Explainer: Euler’s Formula for Trigonometric Identities

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 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 𝜃, 𝑒=𝜃+𝑖𝜃.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 trigonometry—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 𝑒 in terms of sine and cosine.
  2. Given that 𝑒𝑒=1, what trigonometric identity can be derived by expanding the exponentials in terms of trigonometric functions?

Answer

Part 1

Rewriting 𝑒=𝑒,() we can apply Euler’s formula to get 𝑒=(𝜃)+𝑖(𝜃).cossin

Using the odd/even identities for sine and cosine, sinsincoscos(𝜃)=𝜃,(𝜃)=𝜃, we have 𝑒=𝜃𝑖𝜃.cossin

Part 2

Using Euler’s identity and our answer from part 1, we can rewrite 𝑒𝑒=(𝜃+𝑖𝜃)(𝜃𝑖𝜃).cossincossin

Expanding the parentheses, we have 𝑒𝑒=𝜃𝑖𝜃𝜃+𝑖𝜃𝜃𝑖𝜃.coscossincossinsin

Finally, we can simplify this to get 𝑒𝑒=𝜃+𝜃.cossin

Hence, given that 𝑒𝑒=1, we have proven the trigonometric identity sincos𝜃+𝜃1.

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.()

Now that we have two different but equal expressions, we can apply Euler’s formula to each one to get cossincossincossin(𝜃+𝜑)+𝑖(𝜃+𝜑)=(𝜃+𝑖𝜃)(𝜑+𝑖𝜑).

By expanding the parentheses on the right-hand side, we have cossincoscoscossinsincossinsin(𝜃+𝜑)+𝑖(𝜃+𝜑)=𝜃𝜑+𝑖𝜃𝜑+𝑖𝜃𝜑+𝑖𝜃𝜑.

Gathering together the like terms on the right-hand side, we find that cossincoscossinsincossinsincos(𝜃+𝜑)+𝑖(𝜃+𝜑)=(𝜃𝜑𝜃𝜑)+𝑖(𝜃𝜑+𝜃𝜑).

Finally, we can equate the real and imaginary parts to get the following two trigonometric identities: coscoscossinsinsincossinsincos(𝜃+𝜑)=𝜃𝜑𝜃𝜑,(𝜃+𝜑)=𝜃𝜑+𝜃𝜑.

Example 3: Double Angle Formulas from Euler’s Formula

Use Euler’s formula to derive a formula for cos2𝜃 and sin2𝜃 in terms of sin𝜃 and cos𝜃.

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 2𝜃, it makes sense that we would consider the expression 𝑒. 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.

Applying Euler’s formula to both of these expression, we have cossincossin2𝜃+𝑖2𝜃=(𝜃+𝑖𝜃).

By expanding the parentheses, we have cossincoscossinsincossincossin2𝜃+𝑖2𝜃=𝜃+2𝑖𝜃𝜃+𝑖𝜃=𝜃𝜃+2𝑖𝜃𝜃.

Finally, we can equate the real and imaginary parts to get the following two trigonometric identities: coscossinsincossin2𝜃=𝜃𝜃,2𝜃=2𝜃𝜃.

Using a similar method, we can derive multiple angle formulas for sin𝑛𝜃 and cos𝑛𝜃 for larger values of 𝑛. However, to simplify the derivation, we can use the binomial theorem.

The Binomial Theorem

For any integer 𝑛, (𝑎+𝑏)=𝑎+𝐶𝑎𝑏+𝐶𝑎𝑏++𝐶𝑎𝑏++𝐶𝑎𝑏+𝑏, where 𝐶=𝑛!𝑟!(𝑛𝑟)!. Sometimes, 𝐶 is denoted 𝑛𝑟 or 𝐶.

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 cos6𝜃 and sin6𝜃 in terms of sin𝜃 and cos𝜃.
  2. Hence, express tan6𝜃 in terms of tan𝜃.

Answer

Part 1

Using the properties of the exponential function, we can write 𝑒=𝑒.

Using Euler’s formula, we can rewrite this as cossincossin6𝜃+𝑖6𝜃=(𝜃+𝑖𝜃).

Using the binomial theorem, we can expand the parentheses as follows: cossincoscossincossincossincossincossinsin6𝜃+𝑖6𝜃=𝜃+𝐶𝜃(𝑖𝜃)+𝐶𝜃𝑖𝜃+𝐶𝜃𝑖𝜃+𝐶𝜃𝑖𝜃+𝐶𝜃𝑖𝜃+𝑖𝜃.

Substituting in the vales of 𝐶 and simplifying, we have cossincoscossincossincossincossincossinsin6𝜃+𝑖6𝜃=𝜃+6𝑖𝜃𝜃+15𝑖𝜃𝜃+20𝑖𝜃𝜃+15𝑖𝜃𝜃+6𝑖𝜃𝜃+𝑖𝜃.

Evaluating the powers of 𝑖, we get cossincoscossincossincossincossincossinsin6𝜃+𝑖6𝜃=𝜃+6𝑖𝜃𝜃15𝜃𝜃20𝑖𝜃𝜃+15𝜃𝜃+6𝑖𝜃𝜃𝜃.

By equating the real and imaginary parts, we arrive at the following two identities: coscoscossincossinsinsincossincossincossin6𝜃=𝜃15𝜃𝜃+15𝜃𝜃𝜃,6𝜃=6𝜃𝜃20𝜃𝜃+6𝜃𝜃.

Part 2

Using the identities from part 1, we can write tansincoscossincossincossincoscossincossinsin6𝜃=6𝜃6𝜃=6𝜃𝜃20𝜃𝜃+6𝜃𝜃𝜃15𝜃𝜃+15𝜃𝜃𝜃.

To express this in terms of tan𝜃, we need to divide both the numerator and denominator by cos𝜃, which results in tantantantantantantan6𝜃=6𝜃20𝜃+6𝜃115𝜃+15𝜃𝜃.

One of the interesting applications of Euler’s formula is that we can derive an expression for sin𝜃 and cos𝜃 in terms of 𝑒 and 𝑒. In the first example, we saw that we can express 𝑒 in terms of sine and cosine as

𝑒=𝜃𝑖𝜃.cossin(1)

Adding this to Euler’s formula gives 𝑒+𝑒=𝜃+𝑖𝜃+𝜃𝑖𝜃=2𝜃.cossincossincos

Hence, by dividing by 2, we derive the following formula for cosine: cos𝜃=12𝑒+𝑒.

Similarly, we can derive a formula for sine by considering the difference between equation (1) and Euler’s formula as follows: 𝑒𝑒=𝜃+𝑖𝜃(𝜃𝑖𝜃)=2𝑖𝜃.cossincossinsin

Dividing by 2𝑖, we arrive at sin𝜃=12𝑖𝑒𝑒.

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 𝑒𝑒()()?

Answer

Part 1

We begin by applying the multiplicative property of exponentials to rewrite 𝑒+𝑒=𝑒𝑒+𝑒𝑒.()()

Factoring out the common factor of 𝑒, we have 𝑒+𝑒=𝑒𝑒+𝑒.()()

Using the definition of cosine in terms of the exponential function, cos𝜑=12𝑒+𝑒, we can rewrite this as 𝑒+𝑒=2𝜑𝑒.()()cos

We can now use Euler’s formula to rewrite this as cossincossincoscossin(𝜃+𝜑)+𝑖(𝜃+𝜑)+(𝜃𝜑)+𝑖(𝜃𝜑)=2𝜑(𝜃+𝑖𝜃).

Grouping together the real and imaginary parts on both side of the equations yields ((𝜃+𝜑)+(𝜃𝜑))+𝑖((𝜃+𝜑)+(𝜃𝜑))=2𝜃𝜑+2𝑖𝜃𝜑.coscossinsincoscossincos

By equating the real and imaginary parts, we get the following two trigonometric identities: coscoscoscossincossinsin𝜃𝜑=12((𝜃+𝜑)+(𝜃𝜑)),𝜃𝜑=12((𝜃+𝜑)+(𝜃𝜑)).

Part 2

Using a similar method, we can consider the expression 𝑒𝑒()(). Applying the properties of the exponential function, we can rewrite this as 𝑒𝑒=𝑒𝑒𝑒𝑒.()()

By factoring out 𝑒, we can express this as 𝑒𝑒=𝑒𝑒𝑒.()()

Using the definition of sine in terms of the exponential function, sin𝜑=12𝑖𝑒𝑒, we can rewrite this as 𝑒𝑒=2𝑖𝜑𝑒.()()sin

We can now apply Euler’s formula to both sides of the equation to get cossincossinsincossin(𝜃+𝜑)+𝑖(𝜃+𝜑)((𝜃𝜑)+𝑖(𝜃𝜑))=2𝑖𝜑(𝜃+𝑖𝜃).

Grouping together the real and imaginary parts on both side of the equations yields ((𝜃+𝜑)(𝜃𝜑))+𝑖((𝜃+𝜑)(𝜃𝜑))=2𝜃𝜑+2𝑖𝜃𝜑.coscossinsinsinsincossin

By equating the real and imaginary parts, we get the following two trigonometric identities: sinsincoscoscossinsinsin𝜃𝜑=12((𝜃𝜑)(𝜃+𝜑)),𝜃𝜑=12((𝜃+𝜑)(𝜃𝜑)).

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 sincos𝜃𝜃 in the form 𝑎𝜃+𝑏3𝜃+𝑐5𝜃sinsinsin, where 𝑎, 𝑏, and 𝑐 are constants to be found.
  2. Hence, find the solutions of sinsin5𝜃3𝜃=0 in the interval 0𝜃<𝜋. Give your answer in exact form.

Answer

Part 1

Using the definitions of sine and cosine in terms of the complex exponential, sincos𝜃=12𝑖𝑒𝑒,𝜃=12𝑒+𝑒, we can rewrite the expression as sincos𝜃𝜃=12𝑖𝑒𝑒12𝑒+𝑒.

Hence, sincos𝜃𝜃=132𝑖𝑒𝑒𝑒+𝑒.

Using the binomial theorem, we can expand each of the sets of parentheses as follows: sincos𝜃𝜃=132𝑖𝑒3𝑒+3𝑒𝑒𝑒+2+𝑒.

We can now expand the two parentheses to get sincos𝜃𝜃=132𝑖𝑒3𝑒+3𝑒𝑒+2𝑒6𝑒+6𝑒2𝑒+𝑒3𝑒+3𝑒𝑒.

Gathering the 𝑒 terms with 𝑒 terms results in sincos𝜃𝜃=132𝑖𝑒𝑒𝑒𝑒2𝑒𝑒.

Hence, sincos𝜃𝜃=116212𝑖𝑒𝑒+12𝑖𝑒𝑒12𝑖𝑒𝑒.

We can now use the definition of sine in terms of the complex exponential to rewrite this as sincossinsinsin𝜃𝜃=116(2𝜃+3𝜃5𝜃).

Part 2

Using our answer from part 1, we can see that sinsinsinsincos5𝜃3𝜃=2𝜃16𝜃𝜃.

Hence, finding the solutions to sinsin5𝜃3𝜃=0 is equivalent to finding the solutions to 0=2𝜃16𝜃𝜃.sinsincos

We can factor this equation as follows: 0=2𝜃18𝜃𝜃.sinsincos

This has solutions if either sin𝜃=0 or 18𝜃𝜃=0sincos. Considering the first case, given that 0𝜃<𝜋, sin𝜃=0 when 𝜃=0.

We can now consider the case where 18𝜃𝜃=0sincos. Using the double angle formula for sine, sinsincos2𝜃=2𝜃𝜃, we can rewrite this as 122𝜃=0.sin

Hence, sin2𝜃=12.

Taking the square root of both sides of the equation, we get sin2𝜃=±12.

Starting with the positive square root for 𝜃 in the range 0𝜃<𝜋, sin2𝜃=12 when 𝜃=𝜋8 or 3𝜋8. Similarly, for the negative square root, sin2𝜃=12 when 𝜃=5𝜋8 or 7𝜋8.

Hence, the solutions of sinsin5𝜃3𝜃=0 for 𝜃 in the range 0𝜃<𝜋 are 𝜃=0,𝜋8,3𝜋8,5𝜋8,7𝜋8.

Key Points

  1. 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.
  2. We can use Euler’s theorem to express sine and cosine in terms of the complex exponential function as sincos𝜃=12𝑖𝑒𝑒,𝜃=12𝑒+𝑒. 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.
  3. We can use the identities derived from Euler’s formula to help us simplify integrals and solve trigonometric equations.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.