Question

Please help me do this question mathematicians

Answer 1

De Moivre's theorem says

$\left(\cos(\theta) + i \sin(\theta)\right)^n = \cos(n\theta) + i \sin(n\theta)$

This follows from Euler's identity,

$e^{i\theta} = \cos(\theta) + i \sin(\theta)$

and raising both sides to the power n.

Now, $\cos(4\theta)$ is the real part of

$\cos(4\theta) + i \sin(4\theta)$

By de Moivre's theorem,

$\left(\cos(\theta) + i \sin(\theta)\right)^4 = \cos(4\theta) + i \sin(4\theta)$

Both sides are complex numbers, so if they're equal, their real and imaginary parts must be the same. Expand the left side to get

$\left(\cos(\theta) + i\sin(\theta)\right)^4 \\\\ = \cos^4(\theta) + 4i \cos^3(\theta) \sin(\theta) - 6 \cos^2(\theta)\sin^2(\theta) - 4i \cos(\theta) \sin^3(\theta) + \sin^4(\theta)$

Matching up the real parts, it follows that

$\cos(4\theta) = \cos^4(\theta) - 6 \cos^2(\theta) \sin^2(\theta) + \sin^4(\theta)$

Now just put everything in terms of cos(θ), which we do by using the Pythagorean identity

$\cos^2(\theta) + \sin^2(\theta) = 1$

$\implies \cos(4\theta) = \cos^4(\theta) - 6 \cos^2(\theta) (1 - \cos^2(\theta)) + (1-\cos^2(\theta))^2$

$\implies \cos(4\theta) = \cos^4(\theta) - 6 \cos^2(\theta) + 6 \cos^4(\theta) + 1 - 2\cos^2(\theta) + \cos^4(\theta)$

$\implies \cos(4\theta) = 8\cos^4(\theta) - 8 \cos^2(\theta) + 1$

as required.

Answer 2

Using De Moivre's Theorem, we have been able to prove that; cos 4θ = 8 cos⁴θ - 8cos²θ + 1

How to use De Moivre's theorem?

We want to use De Moivre's Theorem to show that;

cos 4θ = 8 cos⁴θ - 8cos²θ + 1

Now, according to De Moivre's Theorem, we know that;

(cos θ + i sinθ)ⁿ = cos(nθ) + i sin(nθ)

From Euler's identity we know that;

e^(iθ) = cos θ + i sinθ

Now, from our question, we can say that;

cos 4θ is the real part of (cos θ + i sinθ)ⁿ  in De Moivre's Theorem. Thus;

(cos θ + i sinθ)⁴ = cos(4θ) + i sin(4θ)

Now, we see that both sides are complex numbers, and since they are equal, it means that their real and imaginary parts must be the same.

We will expand the LHS to get;

(cos θ + i sinθ)⁴ = cos⁴θ + 4i cos³θ*sinθ - 6cos²θ*sin²θ - 4i cosθ*sin³θ + sin⁴θ

Rearranging to reflect real and imaginary parts gives;

(cos θ + i sinθ)⁴ = (cos⁴θ - 6cos²θ*sin²θ + sin⁴θ) + i(4cos³θ*sinθ - 4cosθ*sin³θ)

Thus;

cos(4θ) = cos⁴θ - 6cos²θ*sin²θ + sin⁴θ

We know that from trigonometric identity that;

sin²θ = (1 - cos²θ)

Thus;

cos(4θ) = cos⁴θ - 6cos²θ*(1 - cos²θ) + (1 - cos²θ)(1 - cos²θ)

⇒  cos⁴θ - 6cos²θ +  6cos⁴θ + 1 - 2cos²θ + cos⁴θ

⇒ 8cos⁴θ - 8cos²θ + 1

Read more about De Moivre's Theorem at; https://brainly.com/question/17120893

#SPJ1