Lesson Explainer: Evaluating Trigonometric Functions Using Pythagorean Identities Mathematics

In this explainer, we will learn how to use the Pythagorean identities to find the values of trigonometric functions.

These values of trigonometric functions are often evaluated upon the application of one or more Pythagorean identity, which relate the different trigonometric and reciprocal trigonometric functions.

Trigonometric identities have several real-world applications in various fields such as physics, engineering, architecture, robotics, music theory, and navigation, to name a few. In physics, they can be used in projectile motion, modeling the mechanics of electromagnetic waves, analyzing alternating and direct currents, and finding the trajectory of a mass around a massive body under the force of gravity.

Let’s begin by recalling the trigonometric functions, whose Pythagorean identities we will examine in this explainer. Consider a right triangle:

The trigonometric functions can be expressed in terms of the ratio of the sides of the triangle as sinOHcosAHtanOAπœƒ=,πœƒ=,πœƒ=.

These functions satisfy the following trigonometric identity: tansincosπœƒ=πœƒπœƒ.

We note that these trigonometric ratios are defined for acute angles 0<πœƒ<90∘∘, and the trigonometric functions for all values of πœƒ are defined on the unit circle.

Suppose that a point moves along the unit circle in the counterclockwise direction. At a particular position (π‘₯,𝑦) on the unit circle with angle πœƒ, the sine function is defined as 𝑦=πœƒsin and the cosine function as π‘₯=πœƒcos, as shown in the diagram above. In other words, the trigonometric functions are defined by using the coordinates of the point of intersection of the unit circle with the terminal side of πœƒ in the standard position.

The reciprocal trigonometric equations are defined in terms of the standard trigonometric equations as follows.

Definition: Reciprocal Trigonometric Functions

The cosecant, secant, and cotangent functions are defined as cscsinseccoscottancossinπœƒ=1πœƒ,πœƒ=1πœƒ,πœƒ=1πœƒ=πœƒπœƒ.

Let’s also state the Pythagorean identities, which we will explore in this explainer.

Definition: Pythagorean Identities for Trigonometric Functions

The Pythagorean identities for the trigonometric equations are given by sincoscotcsctansecοŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨πœƒ+πœƒ=1,1+πœƒ=πœƒ,πœƒ+1=πœƒ.

We note that we can manipulate the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1 to derive the other identities for the reciprocal trigonometric functions. In particular, upon dividing by sinοŠ¨πœƒ, we have sincossinsinsinsincossinsincossinsincotcscοŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨πœƒ+πœƒπœƒ=1πœƒπœƒπœƒ+πœƒπœƒ=1πœƒ1+πœƒπœƒ=1πœƒ1+πœƒ=πœƒ.

Similarly, dividing by cosοŠ¨πœƒ, we can obtain the identity relating tanπœƒ and secπœƒ: sincoscoscossincoscoscoscossincoscostansecοŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨πœƒ+πœƒπœƒ=1πœƒπœƒπœƒ+πœƒπœƒ=1πœƒπœƒπœƒ+1=1πœƒπœƒ+1=πœƒ.

This means we do not have to memorize all three Pythagorean identities, as we can derive all the other identities from the most basic one involving the sine and cosine functions.

Let’s find the value of a particular expression by applying an identity, with no other information. This means the values will always be the same, no matter what the angle is or the value of the trigonometric function at that angle.

Example 1: Applying the Pythagorean Identities to Evaluate Some Expressions

Find the value of (π‘₯+π‘₯)+(π‘₯βˆ’π‘₯)cossincossin.

Answer

In this example, we will find the value of a particular trigonometric expression by applying a Pythagorean identity.

Since this example involves sine and cosine functions, we recall the Pythagorean identity sincosπ‘₯+π‘₯=1.

If we distribute the parentheses in the given expression and apply this identity, we obtain (π‘₯+π‘₯)+(π‘₯βˆ’π‘₯)=ο€Ίπ‘₯+π‘₯+2π‘₯π‘₯+ο€Ίπ‘₯+π‘₯βˆ’2π‘₯π‘₯=2π‘₯+2π‘₯=2ο€Ίπ‘₯+π‘₯=2.cossincossincossinsincoscossinsincoscossincossin

The sine function is equivalent to the cosine function by a translation 90∘ to the left, which can be visualized by comparing the plots of both functions.

In particular, we have the following shift identities for angles πœƒ and 90+πœƒβˆ˜: sincoscossin(90+πœƒ)=πœƒ,(90+πœƒ)=βˆ’πœƒ.∘∘

We can also illustrate these on the unit circle as follows:

Similarly, by replacing πœƒ with βˆ’πœƒ, we obtain the following cofunction identities for the complementary angles πœƒ and 90βˆ’πœƒβˆ˜: sincoscossin(90βˆ’πœƒ)=πœƒ,(90βˆ’πœƒ)=πœƒ.∘∘

We can illustrate this as follows:

The figure depicts the right triangle with angle 𝐴𝑂𝐡 in the standard position, which intersects the unit circle at 𝐡(π‘₯,𝑦) and its angle measure is acute, 0<πœƒ<90∘∘.

We can use these cofunction identities along with the Pythagorean identities to simplify trigonometric expressions. For example, if we have the expression sincoscossinsincosπœƒ(90βˆ’πœƒ)+πœƒ(90βˆ’πœƒ)=πœƒ+πœƒ=1,∘∘ we can combine these identities and use them to determine identities for the other trigonometric functions that are defined in terms of the sine and cosine functions.

Definition: Trigonometric Correlated Angle Identities

The trigonometric functions satisfy cofunction identities for all πœƒ in their domains. In particular, we have sincoscossin(90Β±πœƒ)=πœƒ,(90Β±πœƒ)=βˆ“πœƒ.∘∘

We can derive the cofunction identities for other trigonometric functions by relating them to the identities involving sine and cosine. For example, for the tangent function, we have tansincoscossincossintancot(90Β±πœƒ)=(90Β±πœƒ)(90Β±πœƒ)=πœƒβˆ“πœƒ=βˆ“πœƒπœƒ(90Β±πœƒ)=βˆ“πœƒ.∘∘∘∘

All of these identities also hold in radians, in particular, by replacing 90∘ in degrees with πœ‹2 in radians.

Now, let’s consider an example where we use this identity with the sine function to simplify a given trigonometric expression with a Pythagorean identity.

Example 2: Simplifying Trigonometric Expressions Using Cofunction and Pythagorean Identities

Simplify sinsinοŠ¨οŠ¨βˆ˜πœƒ+(90βˆ’πœƒ).

Answer

In this example, we will find the value of a particular trigonometric expression by applying a Pythagorean and cofunction identity.

Since this example involves sine function, we recall the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1 and the cofunction identity sincos(90βˆ’πœƒ)=πœƒ.∘

Using this cofunction identity on the given expression and applying the Pythagorean identity, we obtain sinsinsincosοŠ¨οŠ¨βˆ˜οŠ¨οŠ¨πœƒ+(90βˆ’πœƒ)=πœƒ+πœƒ=1.

We can determine the value of a trigonometric expression from the value of a trigonometric function. In the next example, we will determine the value of an expression using information about the lengths of a right triangle.

Example 3: Evaluating Pythagorean Identities for Angles in Right Triangles

Find 1+𝐴tan, given that 𝐴𝐡𝐢 is a right triangle at 𝐢, where 𝐴𝐡=10cm and 𝐡𝐢=6cm.

Answer

In this example, we will find the value of a particular trigonometric expression by using information about the lengths of a right triangle, to determine the value of a particular trigonometric function, and applying Pythagorean identities.

We can find the value of the desired expression using two different methods. Firstly, using Pythagorean identities, and secondly, using the Pythagorean theorem. Let’s begin with the method of finding this value using Pythagorean identities.

Method 1

Since this example involves the tangent function, we recall the Pythagorean identity 1+𝐴=𝐴.tansec

By definition, the secant function is seccos𝐴=1𝐴.

Side 𝐡𝐢 is opposite to angle 𝐴, and side 𝐴𝐡 is the hypotenuse of the right triangle. Hence, we can find the sine of this angle by taking the ratio of these lengths. We have sinoppositehypotenuse𝐴==𝐡𝐢𝐴𝐡=610=35.

By the Pythagorean identity and the definition of the secant function, 1+𝐴=𝐴=1𝐴.tanseccos

Let’s first determine the denominator of the right-hand side of this expression, which depends on the value of cos𝐴, where we know the value of sin𝐴. Since this involves the sine and cosine functions, recall the Pythagorean identity relating these two values: sincos𝐴+𝐴=1, which we can rearrange as cossin𝐴=1βˆ’π΄ and where we substitute the given value for the sine function to obtain cos𝐴=1βˆ’ο€Ό35=1βˆ’925=25βˆ’925=1625.

Thus, for the given expression, we have 1+𝐴=1𝐴=1=2516.tancos

Method 2

We could also determine this value by finding the adjacent side of the right triangle from the Pythagorean theorem and using this to compute the value of tan𝐴, given as the ratio of the opposite and adjacent sides. By the Pythagorean theorem on the right triangle, we have 𝐡𝐢+𝐴𝐢=𝐴𝐡𝐴𝐢=π΄π΅βˆ’π΅πΆ=10βˆ’6=100βˆ’36=64.

Thus, we have 𝐴𝐢=√64=8cm. Now we can determine the tangent of the angle by the ratio of the opposite side, 𝐡𝐢, and the adjacent side, 𝐴𝐢: tan𝐴=𝐡𝐢𝐴𝐢=68=34.

We note that this is the same as the ratio of the value given by the sine and cosine of the angle, since cos𝐴=1625 implies that cos𝐴=45, as 𝐴 is an acute angle, and hence tansincos𝐴=𝐴𝐴==34.οŠͺ

Now, we can square this value and add 1 to find the value of the required expression: 1+𝐴=1+ο€Ό34=1+916=2516.tan

In the next example, we will determine the value of the square of a trigonometric function, cotοŠ¨πœƒ, from the given value of the square of another trigonometric function, cscοŠ¨πœƒ, by using the Pythagorean identities.

Example 4: Using the Pythagorean Identities to Evaluate a Trigonometric Function

Find the value of cotοŠ¨πœƒ given cscοŠ¨πœƒ=259.

Answer

In this example, we will find the value of the square of a trigonometric function from the value of the square of another trigonometric function.

Since this example involves cotangent and cosecant functions, we recall the Pythagorean identity 1+πœƒ=πœƒ.cotcsc

Thus, if we rearrange this expression to make cotοŠ¨πœƒ the subject and substitute cscοŠ¨πœƒ=259, we have cotcscοŠ¨οŠ¨πœƒ=πœƒβˆ’1=259βˆ’1=25βˆ’99=169.

We can also determine the value of a trigonometric expression from the value of another expression, without the actual values of the functions, by applying the Pythagorean identity. Let’s consider an example of this.

Example 5: Using the Pythagorean Identities to Evaluate a Trigonometric Expression

Find the value of sincosπœƒπœƒ given sincosπœƒ+πœƒ=54.

Answer

In this example, we will find the value of a particular trigonometric expression from the value of another expression.

Since this example involves sine and cosine functions, we recall the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1.

Since the Pythagorean identity involves squares of sine and cosine functions, it make sense to begin by squaring both sides of the equation. If we take the square of the given expression sincosπœƒ+πœƒ=54, distribute the parentheses, and apply this identity, we obtain (πœƒ+πœƒ)=ο€Ό54οˆπœƒ+πœƒ+2πœƒπœƒ=25161+2πœƒπœƒ=25162πœƒπœƒ=2516βˆ’1=25βˆ’1616=916πœƒπœƒ=932.sincossincossincossincossincossincos

Now, let’s determine the value of another trigonometric expression, this time using multiple Pythagorean identities and the definition of the reciprocal trigonometric functions.

Example 6: Evaluating Trigonometric Expressions given Trigonometric Equations

Find the value of tancotοŠ¨οŠ¨πœƒ+πœƒ given tancotπœƒ+πœƒ=17.

Answer

In this example, we will find the value of a particular trigonometric expression from the value of another expression.

We can find the value of the given expression in two different ways: firstly, using Pythagorean identities and, secondly, by squaring the given equation and simplifying the resulting equation. Let’s begin with the method using Pythagorean identities.

Method 1

Since this example involves tangent and cotangent functions, we recall the Pythagorean identities 1+πœƒ=πœƒ,πœƒ+1=πœƒ.cotcsctansec

Note that the cosecant, secant, and cotangent functions are defined as cscsinseccoscotcossinπœƒ=1πœƒ,πœƒ=1πœƒ,πœƒ=πœƒπœƒ.

The left-hand side of the given trigonometric expression can be written as tancotsincoscossinsincossincosπœƒ+πœƒ=πœƒπœƒ+πœƒπœƒ=πœƒ+πœƒπœƒπœƒ.

Since the left-hand side of this expression involves the sine and cosine functions, we recall the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1, which can be used to simplify the numerator of the expression tancotsincosπœƒ+πœƒ=1πœƒπœƒ.

Thus, the given expression is equivalent to 1πœƒπœƒ=17.sincos

Now, for the expression we want to find the value of, we can apply the Pythagorean identities to obtain tancotseccscseccsccossinsincossincossincossincosοŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨πœƒ+πœƒ=ο€Ήπœƒβˆ’1+ο€Ήπœƒβˆ’1=πœƒ+πœƒβˆ’2=1πœƒ+1πœƒβˆ’2=πœƒ+πœƒπœƒπœƒβˆ’2=1πœƒπœƒβˆ’2=ο€Ό1πœƒπœƒοˆβˆ’2=17βˆ’2=287.

Method 2

We note that we could have also found this value without applying the Pythagorean identities by squaring the given expression, tancotπœƒ+πœƒ=17, and using the definition of the cotangent function: (πœƒ+πœƒ)=17πœƒ+2πœƒπœƒ+πœƒ=17πœƒ+2πœƒΓ—1πœƒ+πœƒ=17πœƒ+2+πœƒ=17.tancottantancotcottantantancottancot

Thus, we obtain the value as tancotοŠ¨οŠ¨οŠ¨πœƒ+πœƒ=17βˆ’2=287.

So far, the examples we have considered involved finding the value of a trigonometric expression by applying the Pythagorean identities. For some examples, we also used known values of a trigonometric function or another expression. But what if we want to determine the value of a trigonometric function from the value of another trigonometric expression? In some cases, we may need to know what quadrant the angle lies in because the signs of the trigonometric functions can be different for each quadrant.

The Pythagorean identities involve squares of values of trigonometric functions, so we may need to take the square root to obtain the value of another trigonometric function, which can be positive or negative. The sign of the trigonometric functions will depend on which quadrant we consider. The CAST diagram helps us to remember the signs of the trigonometric functions, for each quadrant.

Let’s recall the CAST diagram.

Rule: The Cast Diagram

  • In the first quadrant, all trigonometric functions are positive.
  • In the second quadrant, the sine function is positive.
  • In the third quadrant, the tangent function is positive.
  • In the fourth quadrant, the cosine function is positive.

The Pythagorean identities allow us to determine the value of a trigonometric function, say cosπœƒ, from given the value of another trigonometric function, sinπœƒ. From the first Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1, we obtain cossincossinοŠ¨οŠ¨οŠ¨πœƒ=1βˆ’πœƒπœƒ=±√1βˆ’πœƒ.

The cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants. In other words, the value of the cosine function will depend on the value of sinπœƒ and which quadrant angle πœƒ lies.

Now, let’s look at an example where we determine the value of the cosine function from the given value of the sine value in a particular quadrant, using the Pythagorean identity.

Example 7: Using the Sine Ratio to Find the Cosine of an Angle

Find cosπœƒ given sinπœƒ=βˆ’35, where 270β‰€πœƒ<360∘∘.

Answer

In this example, we will find the value of a particular trigonometric function from the value of another function, in a particular quadrant.

Since this example involves sine and cosine functions, we recall the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1.

If we substitute sinπœƒ=βˆ’35 and rearrange this, we find ο€Όβˆ’35+πœƒ=1925+πœƒ=1πœƒ=1βˆ’925πœƒ=Β±ο„ž1βˆ’925.coscoscoscos

Since 270β‰€πœƒ<360∘∘, this corresponds to the fourth quadrant. We recall from the CAST diagram that the sign of the cosine function is positive. Thus, taking the positive sign, cosπœƒ=ο„ž1βˆ’925=ο„ž25βˆ’925=ο„ž1625=45.

In fact, we can write the values of the other trigonometric function in terms of sinπœƒ and using the value for the cosine function, which follows from the other Pythagorean identities or the definition of the functions in terms of sine and cosine. For example, for the tangent function, we have tansincossinsinsinsinπœƒ=πœƒπœƒ=πœƒΒ±βˆš1βˆ’πœƒ=Β±πœƒβˆš1βˆ’πœƒ.

The sign of each trigonometric function will depend on the sign of sinπœƒ and the quadrant in which angle πœƒ lies.

In the next example, we will determine the value of the tangent function from the given value of the sine function in a particular quadrant using the Pythagorean identity.

Example 8: Using the Pythagorean Identity to Find the Value of a Trigonometric Function given Another One and the Quadrant of the Angle

Knowing that sinπ‘₯=√137 and πœ‹2≀π‘₯β‰€πœ‹, find tanπ‘₯.

Answer

In this example, we will find the value of a particular trigonometric function from the value of another function, in a particular quadrant.

Since this example involves sine and cosine functions, we recall the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1.

Note that, by definition of the tangent function, we have tansincosπ‘₯=π‘₯π‘₯

Let’s first determine the denominator of this expression. Rearranging the Pythagorean identity and taking the square root, we obtain cossincossinπ‘₯=1βˆ’π‘₯π‘₯=±√1βˆ’π‘₯.

Since πœ‹2≀π‘₯β‰€πœ‹, which is the second quadrant, the cosine function is negative, and we take the negative root: cossinπ‘₯=βˆ’βˆš1βˆ’π‘₯.

Substituting the given value sinπ‘₯=√137, we obtain cosπ‘₯=βˆ’ο„‘ο„£ο„£ο„ 1βˆ’ο€Ώβˆš137=βˆ’ο„ž1βˆ’1349=βˆ’ο„ž49βˆ’1349=βˆ’ο„ž3649=βˆ’67.

Thus, for the tangent function, we obtain tanπ‘₯=βˆ’=βˆ’βˆš136.√

Now, let’s consider an example where we determine the value of the secant function from a given trigonometric expression in the first quadrant, where all the trigonometric functions are positive, by applying a Pythagorean identity.

Example 9: Using the Pythagorean Identities to Evaluate a Trigonometric Function of an Angle

Find the value of secπœƒ given sectanπœƒβˆ’πœƒ=16, where 0<πœƒ<πœ‹2.

Answer

In this example, we will find the value of a particular trigonometric function from the value of a given trigonometric expression, in a particular quadrant.

Since this example involves tangent and secant functions, we recall the Pythagorean identity 1+πœƒ=πœƒ.tansec

We can also rewrite this identity as sectanοŠ¨οŠ¨πœƒβˆ’πœƒ=1. Using the difference of squares formula, π‘Žβˆ’π‘=(π‘Ž+𝑏)(π‘Žβˆ’π‘), we can rewrite this identity as (πœƒ+πœƒ)(πœƒβˆ’πœƒ)=1.sectansectan

If we substitute the given expression, sectanπœƒβˆ’πœƒ=16, we have (πœƒ+πœƒ)ο€Ό16=1πœƒ+πœƒ=6.sectansectan

Now, we can substitute tansecπœƒ=πœƒβˆ’16 in order to eliminate tanπœƒ from the expression and solve for secπœƒ: secsecsecsecsecπœƒ+πœƒβˆ’16=62πœƒ=6+162πœƒ=376πœƒ=3712.

Instead of finding the value of particular trigonometric functions, we can determine the value of a particular expression from a given value or expression.

Now, let’s consider an example where we determine the value of an expression from a given value in the third quadrant.

Example 10: Using the Pythagorean Identities to Evaluate a Trigonometric Expression given a Trigonometric Function and the Quadrant of an Angle

Find the value of cscsintancotcosπœƒπœƒβˆ’πœƒπœƒ+πœƒοŠ¨ given πœƒβˆˆο πœ‹,3πœ‹2 and sinπœƒ=βˆ’45.

Answer

In this example, we will find the value of a particular trigonometric expression from the value of a trigonometric function. We will first determine the value of the sine and cosine functions, separately, then take the product to find the value of the given expression.

Note that the cosecant and cotangent functions are defined as cscsincottanπœƒ=1πœƒ,πœƒ=1πœƒ.

For the given expression, we can substitute these and use the Pythagorean identity to obtain an expression in terms of sinπœƒ as cscsintancotcossinsintantancoscosπœƒπœƒβˆ’πœƒπœƒ+πœƒ=1πœƒΓ—πœƒβˆ’πœƒΓ—1πœƒ+πœƒ=πœƒ.

Since the right-hand side involves a cosine function, and we are given the value of the sine function, we will make use of the Pythagorean identity: sincosοŠ¨οŠ¨πœƒ+πœƒ=1; hence, we obtain cscsintancotcossinπœƒπœƒβˆ’πœƒπœƒ+πœƒ=1βˆ’πœƒ.

Now, we can substitute the given value, sinπœƒ=βˆ’45, to obtain cscsintancotcosπœƒπœƒβˆ’πœƒπœƒ+πœƒ=1βˆ’ο€Όβˆ’45=1βˆ’1625=25βˆ’1625=925.

Finally, let’s consider an example where we find the value of a trigonometric expression from a given value of the tangent function in the third and fourth quadrants by first determining the value of the sine and cosine functions and substituting this into the given expression.

Example 11: Using the Pythagorean Identities to Evaluate a Trigonometric Expression

Find the value of 2πœƒπœƒsincos given 12πœƒ+5=0tan, where 180<πœƒ<360∘∘.

Answer

In this example, we will find the value of a particular trigonometric expression from the value of a trigonometric function. We will first determine the value of the sine and cosine functions, separately, then take the product to find the value of the given expression.

We can write the given expression, 12πœƒ+5=0tan, as a value for the tangent function: tanπœƒ=βˆ’512.

We recall from the CAST diagram that the tangent function is positive in the third quadrant and negative in the fourth quadrant. Since the tangent function takes a negative value and we are given that the angle lies in either the third or the fourth quadrant, corresponding to the range 180<πœƒ<360∘∘, angle πœƒ must lie in the fourth quadrant.

Since we know the value of the tangent function, we can determine the value of the cosine function from the Pythagorean identity involving the tangent and secant functions: tansecοŠ¨οŠ¨πœƒ+1=πœƒ, with the definition of the secant function in terms of the cosine function seccosπœƒ=1πœƒ.

We can rearrange this Pythagorean identity to make secπœƒ the subject as sectansectanοŠ¨οŠ¨οŠ¨πœƒ=1+πœƒπœƒ=±√1+πœƒ.

Since angle πœƒ lies in the fourth quadrant, we remember from the CAST diagram that the cosine function, hence the secant function, is positive in the fourth quadrant. Upon substituting the value of the tangent function from the given expression, we have secπœƒ=ο„Ÿ1+ο€Όβˆ’512=ο„ž1+25144=ο„ž169144=1312.

We can also obtain the value of the cosine function by taking the reciprocal of this to obtain cosπœƒ=1213.

Now, since we know the value of the cosine function, we can determine the value of the sine function using the Pythagorean identity sincosοŠ¨οŠ¨πœƒ+πœƒ=1.

We can rearrange this identity, involving the sine and cosine functions, to make sinπœƒ the subject: sincossincosοŠ¨οŠ¨οŠ¨πœƒ=1βˆ’πœƒπœƒ=±√1βˆ’πœƒ.

Since the sine function is negative in the fourth quadrant, we have sinπœƒ=βˆ’ο„Ÿ1βˆ’ο€Ό1213=βˆ’ο„ž1βˆ’144169=βˆ’ο„ž25169=βˆ’513.

Now, we can determine the value of the trigonometric expression by substituting the values of sine and cosine: 2πœƒπœƒ=2ο€Όβˆ’513οˆο€Ό1213=βˆ’120169.sincos

Let’s finish by recapping a few important key points from this explainer.

Key Points

  • The Pythagorean identities are given by sincoscotcsctansecοŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨πœƒ+πœƒ=1,1+πœƒ=πœƒ,πœƒ+1=πœƒ.
  • The unit circle allows us to determine the correlated angle identities for sine and cosine.
    For instance, the cofunction identities (in radians) are sincoscossinο€»πœ‹2βˆ’πœƒο‡=πœƒ,ο€»πœ‹2βˆ’πœƒο‡=πœƒ. The corresponding identities for the tangent and reciprocal trigonometric functions are found using their definitions in terms of the sine and cosine functions.
  • We may need to apply more than one Pythagorean identity, or type of identity, to simplify a trigonometric expression.
  • Pythagorean identities can only produce the absolute value of a trigonometric function. We need to use the CAST diagram to determine the sign of a trigonometric function within a quadrant.

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