If f(θ)=|( cos )2 θ cos θ sin θ - sin θ cos θ sin θ ( sin )2 θ cos θ sin θ - cos θ 0| then, f((π /6))+f((π /3))+f((π /2))+f((2 π /3))+f((5 π /6))+f(π )+. ldots ldots .+f((53 π /6)) is equal to

Question

If f(θ)=|( cos )2 θ cos θ sin θ - sin θ cos θ sin θ ( sin )2 θ cos θ sin θ - cos θ 0| then, f((π /6))+f((π /3))+f((π /2))+f((2 π /3))+f((5 π /6))+f(π )+. ldots ldots .+f((53 π /6)) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f(θ)= cos 2 θ( cos 2 θ)- cos θ sin θ(- cos θ sin θ)- sin θ(- cos 2 θ sin θ- sin 3 θ) = cos 4 θ+ sin 4 θ+2 cos 2 θ sin 2 θ=( cos 2 θ+ sin 2 θ)2=1 f((π/6))=f((2 π/6))=f((3 π/6))= ldots ldots .=f((53 π/6))=1

Q. If f(θ)=∣∣​(cos)2θcosθsinθsinθ​cosθsinθ(sin)2θ−cosθ​−sinθcosθ0​∣∣​ then, f(6π​)+f(3π​)+f(2π​)+f(32π​)+f(65π​)+f(π)+.…….+f(653π​) is equal to

f(θ)=cos2θ(cos2θ)−cosθsinθ(−cosθsinθ)−sinθ(−cos2θsinθ−sin3θ) =cos4θ+sin4θ+2cos2θsin2θ=(cos2θ+sin2θ)2=1 f(6π​)=f(62π​)=f(63π​)=…….=f(653π​)=1

Q. If $f (θ) = ∣ ∣ (cos)^{2} θ cos θ sin θ sin θ cos θ sin θ (sin)^{2} θ - cos θ - sin θ cos θ 0 ∣ ∣$ then, $f (\frac{π}{6}) + f (\frac{π}{3}) + f (\frac{π}{2}) + f (\frac{2 π}{3}) + f (\frac{5 π}{6}) + f (π) + . \dots\dots . + f (\frac{53 π}{6})$ is equal to