cos^-1(cos((2pi)/3)) + cos^-1(cos((5pi)/3))
cos^-1(cos((2pi)/3)) + cos^-1(cos((5pi)/3))
= cos^-1 (-1/2) + cos^-1 (.5)
= 2π/3 + π/3
= π
We might be tempted to simply say, well since
cos^-1(cos(k)) = k then
cos^-1(cos((2pi)/3)) + cos^-1(cos((5pi)/3))
= 2π/3 + 5π/3
= 7π/3
but all the arc(trigfunctions) have as output a result closest to zero
e.g. cos (5π/3) = cos(300°) = = 1/2
so cos^-1 (1/2) = ± 60°, ± 300°, ± 420°, etc
which explains why cos^-1(cos(5π/3)) = π/3
You can test this with your calculator:
On most press with calculator set to degrees:
cos
300
= ..... to get .5
2ndF
cos
= ..... to get 60°, not the 300° we started with, (5π/3 radians = 300°)