Since arctan and tan are inverse operations, then logically
arctan(tan(-31π/6)) = -31π/6
however .... since it probably wants the angle closest to the origin, and
since tan(-31π/6) = tan(-π/6)
arctan(tan(-31π/6)
= arctan(tan(-π/6)
= -π/6
Evaluate the following expression. Your answer must be in exact form: for example, type pi/6 for π6 or DNE if the expression is undefined.
arctan(tan(−31π6))
4 answers
I am just wondering how you find tan(-π/6) from tan(-31π/6)
one rotation is 2π
so -31π/6 ÷ 2π = -5 - π/6
so we have 5 clockwise rotations and another clockwise of π/6
(remember in trig, counterclockwise is positive and clockwise is negative)
or
consider the rotations in degrees
31π/6 radians = 930° = 5(360°) + 30° , then
-31π/6 radians = -930° = 5(-360°) - 30°
check with your calculator:
tan 930° = tan30°
tan (-930°) = tan (-30°
so -31π/6 ÷ 2π = -5 - π/6
so we have 5 clockwise rotations and another clockwise of π/6
(remember in trig, counterclockwise is positive and clockwise is negative)
or
consider the rotations in degrees
31π/6 radians = 930° = 5(360°) + 30° , then
-31π/6 radians = -930° = 5(-360°) - 30°
check with your calculator:
tan 930° = tan30°
tan (-930°) = tan (-30°
oh i see! thank you!
5 answers