Question

giving medals! And i will fan! help needed :)
cos^-1(cos(-pi/6))

1. What is the value of Cos(-pi/6)

2. To solve the problem cos^-1(cos(-pi/6)) find the angle in the interval [0,pi] whose cosine is Sqrt3/2

Answer 1

not sure what the question is ....

Answer 2

me neither, thats what im trying to figure out

im not sure how to find the value of cos(-pi/6), would that be -sqrt3/2?

Answer 3

@amistre64 ?

Answer 4

well, pi/6 = 30 degrees ... memory stuff. so cos(-pi/6) is the cos of 30 degrees

Answer 5

inverse cosine is restricted, so im not to sure that it will always produce theta

Answer 6

so would it still be pi/6 regardless that is negative? @OOOPS

and that would be at \[\sqrt3/2, 1/2\] @amistre64

Answer 7

cos(pi/6) = cos(-pi/6) since cosine is an even function, but the graph gives a more visual demonstration

Answer 8

one could say that -pi/6 is equal to 2pi - pi/6 in order to keep the angle inside of 0 to 2pi, the range of the inverse

Answer 9

right, im following...
so a. would be Sqrt3/2? as the value
and b would be cos(pi/6)?

Answer 10

seems fair to me. not sure why the interval is from 0 to pi, but yeah