In this explainer, we will learn how to use de Moivreβs theorem to obtain trigonometric identities.

Using the binomial theorem and de Moivreβs theorem, we can express and in terms of powers of and . We begin by recalling de Moivreβs theorem.

### Theorem: De Moivreβs Theorem

For any integer ,

We note that the left-hand side of the equation is a binomial expression, since it is of the form . Thus, let us recall the binomial theorem, which we can use to evaluate expressions of this kind directly.

### Theorem: The Binomial Theorem

For any integer ,

where . Sometimes, is denoted .

By using de Moivreβs theorem and the binomial theorem together, we can express powers of sine and cosine in terms of lower powers. Let us investigate the general form of this method below.

### How To: Deriving Expressions for Powers of Trigonometric Functions

Suppose we want to find a relation between and terms of lower powers of cosine on one side, and on the other side. Then, we can do the following.

- Using de Moivreβs theorem, we have
- We use the binomial theorem on the right-hand side to get
- Multiplying out the terms and moving the -terms to the front, we get
- We use the fact that to evaluate the powers of . This will result in half of the terms in the expansion being real and half being imaginary, as shown:
- We can then equate the real parts and the imaginary parts of the above equation together. Since we only want to find , we only consider the real parts (if we wanted , we would possibly need to consider the imaginary parts, depending on whether is real or imaginary). This gives us
- We use the identity to eliminate the sine terms: We note that the procedure for is almost the same, except we aim to eliminate terms instead.

Having seen the general form that this method takes, let us consider an example where we can demonstrate how the derivation of trigonometric identities works in practice.

### Example 1: Calculating Powers of the Sine Function Using Multiple-Angle Formulas

- Use de Moivreβs theorem to express in terms of powers of .
- By considering the solutions of , find an exact representation for .

### Answer

**Part 1**

Using de Moivreβs theorem, we have

Applying the binomial theorem to the right-hand side, we get

Substituting in the values of and simplifying, we have

Evaluating the powers of , we get

Equating the imaginary parts gives

To eliminate the powers of , we use the identity . Substituting this in, we have

Expanding the parentheses and simplifying give

Gathering like terms, we have

**Part 2**

We begin by considering the solutions of . We know that sine is equal to zero at integer multiples of . Hence, when for . Using our answer from part 1, we have that , when for . Factoring out, we have

We now consider the case where . We know that ; hence, we can conclude that . This is a quadratic in . Hence, setting , we can rewrite this as

Using the quadratic formula,

the solutions are given by

Therefore we have two possibilities, or . Now, we know that is an increasing function for and that in this interval. Therefore, we have

As we know that , this means that

Comparing the two possible answers with , we find that , while , meaning only the first answer is correct. Therefore, we have

We can also use de Moivreβs theorem to derive identities for and . To do this, we start by defining a complex number . Then, we can consider

Applying de Moivreβs theorem, we have

Using the odd/even identities for sine and cosine, and , we can rewrite this as

Therefore,

Similarly, we can consider , for . Using de Moivreβs theorem, we can rewrite this as . In a similar way, we consider

Using de Moivreβs theorem, we can rewrite this as

Applying the odd/even identities for sine and cosine, we get

Hence, adding and subtracting the above derivations, we obtain the following pair of useful identities.

### Identity: Multiple-Angle Formulas in terms of Complex Numbers

Let . Then, for any , we have

These equations are in fact equivalent to the following formulas for expressing sine and cosine in terms of the exponential function:

Using the above identities for sine and cosine in terms of , we can derive many other trigonometric identities. Both the technique used here and the formulas for sine and cosine in terms of should be committed to memory.

Now, in the following few examples, we will demonstrate applications of this identity to various trigonometric problems.

### Example 2: Calculating Powers of the Cosine Function Using Multiple-Angle Identities

Express in terms of , , , , , , and any constant terms.

### Answer

Letting , we can write

Raising both sides to the sixth power, we have that

Hence,

We now apply the binomial theorem to the right-hand side as follows:

Substituting in the values of and simplifying, we have

We can now group terms with terms as follows:

Using , we can express this as

Finally, we simplify to get

Expressing powers of sines and cosines in terms of multiple angles is very useful for evaluating integrals as the next example will demonstrate.

### Example 3: Using de Moivreβs Theorem to Evaluate Trigonometric Integrals

Using de Moivreβs theorem, find the exact value of

### Answer

By applying de Moivreβs theorem, we can express in terms of multiple angles which are simpler to integrate. We begin by setting . Then, using , we can express sine as

Raising both sides to the seventh power, we have that

Since and , we can divide both sides by to get

Applying the binomial theorem, we have

Substituting in the values of and simplifying, we have

We can now group terms with terms as follows:

Using , we can express this as

Simplifying, we get

Substituting this into the integral, we have

Hence,

The applications of de Moivreβs theorem are not limited to simple powers of sine and cosine; we can also find expressions for the product of powers of sine and cosine.

### Example 4: Solving Trigonometric Equations Using Products of Powers of Sine and Cosine Functions

- Express in the form , where , and are constants to be found.
- Hence, find all the solutions of in the interval . Give your answers in exact form.

### Answer

**Part 1**

Letting , we can express sine and cosine in terms of as follows:

Therefore,

Using the binomial theorem, we can expand each parenthesis separately as follows:

Multiplying out the two parentheses gives

Gathering the terms with terms results in

Using , we can express this as

**Part 2**

Using our answer from part 1, we can see that

Hence, is equivalent to

Factoring this expression results in

which is true if either or . For the first case, given that , when .

Now we consider the case when . Using the double-angle formula for sine,

we can rewrite this as

Hence,

Taking the square root of both sides of the equation, we get

Starting with the positive square root, for in the range , when or . Similarly, for the negative square root, when or .

Hence, the solutions of for in the range are

The techniques used for deriving trigonometric identities can also be applied to other trigonometric functions such as the tangent and cotangent functions. The following example will demonstrate how we can find multiple-angle formulas for the tangent function.

### Example 5: Deriving Trigonometric Identities Involving the Tangent Function

- Express in terms of powers of and .
- Express in terms of powers of and .
- Hence, express in terms of powers of .

### Answer

**Part 1**

Using de Moivreβs theorem, we have

Applying the binomial theorem, we get

Substituting in the values of and simplifying, we have

Evaluating the powers of , we get

Equating imaginary parts, we have

**Part 2**

Equating the real parts of equation (1) gives us an equation for

**Part 3**

Using the definition of the tangent function in terms of sine and cosine, we have

Using the answers from parts 1 and 2, we can rewrite this as

Dividing both the numerator and the denominator by , we get

Let us finish by recapping the key points we have learned in this explainer.

### Key Points

- Using de Moivreβs theorem and the binomial theorem, we can derive multiple-angle formulas for different sine, cosine, and tangent functions.
- If we define , we can express sine and cosine in terms of as follows: We can also express and in terms of as Using these equations, we can find expressions for the powers of sine and cosine and even their products.
- Using these techniques for deriving trigonometric identities, we can simplify integrals and solve equations.