How to simplify: \(\displaystyle{\frac{{{4}\sqrt{{{7}}}}}{{{3}}}}{\cos{{\left({\frac{{{1}}}{{{3}}}}{\arccos{{\frac{{{1}}}{{\sqrt{{{28}}}}}}}}\right)}}}+{\frac{{{1}}}{{{3}}}}\)

Arianna Villegas

Arianna Villegas

Answered question

2022-03-29

How to simplify:
473cos(13arccos128)+13

Answer & Explanation

Lesly Fernandez

Lesly Fernandez

Beginner2022-03-30Added 16 answers

Call θ=13arccos128. We can use the formula cos(3θ)=4cos3θ3cosθ to decude
128=4cos3θ3cosθ (1)
The question is asking us to evaluate the quantity
y=473cosθ+13
wit cosθ satisfying (1). Applying the substitution cosθ=3y147 we get
y3y29y=1 (2)
The last step is to show that
y=2(cosπ7+cos2π7+cos3π7)
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
y2=2(cos2π7+cos4π7+2cos2π7cos6π7)+6
and
y3=y2y=24cosπ7+24cos2π7+18cos3π7+4cos4π7+2cos5π7+2cos6π7+6
Now we can compute
y3y29y=2cosπ7+4cos2π7+2cos4π7+2cos5π7+4cos6π7
=2cos2π72cos3π7+2cos6π7=1
from which follows
y3y29y1
The last cosines equality is a well known result and it's the only non trivial step.

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