Resolved: "How to simplify:473cos⁡(13arccos⁡128)+13"

Arianna Villegas

Answered question

2022-03-29

How to simplify:

\frac{4 \sqrt{7}}{3} \cos (\frac{1}{3} \arccos \frac{1}{\sqrt{28}}) + \frac{1}{3}

Lesly Fernandez

Beginner2022-03-30Added 16 answers

Call

θ = \frac{1}{3} \arccos \frac{1}{\sqrt{28}}

. We can use the formula

\cos (3 θ) = 4 \cos^{3} θ - 3 \cos θ

to decude

\frac{1}{\sqrt{28}} = 4 \cos^{3} θ - 3 \cos θ

(1)
The question is asking us to evaluate the quantity

y = \frac{4 \sqrt{7}}{3} \cos θ + \frac{1}{3}

wit

\cos θ

satisfying (1). Applying the substitution

\cos θ = \frac{3 y - 1}{4 \sqrt{7}}

we get

y^{3} - y^{2} - 9 y = - 1

(2)
The last step is to show that

y = 2 (\cos \frac{π}{7} + \cos \frac{2 π}{7} + \cos \frac{3 π}{7})

satisfies (2). This can be done by a straightforward computation, using Werner's formula to deal with the product of cosines. In details we get

y^{2} = 2 (\cos \frac{2 π}{7} + \cos \frac{4 π}{7} + 2 \cos \frac{2 π}{7} - \cos \frac{6 π}{7}) + 6

and

y^{3} = y^{2} \cdot y = 24 \cos \frac{π}{7} + 24 \cos \frac{2 π}{7} + 18 \cos \frac{3 π}{7} + 4 \cos \frac{4 π}{7} + 2 \cos \frac{5 π}{7} + 2 \cos \frac{6 π}{7} + 6

Now we can compute

y^{3} - y^{2} - 9 y = 2 \cos \frac{π}{7} + 4 \cos \frac{2 π}{7} + 2 \cos \frac{4 π}{7} + 2 \cos \frac{5 π}{7} + 4 \cos \frac{6 π}{7}

= 2 \cos \frac{2 π}{7} - 2 \cos \frac{3 π}{7} + 2 \cos \frac{6 π}{7} = - 1

from which follows

y^{3} - y^{2} - 9 y \equiv - 1

The last cosines equality is a well known result and it's the only non trivial step.

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