Explainer: De Moivre’s Theorem for Trigonometric Identities

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 cosπ‘›πœƒ and sinπ‘›πœƒ in terms of powers of cosπœƒ and sinπœƒ. We begin be recalling both de Moivre’s theorem and the binomial theorem.

De Moivre’s Theorem

For any integer 𝑛, (π‘Ÿ(πœƒ+π‘–πœƒ))=π‘Ÿ(π‘›πœƒ+π‘–π‘›πœƒ).cossincossin

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=𝑛!π‘Ÿ!(π‘›βˆ’π‘Ÿ)!. Sometimes, 𝐢 is denoted ο€π‘›π‘ŸοŒ or 𝐢.

In the first example, we will demonstrate the technique by which we can derive these trigonometric identities.

Example 1: Multiple-Angle Formulae

  1. Use de Moivre’s theorem to express sin5πœƒ in terms of powers of sinπœƒ.
  2. By considering the solutions of sin5πœƒ=0, find an exact representation for sinοŠ¨ο€»πœ‹5.

Answer

Part 1

Using de Moivre’s theorem, we have cossincossin5πœƒ+𝑖5πœƒ=(πœƒ+π‘–πœƒ).

Applying the binomial theorem, we get cossincoscossincossincossincossinsin5πœƒ+𝑖5πœƒ=πœƒ+πΆπœƒ(π‘–πœƒ)+πΆπœƒ(π‘–πœƒ)+πΆπœƒ(π‘–πœƒ)+πΆπœƒ(π‘–πœƒ)+(π‘–πœƒ).οŠͺοŠͺοŠͺ

Substituting in the values of 𝐢 and simplifying, we have cossincoscossincossincossincossinsin5πœƒ+𝑖5πœƒ=πœƒ+5π‘–πœƒπœƒ+10π‘–πœƒπœƒ+10π‘–πœƒπœƒ+5π‘–πœƒπœƒ+π‘–πœƒ.οŠͺοŠͺοŠͺ

Evaluating the powers of 𝑖, we get cossincoscossincossincossincossinsin5πœƒ+𝑖5πœƒ=πœƒ+5π‘–πœƒπœƒβˆ’10πœƒπœƒβˆ’10π‘–πœƒπœƒ+5πœƒπœƒ+π‘–πœƒ.οŠͺοŠͺ

Equating the imaginary parts gives sincossincossinsin5πœƒ=5πœƒπœƒβˆ’10πœƒπœƒ+πœƒ.οŠͺ

To eliminate the powers of cosπœƒ, we use the identity cossinοŠ¨οŠ¨πœƒβ‰‘1βˆ’πœƒ. Substituting this in, we have sinsinsinsinsinsin5πœƒ=5ο€Ί1βˆ’πœƒο†πœƒβˆ’10ο€Ί1βˆ’πœƒο†πœƒ+πœƒ.

Expanding the brackets and simplifying give sinsinsinsinsinsinsinsinsinsinsinsinsin5πœƒ=5ο€Ί1βˆ’2πœƒ+πœƒο†πœƒβˆ’10ο€Ίπœƒβˆ’πœƒο†+πœƒ=5πœƒβˆ’10πœƒ+5πœƒβˆ’10πœƒ+10πœƒ+πœƒ.οŠͺ

Gathering like terms, we have sinsinsinsin5πœƒ=16πœƒβˆ’20πœƒ+5πœƒ.

Part 2

We begin by considering the solutions of sin5πœƒ=0. We know that sine is equal to zero at integer multiples of πœ‹. Hence, sin5πœƒ=0 when πœƒ=π‘›πœ‹5 for π‘›βˆˆβ„€. Using our answer from part 1, we have that 16πœƒβˆ’20πœƒ+5πœƒ=0sinsinsin, when πœƒ=π‘›πœ‹5 for π‘›βˆˆβ„€. Factoring sinπœƒ out, we have sinsinsinπœƒο€Ί16πœƒβˆ’20πœƒ+5=0.οŠͺ

We now consider the case where πœƒ=πœ‹5. We know that sinο€»πœ‹5≠0; hence, we can conclude that 16ο€»πœ‹5ο‡βˆ’20ο€»πœ‹5+5=0sinsinοŠͺ. This is a quadratic in sinοŠ¨ο€»πœ‹5. Hence, setting π‘₯=ο€»πœ‹5sin, we can rewrite this as 16π‘₯βˆ’20π‘₯+5=0.

Using the quadratic formula, π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž,

the solutions are given by π‘₯=20±√8032=5±√58.

Therefore, sinοŠ¨ο€»πœ‹5=5+√58 or sinοŠ¨ο€»πœ‹5=5βˆ’βˆš58. By comparing these two values, we can see that sinοŠ¨ο€»πœ‹5=5βˆ’βˆš58.

We can also use de Moivre’s theorem to derive identities for sinοŠπœƒ and cosοŠπœƒ. To do this, we start by defining a complex number 𝑧=πœƒ+π‘–πœƒcossin. Then, we can consider 1𝑧=𝑧=(πœƒ+π‘–πœƒ).cossin

Applying de Moivre’s theorem, we have 1𝑧=(βˆ’πœƒ)+𝑖(βˆ’πœƒ).cossin

Using the odd/even identities for sine and cosine, coscos(βˆ’πœƒ)=πœƒ and sinsin(βˆ’πœƒ)=βˆ’πœƒ, we can rewrite this as 1𝑧=πœƒβˆ’π‘–πœƒ.cossin

Therefore, 𝑧+1𝑧=2πœƒ,π‘§βˆ’1𝑧=2π‘–πœƒ.cossin

Similarly, we can consider 𝑧=(πœƒ+π‘–πœƒ)cossin. Using de Moivre’s theorem, we can rewrite this as 𝑧=π‘›πœƒ+π‘–π‘›πœƒοŠcossin. In a similar way, we consider 1𝑧=𝑧=(πœƒ+π‘–πœƒ).cossin

Using de Moivre’s theorem, we can rewrite this as 1𝑧=(βˆ’π‘›πœƒ)+𝑖(βˆ’π‘›πœƒ).cossin

Applying the odd/even identities for sine and cosine, we get 1𝑧=π‘›πœƒβˆ’π‘–π‘›πœƒ.cossin

Hence, 𝑧+1𝑧=2π‘›πœƒ,π‘§βˆ’1𝑧=2π‘–π‘›πœƒ.cossin

These equations are in fact equivalent to the following formulae for expressing sine and cosine in terms of the exponential function: cossinπ‘›πœƒ=12𝑒+𝑒,π‘›πœƒ=12π‘–ο€Ήπ‘’βˆ’π‘’ο….οƒοŠοΌοŠ±οƒοŠοΌοƒοŠοΌοŠ±οƒοŠοΌ

Using these expressions for sine and cosine in terms of 𝑧, we can derive many trigonometric identities. In the following few examples, we will apply this technique. Both the technique used here and the formulae for sine and cosine in terms of 𝑧 should be committed to memory.

Example 2: Identities for Powers of Cosine

Express cosοŠ¬πœƒ in terms of cos6πœƒ, cos5πœƒ, cos4πœƒ, cos3πœƒ, cos2πœƒ, cosπœƒ, and any constant terms.

Answer

Letting 𝑧=πœƒ+π‘–πœƒcossin, we can write 2πœƒ=𝑧+1𝑧.cos

Raising both sides to the sixth power, we have that 2πœƒ=𝑧+1π‘§οˆ.cos

Hence, cosοŠ¬οŠ¬πœƒ=164𝑧+1π‘§οˆ.

We now apply the binomial theorem to the right-hand side as follows: cosοŠͺοŠͺοŠͺοŠ¬οŠ«οŠ«οŠ¬πœƒ=164𝑧+𝐢𝑧1π‘§οˆ+𝐢𝑧1π‘§οˆ+𝐢𝑧1π‘§οˆ+𝐢𝑧1π‘§οˆ+𝐢𝑧1π‘§οˆ+ο€Ό1π‘§οˆοŠ.

Substituting in the values of 𝐢 and simplifying, we have cosοŠͺοŠͺοŠ¬πœƒ=164𝑧+6𝑧+15𝑧+20+15𝑧+6𝑧+1π‘§οˆ.

We can now group π‘§οŠ terms with 1π‘§οŠ terms as follows: cosοŠͺοŠͺοŠ¨οŠ¨πœƒ=164𝑧+1π‘§οˆ+6𝑧+1π‘§οˆ+15𝑧+1π‘§οˆ+20.

Using 𝑧+1𝑧=2π‘›πœƒοŠοŠcos, we can express this as coscoscoscosοŠ¬πœƒ=164(26πœƒ+6(24πœƒ)+15(22πœƒ)+20).

Finally, we simplify to get coscoscoscosοŠ¬πœƒ=1326πœƒ+3164πœƒ+15322πœƒ+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 ο„Έπœƒπœƒ.ο‘½οŽ‘οŠ¦οŠ­sind

Answer

By applying de Moivre’s theorem, we can express sinοŠ­πœƒ in terms of multiple angles which are simpler to integrate. We begin by setting 𝑧=πœƒ+π‘–πœƒcossin. Then, using 𝑧, we can express sine as 2π‘–πœƒ=π‘§βˆ’1𝑧.sin

Raising both sides to the seventh power, we have that 2π‘–πœƒ=ο€Όπ‘§βˆ’1π‘§οˆ.sin

Since 𝑖=βˆ’π‘–οŠ­ and 2=128, we can divide both sides by βˆ’128𝑖 to get sinοŠ­οŠ­οŠ­πœƒ=βˆ’1128π‘–ο€Όπ‘§βˆ’1π‘§οˆ=𝑖128ο€Όπ‘§βˆ’1π‘§οˆ.

Applying the binomial theorem, we have sinοŠͺοŠͺοŠͺοŠ­οŠ«οŠ¨οŠ«οŠ­οŠ¬οŠ¬οŠ­πœƒ=𝑖128ο€Ύπ‘§βˆ’πΆπ‘§ο€Ό1π‘§οˆ+𝐢𝑧1π‘§οˆβˆ’πΆπ‘§ο€Ό1π‘§οˆ+𝐢𝑧1π‘§οˆβˆ’πΆπ‘§ο€Ό1π‘§οˆ+𝐢𝑧1π‘§οˆβˆ’ο€Ό1π‘§οˆοŠ.

Substituting in the values of 𝐢 and simplifying, we have sinοŠ­οŠ­οŠ«οŠ©οŠ©οŠ«οŠ­πœƒ=𝑖128ο€Όπ‘§βˆ’7𝑧+21π‘§βˆ’35𝑧+35π‘§βˆ’21𝑧+7π‘§βˆ’1π‘§οˆ.

We can now group π‘§οŠ terms with 1π‘§οŠ terms as follows: sinοŠ­οŠ­οŠ­οŠ«οŠ«οŠ©οŠ©πœƒ=𝑖128ο€Όο€Όπ‘§βˆ’1π‘§οˆβˆ’7ο€Όπ‘§βˆ’1π‘§οˆ+21ο€Όπ‘§βˆ’1π‘§οˆβˆ’35ο€Όπ‘§βˆ’1π‘§οˆοˆ.

Using π‘§βˆ’1𝑧=2π‘–π‘›πœƒοŠοŠsin, we can express this as sinsinsinsinsinοŠ­πœƒ=𝑖128(2𝑖7πœƒβˆ’7(2𝑖5πœƒ)+21(2𝑖3πœƒ)βˆ’35(2π‘–πœƒ)).

Simplifying, we get sinsinsinsinsinοŠ­πœƒ=164(35πœƒβˆ’213πœƒ+75πœƒβˆ’7πœƒ).

Substituting this into the integral, we have ο„Έπœƒπœƒ=ο„Έ164(35πœƒβˆ’213πœƒ+75πœƒβˆ’7πœƒ)πœƒ.ο‘½οŽ‘ο‘½οŽ‘οŠ¦οŠ­οŠ¦sindsinsinsinsind

Hence, ο„Έπœƒπœƒ=164ο”ο€Όβˆ’35πœƒ+73πœƒβˆ’755πœƒ+177πœƒοˆο =164ο€Όο€Όβˆ’35πœ‹2+73πœ‹2βˆ’755πœ‹2+177πœ‹2οˆβˆ’ο€Όβˆ’350+70βˆ’750+170=164ο€Ό35βˆ’7+75βˆ’17=1635.ο‘½οŽ‘ο‘½οŽ‘οŠ¦οŠ­οŠ¦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

  1. Express sincosοŠ¨οŠ©πœƒπœƒ in the form π‘Žπœƒ+𝑏3πœƒ+𝑐5πœƒcoscoscos, where π‘Ž,𝑏, and 𝑐 are constants to be found.
  2. Hence, find all the solutions of coscos5πœƒ+3πœƒ=0 in the interval 0β‰€πœƒ<πœ‹. Give your answers in exact form.

Answer

Part 1

Letting 𝑧=πœƒ+π‘–πœƒcossin, we can express sine and cosine in terms of 𝑧 as follows: 2πœƒ=𝑧+1𝑧,2π‘–πœƒ=π‘§βˆ’1𝑧.cossin

Therefore, sincosοŠ¨οŠ©οŠ¨οŠ¨οŠ©οŠ©οŠ¨οŠ©πœƒπœƒ=1(2𝑖)ο€Όπ‘§βˆ’1π‘§οˆο€Ύ12𝑧+1π‘§οˆοŠ=βˆ’132ο€Όπ‘§βˆ’1π‘§οˆο€Όπ‘§+1π‘§οˆ.

Using the binomial theorem, we can expand each bracket separately as follows: sincosοŠ¨οŠ©οŠ¨οŠ¨οŠ©οŠ©πœƒπœƒ=βˆ’132ο€Όπ‘§βˆ’2+1π‘§οˆο€Όπ‘§+3𝑧+3𝑧+1π‘§οˆ.

Multiplying out the two brackets gives sincosοŠ¨οŠ©οŠ«οŠ©οŠ©οŠ©οŠ©οŠ«πœƒπœƒ=βˆ’132𝑧+3𝑧+3𝑧+1π‘§βˆ’2π‘§βˆ’6π‘§βˆ’6π‘§βˆ’2𝑧+𝑧+3𝑧+3𝑧+1π‘§οˆ.

Gathering the π‘§οŠ terms with 1π‘§οŠ terms results in sincosοŠ¨οŠ©οŠ«οŠ«οŠ©οŠ©πœƒπœƒ=βˆ’132𝑧+1π‘§οˆ+𝑧+1π‘§οˆβˆ’2𝑧+1π‘§οˆοˆ.

Using 𝑧+1𝑧=2π‘›πœƒοŠοŠcos, we can express this as sincoscoscoscoscoscoscosοŠ¨οŠ©πœƒπœƒ=βˆ’132(25πœƒ+23πœƒβˆ’2(2πœƒ))=116(2πœƒβˆ’5πœƒβˆ’3πœƒ).

Part 2

Using our answer from part 1, we can see that coscoscossincos5πœƒ+3πœƒ=2πœƒβˆ’16πœƒπœƒ.

Hence, coscos5πœƒ+3πœƒ=0 is equivalent to 0=2πœƒβˆ’16πœƒπœƒ.cossincos

Factoring this expression results in 0=2πœƒο€Ί1βˆ’8πœƒπœƒο†,cossincos

which is true if either cosπœƒ=0 or 1βˆ’8πœƒπœƒ=0sincos. For the first case, given that 0β‰€πœƒ<πœ‹, cosπœƒ=0 when πœƒ=πœ‹2.

Now we consider the case when 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 coscos5πœƒ+3πœƒ=0 for πœƒ in the range 0β‰€πœƒ<πœ‹ are πœƒ=πœ‹8,3πœ‹8,πœ‹2,5πœ‹8,7πœ‹8.

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 formulae for the tangent function.

Example 5: Deriving Trigonometric Identities Involving Tangent

  1. Express sin6πœƒ in terms of powers of sinπœƒ and cosπœƒ.
  2. Express cos6πœƒ in terms of powers of sinπœƒ and cosπœƒ.
  3. Hence, express tan6πœƒ in terms of powers of tanπœƒ.

Answer

Part 1

Using de Moivre’s theorem, we have cossincossin6πœƒ+𝑖6πœƒ=(πœƒ+π‘–πœƒ).

Applying the binomial theorem, we get cossincoscossincossincossincossincossinsin6πœƒ+𝑖6πœƒ=πœƒ+πΆπœƒ(π‘–πœƒ)+πΆπœƒ(π‘–πœƒ)+πΆπœƒ(π‘–πœƒ)+πΆπœƒ(π‘–πœƒ)+πΆπœƒ(π‘–πœƒ)+(π‘–πœƒ).οŠͺοŠͺοŠͺ

Substituting in the values 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(1)

Equating imaginary parts, we have sincossincossincossin6πœƒ=6πœƒπœƒβˆ’20πœƒπœƒ+6πœƒπœƒ.

Part 2

Equating the real parts of equation (1) gives an equation for coscoscossincossinsin6πœƒ=πœƒβˆ’15πœƒπœƒ+15πœƒπœƒβˆ’πœƒ.οŠͺοŠͺ

Part 3

Using the definition of the tangent function in terms of sine and cosine, we have tansincos6πœƒ=6πœƒ6πœƒ.

Using the answers from parts 1 and 2, we can rewrite this as tancossincossincossincoscossincossinsin6πœƒ=6πœƒπœƒβˆ’20πœƒπœƒ+6πœƒπœƒπœƒβˆ’15πœƒπœƒ+15πœƒπœƒβˆ’πœƒ.οŠͺοŠͺ

Dividing both the numerator and the denominator by cosοŠ¬πœƒ, we get tantantantantantantan6πœƒ=6πœƒβˆ’20πœƒ+6πœƒ1βˆ’15πœƒ+15πœƒβˆ’πœƒ.οŠͺ

Key Points

  1. Using de Moivre’s theorem and the binomial theorem, we can derive multiple-angle formulae for different sine, cosine, and tangent functions.
  2. If we define 𝑧=πœƒ+π‘–πœƒcossin, we can express sine and cosine in terms of 𝑧 as follows: 𝑧+1𝑧=2πœƒ,π‘§βˆ’1𝑧=2π‘–πœƒ.cossin We can also express sinπ‘›πœƒ and cosπ‘›πœƒ in terms of 𝑧 as 𝑧+1𝑧=2π‘›πœƒ,π‘§βˆ’1𝑧=2π‘–π‘›πœƒ.cossin Using these equations, we can find expressions for the powers of sine and cosine and even their products.
  3. Using these techniques for deriving trigonometric identities, we can simplify integrals and solve equations.

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