Lesson Explainer: Euler’s Formula for Trigonometric Identities | Nagwa Lesson Explainer: Euler’s Formula for Trigonometric Identities | Nagwa

Lesson Explainer: Euler’s Formula for Trigonometric Identities Mathematics

In this explainer, we will learn how to use Euler’s formula to prove trigonometric identities like double angle and half angle.

When we first learn about trigonometric functions and the exponential functions, they seem to have little, to nothing, in common. Trigonometric functions are periodic, and, in the case of sine and cosine, are bounded above and below by 1 and βˆ’1, whereas the exponential function is nonperiodic and has no upper bound. However, Euler’s formula demonstrates that, through the introduction of complex numbers, these seemingly unconnected ideas are, in fact, intimately connected and, in many ways, are like two sides of a coin.

Definition: Euler’s Formula

Euler’s formula states that for any real number πœƒ, 𝑒=πœƒ+π‘–πœƒ.cossin

This formula is alternatively referred to as Euler’s relation.

Euler’s formula has applications in many area of mathematics, such as functional analysis, differential equations, and Fourier analysis. Furthermore, its applications extend into physics and engineering in such diverse areas as signal processing, electrical engineering, and quantum mechanics. In this explainer, we will focus on the its applications in trigonometryβ€”in particular the derivations of trigonometric identities.

In the first example, we will demonstrate how we can derive trigonometric identities by using one of the properties of the exponential function and then applying Euler’s formula.

Example 1: Using Euler’s Formula to Derive Trigonometric Identities

  1. Use Euler’s formula to express π‘’οŠ±οƒοΌ in terms of sine and cosine.
  2. Given that 𝑒𝑒=1οƒοΌοŠ±οƒοΌ, what trigonometric identity can be derived by expanding the exponentials in terms of trigonometric functions?

Answer

Part 1

Rewriting 𝑒=𝑒,οŠ±οƒοΌοƒ() we can apply Euler’s formula to get 𝑒=(βˆ’πœƒ)+𝑖(βˆ’πœƒ).οŠ±οƒοΌcossin

Using the odd/even identities for sine and cosine, sinsincoscos(βˆ’πœƒ)=βˆ’πœƒ,(βˆ’πœƒ)=πœƒ, we have 𝑒=πœƒβˆ’π‘–πœƒ.οŠ±οƒοΌcossin

Part 2

Using Euler’s identity and our answer from part 1, we can rewrite 𝑒𝑒=(πœƒ+π‘–πœƒ)(πœƒβˆ’π‘–πœƒ).οƒοΌοŠ±οƒοΌcossincossin

Expanding the parentheses, we have 𝑒𝑒=πœƒβˆ’π‘–πœƒπœƒ+π‘–πœƒπœƒβˆ’π‘–πœƒ.οƒοΌοŠ±οƒοΌοŠ¨οŠ¨οŠ¨coscossincossinsin

Finally, we can simplify this to get 𝑒𝑒=πœƒ+πœƒ.οƒοΌοŠ±οƒοΌοŠ¨οŠ¨cossin

Hence, given that 𝑒𝑒=1οƒοΌοŠ±οƒοΌ, we have proven the trigonometric identity sincosοŠ¨οŠ¨πœƒ+πœƒβ‰‘1.

As we saw in the last example, by using the properties of the exponential function and applying Euler’s formula, we can derive trigonometric identities. The next example will demonstrate the versatility of this method.

Example 2: Trigonometric Identities from Euler’s Formula

What trigonometric identities can be derived by applying Euler’s identity to 𝑒()?

Answer

The first thing we need to do is apply one of the properties of the exponential function. The obvious choice here is the multiplicative property: 𝑒=𝑒𝑒.()οƒοΌοƒοŽŠ

Now that we have two different but equal expressions, we can apply Euler’s formula to each one to get cossincossincossin(πœƒ+πœ‘)+𝑖(πœƒ+πœ‘)=(πœƒ+π‘–πœƒ)(πœ‘+π‘–πœ‘).

By expanding the parentheses on the right-hand side, we have cossincoscoscossinsincossinsin(πœƒ+πœ‘)+𝑖(πœƒ+πœ‘)=πœƒπœ‘+π‘–πœƒπœ‘+π‘–πœƒπœ‘+π‘–πœƒπœ‘.

Gathering together the like terms on the right-hand side, we find that cossincoscossinsincossinsincos(πœƒ+πœ‘)+𝑖(πœƒ+πœ‘)=(πœƒπœ‘βˆ’πœƒπœ‘)+𝑖(πœƒπœ‘+πœƒπœ‘).

Finally, we can equate the real and imaginary parts to get the following two trigonometric identities: coscoscossinsinsincossinsincos(πœƒ+πœ‘)=πœƒπœ‘βˆ’πœƒπœ‘,(πœƒ+πœ‘)=πœƒπœ‘+πœƒπœ‘.

Example 3: Double Angle Formulas from Euler’s Formula

Use Euler’s formula to derive a formula for cos2πœƒ and sin2πœƒ in terms of sinπœƒ and cosπœƒ.

Answer

The first thing we need to consider is what property of the exponential function we can apply to get two different but equal expressions. Since we are looking for identities in 2πœƒ, it makes sense that we would consider the expression π‘’οŠ¨οΌοƒ. At this point, there are two obvious choices for properties of the exponential function to use, either the power rule or the multiplicative rule. In both of these cases, we will get the same result: 𝑒=𝑒.οŠ¨οΌοƒοƒοΌοŠ¨

Applying Euler’s formula to both of these expression, we have cossincossin2πœƒ+𝑖2πœƒ=(πœƒ+π‘–πœƒ).

By expanding the parentheses, we have cossincoscossinsincossincossin2πœƒ+𝑖2πœƒ=πœƒ+2π‘–πœƒπœƒ+π‘–πœƒ=ο€Ίπœƒβˆ’πœƒο†+2π‘–πœƒπœƒ.

Finally, we can equate the real and imaginary parts to get the following two trigonometric identities: coscossinsincossin2πœƒ=πœƒβˆ’πœƒ,2πœƒ=2πœƒπœƒ.

Using a similar method, we can derive multiple angle formulas for sinπ‘›πœƒ and cosπ‘›πœƒ for larger values of 𝑛. However, to simplify the derivation, we can use the binomial theorem.

The Binomial Theorem

For any integer 𝑛, (π‘Ž+𝑏)=π‘Ž+πΆπ‘Žπ‘+πΆπ‘Žπ‘+β‹―+πΆπ‘Žπ‘+β‹―+πΆπ‘Žπ‘+𝑏, where 𝐢=π‘›π‘Ÿπ‘›βˆ’π‘Ÿ. Sometimes, 𝐢 is denoted ο€Ώπ‘›π‘Ÿο‹ or 𝐢.

In the next example, we will demonstrate how we can use the binomial theorem to simplify the derivation of multiple angle formulas for sine and cosine.

Example 4: Multiple Angle Formulas from Euler’s Formula

  1. Use Euler’s formula to derive a formula for cos6πœƒ and sin6πœƒ in terms of sinπœƒ and cosπœƒ.
  2. Hence, express tan6πœƒ in terms of tanπœƒ.

Answer

Part 1

Using the properties of the exponential function, we can write 𝑒=𝑒.οŠ¬οƒοΌοƒοΌοŠ¬

Using Euler’s formula, we can rewrite this as cossincossin6πœƒ+𝑖6πœƒ=(πœƒ+π‘–πœƒ).

Using the binomial theorem, we can expand the parentheses as follows: cossincoscossincossincossincossincossinsin6πœƒ+𝑖6πœƒ=πœƒ+πΆπœƒ(π‘–πœƒ)+πΆπœƒο€Ίπ‘–πœƒο†+πΆπœƒο€Ίπ‘–πœƒο†+πΆπœƒο€Ίπ‘–πœƒο†+πΆπœƒο€Ίπ‘–πœƒο†+π‘–πœƒ.οŠͺοŠͺοŠͺοŠͺ

Substituting in the vales of 𝐢 and simplifying, we have cossincoscossincossincossincossincossinsin6πœƒ+𝑖6πœƒ=πœƒ+6π‘–πœƒπœƒ+15π‘–πœƒπœƒ+20π‘–πœƒπœƒ+15π‘–πœƒπœƒ+6π‘–πœƒπœƒ+π‘–πœƒ.οŠͺοŠͺοŠͺ

Evaluating the powers of 𝑖, we get cossincoscossincossincossincossincossinsin6πœƒ+𝑖6πœƒ=πœƒ+6π‘–πœƒπœƒβˆ’15πœƒπœƒβˆ’20π‘–πœƒπœƒ+15πœƒπœƒ+6π‘–πœƒπœƒβˆ’πœƒ.οŠͺοŠͺ

By equating the real and imaginary parts, we arrive at the following two identities: coscoscossincossinsinsincossincossincossin6πœƒ=πœƒβˆ’15πœƒπœƒ+15πœƒπœƒβˆ’πœƒ,6πœƒ=6πœƒπœƒβˆ’20πœƒπœƒ+6πœƒπœƒ.οŠͺοŠͺ

Part 2

Using the identities from part 1, we can write tansincoscossincossincossincoscossincossinsin6πœƒ=6πœƒ6πœƒ=6πœƒπœƒβˆ’20πœƒπœƒ+6πœƒπœƒπœƒβˆ’15πœƒπœƒ+15πœƒπœƒβˆ’πœƒ.οŠͺοŠͺ

To express this in terms of tanπœƒ, we need to divide both the numerator and denominator by cosοŠ¬πœƒ, which results in tantantantantantantan6πœƒ=6πœƒβˆ’20πœƒ+6πœƒ1βˆ’15πœƒ+15πœƒβˆ’πœƒ.οŠͺ

One of the interesting applications of Euler’s formula is that we can derive an expression for sinπœƒ and cosπœƒ in terms of 𝑒 and π‘’οŠ±οƒοΌ. In the first example, we saw that we can express π‘’οŠ±οƒοΌ in terms of sine and cosine as

𝑒=πœƒβˆ’π‘–πœƒ.οŠ±οƒοΌcossin(1)

Adding this to Euler’s formula gives 𝑒+𝑒=πœƒ+π‘–πœƒ+πœƒβˆ’π‘–πœƒ=2πœƒ.οƒοΌοŠ±οƒοΌcossincossincos

Hence, by dividing by 2, we derive the following formula for cosine: cosπœƒ=12𝑒+𝑒.οƒοΌοŠ±οƒοΌ

Similarly, we can derive a formula for sine by considering the difference between equation (1) and Euler’s formula as follows: π‘’βˆ’π‘’=πœƒ+π‘–πœƒβˆ’(πœƒβˆ’π‘–πœƒ)=2π‘–πœƒ.οƒοΌοŠ±οƒοΌcossincossinsin

Dividing by 2𝑖, we arrive at sinπœƒ=12π‘–ο€Ήπ‘’βˆ’π‘’ο….οƒοΌοŠ±οƒοΌ

These two formulas are certainly interesting in their own right. However, for the purpose of this explainer, we will demonstrate how they can be used to derive further trigonometric identities.

Example 5: Product-to-Sum Formulas from Euler’s Formula

  1. Derive two trigonometric identities by considering the real and imaginary parts of 𝑒+𝑒()().
  2. Similarly, what two trigonometric identities can be derived by considering the real and imaginary parts of π‘’βˆ’π‘’οƒ()()?

Answer

Part 1

We begin by applying the multiplicative property of exponentials to rewrite 𝑒+𝑒=𝑒𝑒+𝑒𝑒.()()οƒοΌοƒοŽŠοƒοΌοŠ±οƒοŽŠ

Factoring out the common factor of 𝑒, we have 𝑒+𝑒=𝑒𝑒+𝑒.()()οƒοΌοƒοŽŠοŠ±οƒοŽŠ

Using the definition of cosine in terms of the exponential function, cosπœ‘=12𝑒+𝑒,οƒοŽŠοŠ±οƒοŽŠ we can rewrite this as 𝑒+𝑒=2πœ‘ο€Ήπ‘’ο….()()cos

We can now use Euler’s formula to rewrite this as cossincossincoscossin(πœƒ+πœ‘)+𝑖(πœƒ+πœ‘)+(πœƒβˆ’πœ‘)+𝑖(πœƒβˆ’πœ‘)=2πœ‘(πœƒ+π‘–πœƒ).

Grouping together the real and imaginary parts on both side of the equations yields ((πœƒ+πœ‘)+(πœƒβˆ’πœ‘))+𝑖((πœƒ+πœ‘)+(πœƒβˆ’πœ‘))=2πœƒπœ‘+2π‘–πœƒπœ‘.coscossinsincoscossincos

By equating the real and imaginary parts, we get the following two trigonometric identities: coscoscoscossincossinsinπœƒπœ‘=12((πœƒ+πœ‘)+(πœƒβˆ’πœ‘)),πœƒπœ‘=12((πœƒ+πœ‘)+(πœƒβˆ’πœ‘)).

Part 2

Using a similar method, we can consider the expression π‘’βˆ’π‘’οƒ()(). Applying the properties of the exponential function, we can rewrite this as π‘’βˆ’π‘’=π‘’π‘’βˆ’π‘’π‘’.()()οƒοΌοƒοŽŠοƒοΌοŠ±οƒοŽŠ

By factoring out 𝑒, we can express this as π‘’βˆ’π‘’=π‘’ο€Ήπ‘’βˆ’π‘’ο….()()οƒοΌοƒοŽŠοŠ±οƒοŽŠ

Using the definition of sine in terms of the exponential function, sinπœ‘=12π‘–ο€Ήπ‘’βˆ’π‘’ο…,οƒοŽŠοŠ±οƒοŽŠ we can rewrite this as π‘’βˆ’π‘’=2π‘–πœ‘ο€Ήπ‘’ο….()()sin

We can now apply Euler’s formula to both sides of the equation to get cossincossinsincossin(πœƒ+πœ‘)+𝑖(πœƒ+πœ‘)βˆ’((πœƒβˆ’πœ‘)+𝑖(πœƒβˆ’πœ‘))=2π‘–πœ‘(πœƒ+π‘–πœƒ).

Grouping together the real and imaginary parts on both side of the equations yields ((πœƒ+πœ‘)βˆ’(πœƒβˆ’πœ‘))+𝑖((πœƒ+πœ‘)βˆ’(πœƒβˆ’πœ‘))=βˆ’2πœƒπœ‘+2π‘–πœƒπœ‘.coscossinsinsinsincossin

By equating the real and imaginary parts, we get the following two trigonometric identities: sinsincoscoscossinsinsinπœƒπœ‘=12((πœƒβˆ’πœ‘)βˆ’(πœƒ+πœ‘)),πœƒπœ‘=12((πœƒ+πœ‘)βˆ’(πœƒβˆ’πœ‘)).

In this final example, we use the definitions of sine and cosine in terms of the complex exponential function to express powers of sine and cosine in terms of multiple angle formulae.

Example 6: Powers in terms of Multiple Angles from Euler’s Formula

  1. Use Euler’s formula to express sincosοŠ©οŠ¨πœƒπœƒ in the form π‘Žπœƒ+𝑏3πœƒ+𝑐5πœƒsinsinsin, where π‘Ž, 𝑏, and 𝑐 are constants to be found.
  2. Hence, find the solutions of sinsin5πœƒβˆ’3πœƒ=0 in the interval 0β‰€πœƒ<πœ‹. Give your answer in exact form.

Answer

Part 1

Using the definitions of sine and cosine in terms of the complex exponential, sincosπœƒ=12π‘–ο€Ήπ‘’βˆ’π‘’ο…,πœƒ=12𝑒+𝑒,οƒοΌοŠ±οƒοΌοƒοΌοŠ±οƒοΌ we can rewrite the expression as sincosοŠ©οŠ¨οƒοΌοŠ±οƒοΌοŠ©οƒοΌοŠ±οƒοΌοŠ¨πœƒπœƒ=ο€Ό12π‘–ο€Ήπ‘’βˆ’π‘’ο…οˆο€Ό12𝑒+π‘’ο…οˆ.

Hence, sincosοŠ©οŠ¨οƒοΌοŠ±οƒοΌοŠ©οƒοΌοŠ±οƒοΌοŠ¨πœƒπœƒ=βˆ’132π‘–ο€Ήπ‘’βˆ’π‘’ο…ο€Ήπ‘’+𝑒.

Using the binomial theorem, we can expand each of the sets of parentheses as follows: sincosοŠ©οŠ¨οŠ©οƒοΌοƒοΌοŠ±οƒοΌοŠ±οŠ©οƒοΌοŠ¨οƒοΌοŠ±οŠ¨οƒοΌπœƒπœƒ=βˆ’132π‘–ο€Ήπ‘’βˆ’3𝑒+3π‘’βˆ’π‘’ο…ο€Ήπ‘’+2+𝑒.

We can now expand the two parentheses to get sincosοŠ©οŠ¨οŠ«οƒοΌοŠ©οƒοΌοƒοΌοŠ±οƒοΌοŠ©οƒοΌοƒοΌοŠ±οƒοΌοŠ±οŠ©οƒοΌοƒοΌοŠ±οƒοΌοŠ±οŠ©οƒοΌοŠ±οŠ«οƒοΌπœƒπœƒ=βˆ’132π‘–ο€Ήπ‘’βˆ’3𝑒+3π‘’βˆ’π‘’+2π‘’βˆ’6𝑒+6π‘’βˆ’2𝑒+π‘’βˆ’3𝑒+3π‘’βˆ’π‘’ο….

Gathering the π‘’οƒοŠοΌ terms with π‘’οŠ±οƒοŠοΌ terms results in sincosοŠ©οŠ¨οŠ«οƒοΌοŠ±οŠ«οƒοΌοŠ©οƒοΌοŠ±οŠ©οƒοΌοƒοΌοŠ±οƒοΌπœƒπœƒ=βˆ’132π‘–ο€Ήο€Ήπ‘’βˆ’π‘’ο…βˆ’ο€Ήπ‘’βˆ’π‘’ο…βˆ’2ο€Ήπ‘’βˆ’π‘’ο…ο….

Hence, sincosοŠ©οŠ¨οƒοΌοŠ±οƒοΌοŠ©οƒοΌοŠ±οŠ©οƒοΌοŠ«οƒοΌοŠ±οŠ«οƒοΌπœƒπœƒ=116ο€Ό2ο€Ό12π‘–ο€Ήπ‘’βˆ’π‘’ο…οˆ+12π‘–ο€Ήπ‘’βˆ’π‘’ο…βˆ’12π‘–ο€Ήπ‘’βˆ’π‘’ο…οˆ.

We can now use the definition of sine in terms of the complex exponential to rewrite this as sincossinsinsinοŠ©οŠ¨πœƒπœƒ=116(2πœƒ+3πœƒβˆ’5πœƒ).

Part 2

Using our answer from part 1, we can see that sinsinsinsincos5πœƒβˆ’3πœƒ=2πœƒβˆ’16πœƒπœƒ.

Hence, finding the solutions to sinsin5πœƒβˆ’3πœƒ=0 is equivalent to finding the solutions to 0=2πœƒβˆ’16πœƒπœƒ.sinsincos

We can factor this equation as follows: 0=2πœƒο€Ί1βˆ’8πœƒπœƒο†.sinsincos

This has solutions if either sinπœƒ=0 or 1βˆ’8πœƒπœƒ=0sincos. Considering the first case, given that 0β‰€πœƒ<πœ‹, sinπœƒ=0 when πœƒ=0.

We can now consider the case where 1βˆ’8πœƒπœƒ=0sincos. Using the double angle formula for sine, sinsincos2πœƒ=2πœƒπœƒ, we can rewrite this as 1βˆ’22πœƒ=0.sin

Hence, sin2πœƒ=12.

Taking the square root of both sides of the equation, we get sin2πœƒ=Β±1√2.

Starting with the positive square root for πœƒ in the range 0β‰€πœƒ<πœ‹, sin2πœƒ=1√2 when πœƒ=πœ‹8 or 3πœ‹8. Similarly, for the negative square root, sin2πœƒ=βˆ’1√2 when πœƒ=5πœ‹8 or 7πœ‹8.

Hence, the solutions of sinsin5πœƒβˆ’3πœƒ=0 for πœƒ in the range 0β‰€πœƒ<πœ‹ are πœƒ=0,πœ‹8,3πœ‹8,5πœ‹8,7πœ‹8.

Key Points

  • Using Euler’s formula in conjunction with the properties of the exponential function, we can derive many trigonometric identities such as the Pythagorean identity, sum and difference formulas, and multiple angle formulas.
  • We can use Euler’s theorem to express sine and cosine in terms of the complex exponential function as sincosπœƒ=12π‘–ο€Ήπ‘’βˆ’π‘’ο…,πœƒ=12𝑒+𝑒.οƒοΌοŠ±οƒοΌοƒοΌοŠ±οƒοΌ Using these formulas, we can derive further trigonometric identities, such as the sum to product formulas and formulas for expressing powers of sine and cosine and products of the two in terms of multiple angles.
  • We can use the identities derived from Euler’s formula to help us simplify integrals and solve trigonometric equations.

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