شارح الدرس: De Moivre’s Theorem for Trigonometric Identities | نجوى شارح الدرس: De Moivre’s Theorem for Trigonometric Identities | نجوى

شارح الدرس: De Moivre’s Theorem for Trigonometric Identities الرياضيات

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

Using the binomial theorem and de Moivre’s theorem, we can express cos𝑛𝜃 and sin𝑛𝜃 in terms of powers of cos𝜃 and sin𝜃. We begin by recalling de Moivre’s theorem.

Theorem: De Moivre’s Theorem

For any integer 𝑛, (𝑟(𝜃+𝑖𝜃))=𝑟(𝑛𝜃+𝑖𝑛𝜃).cossincossin

We note that the left-hand side of the equation is a binomial expression, since it is of the form (𝑎+𝑏). 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 𝑛, (𝑎+𝑏)=𝑎+𝐶𝑎𝑏+𝐶𝑎𝑏++𝐶𝑎𝑏++𝐶𝑎𝑏+𝑏,

where 𝐶=𝑛𝑟(𝑛𝑟). Sometimes, 𝐶 is denoted 𝑛𝑟.

By using de Moivre’s 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 cos(𝜃) and terms of lower powers of cosine on one side, and cos𝑛𝜃 on the other side. Then, we can do the following.

  1. Using de Moivre’s theorem, we have cossincossin𝑛𝜃+𝑖𝑛𝜃=(𝜃+𝑖𝜃).
  2. We use the binomial theorem on the right-hand side to get cossincoscossincossincossinsin𝑛𝜃+𝑖𝑛𝜃=𝜃+𝐶𝜃(𝑖𝜃)+𝐶𝜃(𝑖𝜃)++𝐶𝜃(𝑖𝜃)+(𝑖𝜃).
  3. Multiplying out the (𝑖𝜃)sin terms and moving the 𝑖-terms to the front, we get cossincoscossincossincossinsin𝑛𝜃+𝑖𝑛𝜃=𝜃+𝑖𝐶𝜃𝜃+𝑖𝐶𝜃𝜃++𝑖𝐶𝜃𝜃+𝑖𝜃.
  4. We use the fact that 𝑖=1 to evaluate the powers of 𝑖. This will result in half of the terms in the expansion being real and half being imaginary, as shown: cossincoscossincossincossincossin𝑛𝜃+𝑖𝑛𝜃=𝜃𝐶𝜃𝜃+𝐶𝜃𝜃++𝑖𝐶𝜃𝜃𝐶𝜃𝜃+.
  5. We can then equate the real parts and the imaginary parts of the above equation together. Since we only want to find cos𝜃, we only consider the real parts (if we wanted sin𝜃, we would possibly need to consider the imaginary parts, depending on whether 𝑖 is real or imaginary). This gives us coscoscossincossin𝑛𝜃=𝜃𝐶𝜃𝜃+𝐶𝜃𝜃+.
  6. We use the identity sincos𝜃1𝜃 to eliminate the sine terms: coscoscoscoscoscoscoscoscos𝑛𝜃=𝜃𝐶𝜃1𝜃+𝐶𝜃1𝜃+=1+𝐶𝜃𝐶+𝐶𝜃+𝐶+𝐶𝜃+. We note that the procedure for sin𝜃 is almost the same, except we aim to eliminate cos 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

  1. Use de Moivre’s theorem to express sin5𝜃 in terms of powers of sin𝜃.
  2. By considering the solutions of sin5𝜃=0, find an exact representation for sin𝜋5.

Answer

Part 1

Using de Moivre’s theorem, we have cossincossin5𝜃+𝑖5𝜃=(𝜃+𝑖𝜃).

Applying the binomial theorem to the right-hand side, we get cossincoscossincossincossincossinsin5𝜃+𝑖5𝜃=𝜃+𝐶𝜃(𝑖𝜃)+𝐶𝜃(𝑖𝜃)+𝐶𝜃(𝑖𝜃)+𝐶𝜃(𝑖𝜃)+(𝑖𝜃).

Substituting in the values of 𝐶 and simplifying, we have cossincoscossincossincossincossinsin5𝜃+𝑖5𝜃=𝜃+5𝑖𝜃𝜃+10𝑖𝜃𝜃+10𝑖𝜃𝜃+5𝑖𝜃𝜃+𝑖𝜃.

Evaluating the powers of 𝑖, we get cossincoscossincossincossincossinsin5𝜃+𝑖5𝜃=𝜃+5𝑖𝜃𝜃10𝜃𝜃10𝑖𝜃𝜃+5𝜃𝜃+𝑖𝜃.

Equating the imaginary parts gives sincossincossinsin5𝜃=5𝜃𝜃10𝜃𝜃+𝜃.

To eliminate the powers of cos𝜃, we use the identity cossin𝜃1𝜃. Substituting this in, we have sinsinsinsinsinsin5𝜃=51𝜃𝜃101𝜃𝜃+𝜃.

Expanding the parentheses and simplifying give sinsinsinsinsinsinsinsinsinsinsinsinsin5𝜃=512𝜃+𝜃𝜃10𝜃𝜃+𝜃=5𝜃10𝜃+5𝜃10𝜃+10𝜃+𝜃.

Gathering like terms, we have sinsinsinsin5𝜃=16𝜃20𝜃+5𝜃.

Part 2

We begin by considering the solutions of sin5𝜃=0. We know that sine is equal to zero at integer multiples of 𝜋. Hence, sin5𝜃=0 when 𝜃=𝑛𝜋5 for 𝑛. Using our answer from part 1, we have that 16𝜃20𝜃+5𝜃=0sinsinsin, when 𝜃=𝑛𝜋5 for 𝑛. Factoring sin𝜃 out, we have sinsinsin𝜃16𝜃20𝜃+5=0.

We now consider the case where 𝜃=𝜋5. We know that sin𝜋50; hence, we can conclude that 16𝜋520𝜋5+5=0sinsin. This is a quadratic in sin𝜋5. Hence, setting 𝑥=𝜋5sin, we can rewrite this as 16𝑥20𝑥+5=0.

Using the quadratic formula, 𝑥=𝑏±𝑏4𝑎𝑐2𝑎,

the solutions are given by 𝑥=20±8032=5±58.

Therefore we have two possibilities, sin𝜋5=5+58 or sin𝜋5=558. Now, we know that sin𝜃 is an increasing function for 𝜃0,𝜋2 and that 0𝜃1sin in this interval. Therefore, we have sinsin𝜋5<𝜋4.

As we know that sin𝜋4=12, this means that sin𝜋5<12.

Comparing the two possible answers with 12, we find that 5580.35<12, while 5+580.90>12, meaning only the first answer is correct. Therefore, we have sin𝜋5=558.

We can also use de Moivre’s theorem to derive identities for sin𝑛𝜃 and cos𝑛𝜃. To do this, we start by defining a complex number 𝑧=𝜃+𝑖𝜃cossin. Then, we can consider 1𝑧=𝑧=(𝜃+𝑖𝜃).cossin

Applying de Moivre’s theorem, we have 1𝑧=(𝜃)+𝑖(𝜃).cossin

Using the odd/even identities for sine and cosine, coscos(𝜃)=𝜃 and sinsin(𝜃)=𝜃, we can rewrite this as 1𝑧=𝜃𝑖𝜃.cossin

Therefore, 𝑧+1𝑧=2𝜃,𝑧1𝑧=2𝑖𝜃.cossin

Similarly, we can consider 𝑧=(𝜃+𝑖𝜃)cossin, for 𝑛𝑍. Using de Moivre’s theorem, we can rewrite this as 𝑧=𝑛𝜃+𝑖𝑛𝜃cossin. In a similar way, we consider 1𝑧=𝑧=(𝜃+𝑖𝜃).cossin

Using de Moivre’s theorem, we can rewrite this as 1𝑧=(𝑛𝜃)+𝑖(𝑛𝜃).cossin

Applying the odd/even identities for sine and cosine, we get 1𝑧=𝑛𝜃𝑖𝑛𝜃.cossin

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 𝑧=𝜃+𝑖𝜃cossin. Then, for any 𝑛, we have 𝑧+1𝑧=2𝑛𝜃,𝑧1𝑧=2𝑖𝑛𝜃.cossin

These equations are in fact equivalent to the following formulas for expressing sine and cosine in terms of the exponential function: cossin𝑛𝜃=12𝑒+𝑒,𝑛𝜃=12𝑖𝑒𝑒.

Using the above identities for sine and cosine in terms of 𝑧, we can derive many other trigonometric identities. Both the technique used here and the formulas for sine and cosine in terms of 𝑧 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 cos𝜃 in terms of cos6𝜃, cos5𝜃, cos4𝜃, cos3𝜃, cos2𝜃, cos𝜃, and any constant terms.

Answer

Letting 𝑧=𝜃+𝑖𝜃cossin, we can write 2𝜃=𝑧+1𝑧.cos

Raising both sides to the sixth power, we have that 2𝜃=𝑧+1𝑧.cos

Hence, cos𝜃=164𝑧+1𝑧.

We now apply the binomial theorem to the right-hand side as follows: cos𝜃=164𝑧+𝐶𝑧1𝑧+𝐶𝑧1𝑧+𝐶𝑧1𝑧+𝐶𝑧1𝑧+𝐶𝑧1𝑧+1𝑧.

Substituting in the values of 𝐶 and simplifying, we have cos𝜃=164𝑧+6𝑧+15𝑧+20+15𝑧+6𝑧+1𝑧.

We can now group 𝑧 terms with 1𝑧 terms as follows: cos𝜃=164𝑧+1𝑧+6𝑧+1𝑧+15𝑧+1𝑧+20.

Using 𝑧+1𝑧=2𝑛𝜃cos, we can express this as coscoscoscos𝜃=164(26𝜃+6(24𝜃)+15(22𝜃)+20).

Finally, we simplify to get coscoscoscos𝜃=1326𝜃+3164𝜃+15322𝜃+516.

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 Moivre’s Theorem to Evaluate Trigonometric Integrals

Using de Moivre’s theorem, find the exact value of 𝜃𝜃.sind

Answer

By applying de Moivre’s theorem, we can express sin𝜃 in terms of multiple angles which are simpler to integrate. We begin by setting 𝑧=𝜃+𝑖𝜃cossin. Then, using 𝑧, we can express sine as 2𝑖𝜃=𝑧1𝑧.sin

Raising both sides to the seventh power, we have that 2𝑖𝜃=𝑧1𝑧.sin

Since 𝑖=𝑖 and 2=128, we can divide both sides by 128𝑖 to get sin𝜃=1128𝑖𝑧1𝑧=𝑖128𝑧1𝑧.

Applying the binomial theorem, we have sin𝜃=𝑖128𝑧𝐶𝑧1𝑧+𝐶𝑧1𝑧𝐶𝑧1𝑧+𝐶𝑧1𝑧𝐶𝑧1𝑧+𝐶𝑧1𝑧1𝑧.

Substituting in the values of 𝐶 and simplifying, we have sin𝜃=𝑖128𝑧7𝑧+21𝑧35𝑧+35𝑧21𝑧+7𝑧1𝑧.

We can now group 𝑧 terms with 1𝑧 terms as follows: sin𝜃=𝑖128𝑧1𝑧7𝑧1𝑧+21𝑧1𝑧35𝑧1𝑧.

Using 𝑧1𝑧=2𝑖𝑛𝜃sin, we can express this as sinsinsinsinsin𝜃=𝑖128(2𝑖7𝜃7(2𝑖5𝜃)+21(2𝑖3𝜃)35(2𝑖𝜃)).

Simplifying, we get sinsinsinsinsin𝜃=164(35𝜃213𝜃+75𝜃7𝜃).

Substituting this into the integral, we have 𝜃𝜃=164(35𝜃213𝜃+75𝜃7𝜃)𝜃.sindsinsinsinsind

Hence, 𝜃𝜃=16435𝜃+73𝜃755𝜃+177𝜃=16435𝜋2+73𝜋2755𝜋2+177𝜋2350+70750+170=164357+7517=1635.sindcoscoscoscoscoscoscoscoscoscoscoscos

The applications of de Moivre’s 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

  1. Express sincos𝜃𝜃 in the form 𝑎𝜃+𝑏3𝜃+𝑐5𝜃coscoscos, where 𝑎,𝑏, and 𝑐 are constants to be found.
  2. Hence, find all the solutions of coscos5𝜃+3𝜃=0 in the interval 0𝜃<𝜋. Give your answers in exact form.

Answer

Part 1

Letting 𝑧=𝜃+𝑖𝜃cossin, we can express sine and cosine in terms of 𝑧 as follows: 2𝜃=𝑧+1𝑧,2𝑖𝜃=𝑧1𝑧.cossin

Therefore, sincos𝜃𝜃=1(2𝑖)𝑧1𝑧12𝑧+1𝑧=132𝑧1𝑧𝑧+1𝑧.

Using the binomial theorem, we can expand each parenthesis separately as follows: sincos𝜃𝜃=132𝑧2+1𝑧𝑧+3𝑧+3𝑧+1𝑧.

Multiplying out the two parentheses gives sincos𝜃𝜃=132𝑧+3𝑧+3𝑧+1𝑧2𝑧6𝑧6𝑧2𝑧+𝑧+3𝑧+3𝑧+1𝑧.

Gathering the 𝑧 terms with 1𝑧 terms results in sincos𝜃𝜃=132𝑧+1𝑧+𝑧+1𝑧2𝑧+1𝑧.

Using 𝑧+1𝑧=2𝑛𝜃cos, we can express this as sincoscoscoscoscoscoscos𝜃𝜃=132(25𝜃+23𝜃2(2𝜃))=116(2𝜃5𝜃3𝜃).

Part 2

Using our answer from part 1, we can see that coscoscossincos5𝜃+3𝜃=2𝜃16𝜃𝜃.

Hence, coscos5𝜃+3𝜃=0 is equivalent to 0=2𝜃16𝜃𝜃.cossincos

Factoring this expression results in 0=2𝜃18𝜃𝜃,cossincos

which is true if either cos𝜃=0 or 18𝜃𝜃=0sincos. For the first case, given that 0𝜃<𝜋, cos𝜃=0 when 𝜃=𝜋2.

Now we consider the case when 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 coscos5𝜃+3𝜃=0 for 𝜃 in the range 0𝜃<𝜋 are 𝜃=𝜋8,3𝜋8,𝜋2,5𝜋8,7𝜋8.

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

  1. Express sin6𝜃 in terms of powers of sin𝜃 and cos𝜃.
  2. Express cos6𝜃 in terms of powers of sin𝜃 and cos𝜃.
  3. Hence, express tan6𝜃 in terms of powers of tan𝜃.

Answer

Part 1

Using de Moivre’s theorem, we have cossincossin6𝜃+𝑖6𝜃=(𝜃+𝑖𝜃).

Applying the binomial theorem, we get cossincoscossincossincossincossincossinsin6𝜃+𝑖6𝜃=𝜃+𝐶𝜃(𝑖𝜃)+𝐶𝜃(𝑖𝜃)+𝐶𝜃(𝑖𝜃)+𝐶𝜃(𝑖𝜃)+𝐶𝜃(𝑖𝜃)+(𝑖𝜃).

Substituting in the values 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.(1)

Equating imaginary parts, we have sincossincossincossin6𝜃=6𝜃𝜃20𝜃𝜃+6𝜃𝜃.

Part 2

Equating the real parts of equation (1) gives us an equation for coscoscossincossinsin6𝜃=𝜃15𝜃𝜃+15𝜃𝜃𝜃.

Part 3

Using the definition of the tangent function in terms of sine and cosine, we have tansincos6𝜃=6𝜃6𝜃.

Using the answers from parts 1 and 2, we can rewrite this as tancossincossincossincoscossincossinsin6𝜃=6𝜃𝜃20𝜃𝜃+6𝜃𝜃𝜃15𝜃𝜃+15𝜃𝜃𝜃.

Dividing both the numerator and the denominator by cos𝜃, we get tantantantantantantan6𝜃=6𝜃20𝜃+6𝜃115𝜃+15𝜃𝜃.

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

Key Points

  • Using de Moivre’s theorem and the binomial theorem, we can derive multiple-angle formulas for different sine, cosine, and tangent functions.
  • If we define 𝑧=𝜃+𝑖𝜃cossin, we can express sine and cosine in terms of 𝑧 as follows: 𝑧+1𝑧=2𝜃,𝑧1𝑧=2𝑖𝜃.cossin We can also express sin𝑛𝜃 and cos𝑛𝜃 in terms of 𝑧 as 𝑧+1𝑧=2𝑛𝜃,𝑧1𝑧=2𝑖𝑛𝜃.cossin 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.

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