2(\\theta)+x\\tan(\\theta)"]}}

Discussion

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.

Rule

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 $ℓ.$

At the same time, the sine and cosine of $θ$ can be written as follows.$sinθ=mk cosθ=mℓ $

By manipulating the right-hand side of the equation $tanθ=ℓk ,$ the tangent can be expressed as the sine over the cosine of $θ.$
$tanθ=ℓk $

ReduceFrac

$ba =b/ma/m $

$tanθ=ℓ/mk/m $

SubstituteII

$k/m=sin(θ)$, $ℓ/m=cos(θ)$

$tanθ=cosθsinθ $

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 $cosθ.$

As shown previously, the tangent of a right triangle is defined as the ratio of the length of the opposite side — in this case, $sinθ$ — to the length of the adjacent side, which here is $cosθ.$$tanθ=cosθsinθ ✓ $

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 $k.$

Additionally, the sine and cosine of $θ$ can be written as follows.$sinθ=mk cosθ=mℓ $

By manipulating the right-hand side of the equation $cotθ=kℓ ,$ the cotangent can be expressed as the cosine over the sine of $θ.$
$cotθ=kℓ $

ReduceFrac

$ba =b/ma/m $

$cotθ=k/mℓ/m $

SubstituteII

$ℓ/m=cos(θ)$, $k/m=sin(θ)$

$cotθ=sinθcosθ $

$tanθ=cosθsinθ $

By manipulating the above equation, it can be shown that the cotangent of $θ$ is the ratio of the cosine of $θ$ to the sine of $θ.$
$tanθ=cosθsinθ $

MultEqn

$LHS⋅sinθcosθ =RHS⋅sinθcosθ $

$tanθ(sinθcosθ )=1$

DivEqn

$LHS/tanθ=RHS/tanθ$

$sinθcosθ =tanθ1 $

$cot(θ)=tan(θ)1 $

$sinθcosθ =cotθ$

RearrangeEqn

Rearrange equation

$cotθ=sinθcosθ ✓$

There are also trigonometric identities which show that some trigonometric functions are reciprocals of others.

Rule

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 $θ.$

Next, the sine, cosine, tangent, cosecant, secant, and cotangent ratios are written.$sinθ=hypopp cosθ=hypadj tanθ=adjopp cscθ=opphyp secθ=adjhyp cotθ=oppadj $

The reciprocal of the sine ratio will now be calculated.
$sinθ=hypopp $

▼

Solve for $sinθ1 $

MultEqn

$LHS⋅sinθ1 =RHS⋅sinθ1 $

$1=hypopp (sinθ1 )$

MultEqn

$LHS⋅opphyp =RHS⋅opphyp $

$opphyp =sinθ1 $

RearrangeEqn

Rearrange equation

$sinθ1 =opphyp $

$⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧ sinθ1 =opphyp .cscθ=opphyp ⇓cscθ=sinθ1 $

By following a similar procedure, the other two identities for secant and cotangent can be proven.
Example

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.

a $tan_{2}θsin_{2}θ $

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b $cotθsecθ$

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c $cscθtanθ+3secθ$

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a Recall the Tangent Identity.

b Use the Cotangent Identity and the Reciprocal Identity for secant.

a The first expression contains two different trigonometric functions — sine and tangent. This means the Tangent Identity could be useful here.

$tanθ=cosθsinθ $

Substitute $cosθsinθ $ for $tanθ$ into the expression and simplify.
$tan_{2}θsin_{2}θ $

Substitute

$tanθ=cosθsinθ $

$(cosθsinθ )_{2}sin_{2}θ $

PowQuot

$(ba )_{m}=b_{m}a_{m} $

$cos_{2}θsin_{2}θ sin_{2}θ $

DivByFracD

$b/ca =ba⋅c $

$sin_{2}θsin_{2}θ⋅cos_{2}θ $

▼

Reduce by $sin_{2}θ$

CrossCommonFac

Cross out common factors

$sin_{2}θsin_{2}θ⋅cos_{2}θ $

CancelCommonFac

Cancel out common factors

$11⋅cos_{2}θ $

OneMult

$1⋅a=a$

$1cos_{2}θ $

DivByOne

$1a =a$

$cos_{2}θ$

b To simplify the second expression, recall the Cotangent Identity and the Reciprocal Identity for secant.

$cotθ=sinθcosθ secθ=cosθ1 $

Next, substitute the expressions for $cotθ$ and $secθ$ and simplify.
$cotθsecθ$

SubstituteII

$cotθ=sinθcosθ $, $secθ=cosθ1 $

$sinθcosθ ⋅cosθ1 $

MultFrac

Multiply fractions

$sinθ⋅cosθcosθ⋅1 $

ReduceFrac

$ba =b/cosθa/cosθ $

$sinθ⋅11⋅1 $

MultByOne

$a⋅1=a$

$sinθ1 $

$sinθ1 =cscθ $

Therefore, the last expression simplifies to $cscθ.$
$cotθsecθ=cscθ $

c The last expression is the sum of two terms.

$cscθtanθ+3secθ $

Begin by simplifying the $first term.$ In order to do this, review the Reciprocal Identity for cosecant and the Tangent Identity.
$cscθ=sinθ1 tanθ=cosθsinθ $

Next, use these identities to find a simpler form of the first term.
$cscθtanθ$

SubstituteII

$cscθ=sinθ1 $, $tanθ=cosθsinθ $

$sinθ1 ⋅cosθsinθ $

MultFrac

Multiply fractions

$sinθcosθsinθ $

CrossCommonFac

Cross out common factors

$sinθcosθsinθ $

CancelCommonFac

Cancel out common factors

$cosθ1 $

$cosθ1 =secθ $

This means that the first term can be simplified to $secθ.$
Therefore, $4secθ$ is the most simplified form of the given expression.
Discussion

Some techniques, like the following, are helpful when verifying if trigonometric identities are true.

- Substitute one or more basic trigonometric identities to simplify the expression.
- Factor or multiply as necessary. Sometimes it is necessary to multiply both the numerator and denominator by the same trigonometric expression.
- Write each side of the identity in terms of sine and cosine only. Then simplify each side as much as possible.

Remember that the properties of equality do not apply to identities as they do with equations. It is not possible to perform operations to the quantities on each side of an unverified identity.

Discussion

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.

Rule

For any angle $θ,$ the following trigonometric identities hold true.

$sin_{2}θ+cos_{2}θ=1$

$1+tan_{2}θ=sec_{2}θ$

$1+cot_{2}θ=csc_{2}θ$

Consider a right triangle with a hypotenuse of $1.$

Since $cosθ$ represents a side length, it is **not** $0.$ Therefore, by diving both sides of the above equation by $cos_{2}θ,$ the second identity can be obtained.
The second identity was obtained.
Since $sinθ$ represents a side length, it is **not** $0.$ Therefore, by dividing both sides of $sin_{2}θ+cos_{2}θ=1$ by $sin_{2}θ,$ the third identity can be proven.
Finally, the third identity was obtained.

By recalling the sine and cosine ratios, the lengths of the opposite and adjacent sides to $∠θ$ can be expressed in terms of the angle.

Definition | Substitute | Simplify | |
---|---|---|---|

$sinθ$ | $HypotenuseLength ofoppositeside to∠θ $ | $1opp $ | $opp$ |

$cosθ$ | $HypotenuseLength ofadjacentside 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 $1.$

$sin_{2}θ+cos_{2}θ⇓sin_{2}θ+cos_{2}θ =1_{2}=1 $

$sin_{2}θ+cos_{2}θ=1$

DivEqn

$LHS/cos_{2}θ=RHS/cos_{2}θ$

$cos_{2}θsin_{2}θ+cos_{2}θ =cos_{2}θ1 $

▼

Simplify

WriteSumFrac

Write as a sum of fractions

$cos_{2}θsin_{2}θ +cos_{2}θcos_{2}θ =cos_{2}θ1 $

QuotOne

$aa =1$

$cos_{2}θsin_{2}θ +1=cos_{2}θ1 $

WritePow

Write as a power

$cos_{2}θsin_{2}θ +1=cos_{2}θ1_{2} $

$b_{m}a_{m} =(ba )_{m}$

$(cosθsinθ )_{2}+1=(cosθ1 )_{2}$

$cosθsinθ =tanθ$

$tan_{2}θ+1=(cosθ1 )_{2}$

$cosθ1 =secθ$

$tan_{2}θ+1=sec_{2}θ$

CommutativePropAdd

Commutative Property of Addition

$1+tan_{2}θ=sec_{2}θ$

$1+tan_{2}θ=sec_{2}θ$

$sin_{2}θ+cos_{2}θ=1$

DivEqn

$LHS/sin_{2}θ=RHS/sin_{2}θ$

$sin_{2}θsin_{2}θ+cos_{2}θ =sin_{2}θ1 $

▼

Simplify

WriteSumFrac

Write as a sum of fractions

$sin_{2}θsin_{2}θ +sin_{2}θcos_{2}θ =sin_{2}θ1 $

QuotOne

$aa =1$

$1+sin_{2}θcos_{2}θ =sin_{2}θ1 $

WritePow

Write as a power

$1+sin_{2}θcos_{2}θ =sin_{2}θ1_{2} $

$b_{m}a_{m} =(ba )_{m}$

$1+(sinθcosθ )_{2}=(sinθ1 )_{2}$

$sinθcosθ =cotθ$

$1+cot_{2}θ=(sinθ1 )_{2}$

$sinθ1 =cscθ$

$1+cot_{2}θ=csc_{2}θ$

$1+cot_{2}θ=csc_{2}θ$

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 $θ.$

By the Pythagorean Theorem, the sum of the squares of $x$ and $y$ equals $1.$$x_{2}+y_{2}=1 $

In fact, this is true not only for points in the first quadrant, but for $cos_{2}θ+sin_{2}θ=1$

Example

Enjoying the graduation party, Maya and Paulina shared some memories about some of the fun they had exploring their neighborhood as kids. After school they would buy junk food at the $24-$hour store, bird watch in the trees of the park, and play around the lake.
The sine of the angle $θ$ formed by the road connecting Paulina's and Maya's houses to the store is $54 .$ The cosine of the angle $β$ formed by the road connecting Paulina's and Maya's houses to the school is $-87 .$ Interact with the map to view these angles.
### Hint

### Solution

External credits: Hari Panicker

a What is the cosine of $θ?$

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b What is the tangent of $θ?$

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c What is the sine of $β?$

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d What is the cotangent of $β?$

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a Apply the Pythagorean Identity.

b Recall the Tangent Identity.

c To determine the sign of the sine, identify in which quadrant $β$ is located.

d Use the Cotangent Identity.

a It is given that the sine of $θ$ is $54 .$ To find the value of the cosine of $θ,$ the Pythagorean Identity can be used.

$sin_{2}θ+cos_{2}θ=1 $

Substitute the known value of $sinθ$ and solve for $cosθ.$
To determine the sign of $cosθ,$ identify in which quadrant the angle is situated. External credits: Hari Panicker

The angle is in the first quadrant, where cosine is positive. Therefore, it can be concluded that the cosine of $θ$ is $53 .$

b To calculate the tangent of $θ,$ the Tangent Identity can be used.

$tanθ=cosθsinθ $

Since both $sinθ$ and $cosθ$ are known, substitute their values and solve for $tanθ.$
$tanθ=cosθsinθ $

SubstituteII

$sinθ=54 $, $cosθ=53 $

$tanθ=53 54 $

▼

Simplify right-hand side

DivFracByFracSL

$ba /dc =ba ⋅cd $

$tanθ=54 ⋅35 $

MultFrac

Multiply fractions

$tanθ=5⋅34⋅5 $

CrossCommonFac

Cross out common factors

$tanθ=5 ⋅34⋅5 $

CancelCommonFac

Cancel out common factors

$tanθ=34 $

c This time the sine of the angle is known, while the cosine should be found. Again, the Pythagorean Theorem can be used. Substitute $-87 $ for $cosβ$ and solve for $sinβ.$