If cos θ, cos (60°-θ), cos (60°+θ) in order are three consecutive terms of A.P, then the value of cos 2 θ is equal to

Question

If cos θ, cos (60°-θ), cos (60°+θ) in order are three consecutive terms of A.P, then the value of cos 2 θ is equal to (A) (11/12) (B) (12/13) (C) (13/14) (D) (14/15)

Tardigrade Admin · Accepted Answer

2 cos (60-θ)= cos θ+ cos (60+θ) 2[( cos θ/2)+(√3/2) sin θ]= cos θ+( cos θ/2)-(√3/2) sin θ ⇒ tan θ=(1/3 √3) ⇒ cos 2 θ=(1- tan 2 θ/1+ tan 2 θ)=(27-1/27+1)=(13/14)

Q. If $cos θ, cos (6 0^{\circ} - θ), cos (6 0^{\circ} + θ)$ in order are three consecutive terms of A.P, then the value of $cos 2 θ$ is equal to

$\frac{11}{12}$

$\frac{12}{13}$

$\frac{13}{14}$

$\frac{14}{15}$

$2 cos (60 - θ) = cos θ + cos (60 + θ)$
$2 [\frac{c o s θ}{2} + \frac{3}{2} sin θ] = cos θ + \frac{c o s θ}{2} - \frac{3}{2} sin θ$
$\Rightarrow tan θ = \frac{1}{3 3}$
$\Rightarrow cos 2 θ = \frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ} = \frac{27 - 1}{27 + 1} = \frac{13}{14}$

Q. If cosθ,cos(60∘−θ),cos(60∘+θ) in order are three consecutive terms of A.P, then the value of cos2θ is equal to

1211​

1312​

1413​

1514​