Find the distance between the points with [a cos[theta + 2pi/3], acos[theta+2pi/3], and[a cos[theta + pi/3] a sin [theta + pi/3
Answers
we have to find the distance between the points
[a cos(θ + 2π/3), a sin(θ + 2π/3) ] and [a cos(θ + π/3), a sin(θ + π/3)]
To find : using distance formula,
distance between the points = √[a²{cos(θ + 2π/3)-cos(θ + π/3)}² + a²{sin(θ + 2π/3) - sin(θ + π/3)}²]
= a√[cos²(θ + 2π/3) + cos²(θ + π/3) - 2cos(θ + 2π/3) cos(θ + π/3) + sin²(θ + 2π/3) + sin²(θ + π/3) - 2sin(θ + 2π/3) sin(θ + π/3)]
= a√[{cos²(θ + 2π/3) + sin²(θ + 2π/3)} + {cos²(θ + π/3) + sin²(θ + π/3) - 2{cos(θ + 2π/3) cos(θ + π/3) + sin(θ + 2π/3) sin(θ + π/3)}]
[ highlighted term is similar as cosA cosB + sinA sinB. but we know, cos(A - B) = cosA cosB + sinA sin B so, cos(θ + 2π/3) cos(θ + π/3) + sin(θ + 2π/3) sin(θ + π/3) = cos(θ + 2π/3 - θ - π/3) = cos(π/3) ]
= a√[ 1 + 1 + 2cos(π/3) ]
= 2a
Therefore the distance between the given points is 2 units.