Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

So, by de Moivre's we know: \[ (\cos \theta+i \sin\theta)^n=\cos(n\theta)+i\sin(n\theta) \]Therefore, we know: \[ \Re\left(\left(\cos\theta+i\sin\theta\right)^5\right)=\cos(5\theta)\\ \Im\left(\left(\cos\theta+i\sin\theta\right)^5\right)=\sin(5\theta) \]Expand the previous identities, and find the real part to give a closed-form expression for \(\cos\theta\), it follows similarly (for the imaginary part) for \(\sin\theta\).
Would my answer then be (cos θ + i sin θ)^5 = cos 5θ + i sin 5θ?
Although the original answer I gave *is* a correct answer. It should not be the final statement, it was more of a hint. Let's solve for \(\cos(\theta)\) and leave the other for you. Using binomial theorem: \[ \left(\cos\theta+i\sin(\theta)\right)^5=\\ \cos^5\theta+C_1i(\sin\theta)(\cos^4\theta)+C_2(i^2\sin^2\theta)(\cos^3\theta)+\cdots+C_5(i^5\sin^5\theta) \]We can ignore all of the odd exponents (excluding the first), since we are looking for the real part, and, thus, we get: \[ \cos^5\theta-10(\sin^2\theta)(\cos^3\theta)+5(\sin^4\theta)(\cos\theta)=\cos(5\theta) \]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question