33 - Trigonometry Questions Answers

let z = r(cosθ+i sinθ) and it is given that z¹⁵ = 1  =>  r¹⁵(cosθ+i sinθ)¹⁵ = 1

or r¹⁵(cos15θ+i sin15θ) = 1+0i                   (Here we use Demoivre's theorem)

on comparison we get sin 15θ = 0 => 15θ = nπ          here n is an integer

for n = -7 to +7 we will get 15 answers for which θ will lie between -π/2 to +π/2

max of sin and cos are 1 so max of left is 2

but x² + 1/x² ≥ 2x(1/x) = 2

so only solution will be obtained for 2cos²(x/2)sin²(x) = 2

so (1+cosx)(1-cos²x) = 2 which is impossible,so i think no solution

I think the last term would be cos(14π/15) 

We know that cos(14π/15) = -cos(π/15)

so     16cos(2π/15)cos(4π/15)cos(8π/15)cos(14π/15)

   = -16cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)

   = -[8/sin(π/15)]2sin(π/15)cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)

  =  -[8/sin(π/15)]sin(2π/15)cos(2π/15)cos(4π/15)cos(8π/15)   

take the similar steps