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------------------------------------------------ By using de Moivre's Theorem, prove that [tex]\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta[/tex].

Hence, find all the roots of the equation [tex]48x^5 - 60x^3 + 15x + 2 = 0[/tex].

To prove the equation cos 5θ = 16 cos^5θ - 20 cos³θ + 5 cosθ using de Moivre's Theorem, we start by representing cos 5θ and cos θ in terms of complex numbers.

Let z = cos θ + i sin θ be a complex number on the unit circle. Applying de Moivre's Theorem, we have z^5 = (cos θ + i sin θ)^5. Expanding this expression and equating the real parts, we obtain the equation cos 5θ = 16 cos^5θ - 20 cos³θ + 5 cosθ. Using de Moivre's Theorem, we can show that cos 5θ is equal to the expression 16 cos^5θ - 20 cos³θ + 5 cosθ. By representing cos θ as a complex number and applying the theorem, we derive the desired equation. This result is useful in finding the roots of the equation 48x^5 - 60x³ + 15x + 2 = 0.

Learn more about Moivre's Theorem here : brainly.com/question/29750103
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