Question

Use De Moivre's theorem to write the complex number in trigonometric form: (cos(3pi/12) + i sin (3pi/12))^5

Answer 1

$\boxed{\text{De Moivre's Theorem: } (cos\theta + isin\theta )^{n} = cosn\theta + isin(n\theta )}$

$(cos\frac{3\pi}{12} + isin\frac{3\pi}{12})^{5} = cos\frac{5 \cdot 3\pi}{12} + isin\frac{5 \cdot 3\pi}{12}$
$= cos\frac{15\pi}{12} + isin\frac{15\pi}{12} \text{ or (A)}$

Answer 2

(cos (3π/12) + i sin (3π/12))⁵ = (cos (15π/12) + i sin (15π/12))

Further explanation

There are many types of numbers in mathematics such as :

• Natural Numbers : 1 , 2 , 3 , 4 , 5 , . . .
• Whole Numbers : 0 , 1 , 2 , 3 , 5 , . . .
• Integers : . . . , - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 , 4 , . . .
• etc

Complex Number consist of Real Number and Imaginary Number and can be expressed as :

$Z = a + b ~ i$

The absolute value of complex number is also called Modulus and can be calculated using this formula :

$|Z| = \sqrt { a^2 + b^2 }$

Let us tackle the problem.

De Moivre's formula for complex numbers is as follows :

$\large {\boxed {[r( \cos \theta + i\sin \theta)]^n = r^n(\cos n\theta + i\sin n\theta)} }$

Using the formula above, we can solve the problem in the following way

$[\cos (\frac{3 \pi}{12}) + i\sin (\frac{3 \pi}{12})]^5 = \cos (5 \times \frac{3 \pi}{12}) + i\sin (5 \times \frac{3 \pi}{12})$

$\large {\boxed {[\cos (\frac{3 \pi}{12}) + i\sin (\frac{3 \pi}{12})]^5 = \cos (\frac{15 \pi}{12}) + i\sin (\frac{15 \pi}{12})}}$

Learn more

• Complex Numbers : brainly.com/question/5056377
• Match Each Product of Complex Numbers : brainly.com/question/1514840
• Graph of The Complex Plane : brainly.com/question/10662770

Answer details

Grade: High School

Subject: Mathematics

Chapter: Complex Numbers

Keywords: Complex , Number , Real , Imaginary , Whole , Natural , Integers