Lesson Explainer: De Moivre’s Theorem for Trigonometric Identities Mathematics

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 by recalling de Moivre’s theorem.

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

We note that the left-hand side of the equation is a binomial expression, since it is of the form (π‘Ž+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, (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.

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(πœƒ) and terms of lower powers of cosine on one side, and cosπ‘›πœƒ on the other side. Then, we can do the following.

  1. Using de Moivre’s theorem, we have cossincossinπ‘›πœƒ+π‘–π‘›πœƒ=(πœƒ+π‘–πœƒ).
  2. We use the binomial theorem on the right-hand side to get cossincoscossincossincossinsinπ‘›πœƒ+π‘–π‘›πœƒ=πœƒ+πΆπœƒ(π‘–πœƒ)+πΆπœƒ(π‘–πœƒ)+β‹―β‹―+πΆπœƒ(π‘–πœƒ)+(π‘–πœƒ).
  3. Multiplying out the (π‘–πœƒ)sin terms and moving the 𝑖-terms to the front, we get cossincoscossincossincossinsinπ‘›πœƒ+π‘–π‘›πœƒ=πœƒ+π‘–β‹…πΆπœƒπœƒ+π‘–β‹…πΆπœƒπœƒ+β‹―β‹―+π‘–β‹…πΆπœƒπœƒ+π‘–πœƒ.
  4. We use the fact that 𝑖=βˆ’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: cossincoscossincossincossincossinπ‘›πœƒ+π‘–π‘›πœƒ=πœƒβˆ’πΆπœƒπœƒ+πΆπœƒπœƒ+β‹―+π‘–ο€ΊπΆπœƒπœƒβˆ’πΆπœƒπœƒ+⋯.οŠͺοŠͺοŠͺ
  5. We can then equate the real parts and the imaginary parts of the above equation together. Since we only want to find cosοŠπœƒ, we only consider the real parts (if we wanted sinοŠπœƒ, we would possibly need to consider the imaginary parts, depending on whether π‘–οŠ is real or imaginary). This gives us coscoscossincossinπ‘›πœƒ=πœƒβˆ’πΆπœƒπœƒ+πΆπœƒπœƒ+β‹―.οŠͺοŠͺοŠͺ
  6. We use the identity sincosοŠ¨οŠ¨πœƒβ‰‘1βˆ’πœƒ to eliminate the sine terms: coscoscoscoscoscoscoscoscosπ‘›πœƒ=πœƒβˆ’πΆπœƒο€Ή1βˆ’πœƒο…+πΆπœƒο€Ή1βˆ’πœƒο…+β‹―=ο€Ή1+πΆο…πœƒβˆ’ο€ΉπΆ+πΆο…πœƒ+𝐢+πΆο…πœƒ+β‹―.οŠͺοŠͺοŠͺοŠͺοŠͺ We note that the procedure for sinοŠπœƒ 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 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 to the right-hand side, 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 parentheses 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 we have two possibilities, sinοŠ¨ο€»πœ‹5=5+√58 or sinοŠ¨ο€»πœ‹5=5βˆ’βˆš58. Now, we know that sinπœƒ is an increasing function for πœƒβˆˆο“0,πœ‹2 and that 0β‰€πœƒβ‰€1sin in this interval. Therefore, we have sinsinοŠ¨οŠ¨ο€»πœ‹5<ο€»πœ‹4.

As we know that sinοŠ¨ο€»πœ‹4=12, this means that sinοŠ¨ο€»πœ‹5<12.

Comparing the two possible answers with 12, we find that 5βˆ’βˆš58β‰ˆ0.35<12, while 5+√58β‰ˆ0.90>12, meaning only the first answer is correct. Therefore, we have sinοŠ¨ο€»πœ‹5=5βˆ’βˆš58.

We can also use de Moivre’s theorem to derive identities for sinπ‘›πœƒ and cosπ‘›π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, for π‘›βˆˆπ‘. 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, 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=πœƒ+π‘–πœƒcossin. Then, for any π‘›βˆˆβ„€, we have 𝑧+1𝑧=2π‘›πœƒ,π‘§βˆ’1𝑧=2π‘–π‘›πœƒ.cossin

These equations are in fact equivalent to the following formulas 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 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 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 Using Products of Powers of Sine and Cosine Functions

  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 parenthesis 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 parentheses 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 formulas for the tangent function.

Example 5: Deriving Trigonometric Identities Involving the Tangent Function

  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 us 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

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=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.
  • Using these techniques for deriving trigonometric identities, we can simplify integrals and solve equations.

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