Question

show that 2cos 2π/7 is a root of X³+x²-2x-1=0​

Answer 1

Answer:

We have

cos2θ=2cos2θ−1cos3θ=cosθ(2cos2θ−1)−2sin2θcosθ=4cos3θ−3cosθ,

so

cos2θ=12(cos2θ+1)cos3θ=14(cos3θ+3cosθ)

Putting these into the equation gives

8cos3θ−4cos2θ−4cosθ+1=2cos3θ+6cosθ−2cos2θ−2−4cosθ+1=−1+2(cos3θ−cos2θ+cosθ)=cos(7x/2)cos(x/2),

the last part of which comes from the formula

∑k=−nn(−1)kcoskx=(−1)ncos(n+1/2)xcos(x/2),

which can be proven by induction. It's then clear that this is zero if x is a zero of cos(7x/2), but not cos(x/2), and the first one of these is θ=π/7.

Step-by-step explanation:

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