Question

Find the exact real value of arccos(-√2/√2)? Dont want the anwser just need an explanation on how to do it. Thank you!

Answer 1

Tell may what teta should be so as to have for cos the value \[\sqrt{2}/2\]

Answer 2

Tell me ..

Answer 3

Thats the problem I dont get this or to aleast start it off

Answer 4

\[y = \cos \Theta \rightarrow \Theta=\cos^{-1} y \]

Answer 5

Is it sqrt(2)/2 or sqrt(2)/sqrt(2)?

Answer 6

Its sqrt -2/ 2 and i have been using this website for refrence I just dont know what to with the numbers given http://www.math.com/tables/algebra/functions/trig/functions.htm

Answer 7

i mean -sqrt(2) / 2

Answer 8

Yes sir

Answer 9

ok so let's assume that arccos(-sqrt(2) / 2) = x, for some value 'x'. We also know that arccos(x) is the inverse function of cos(x) and has a domain of [-1,1] with a range of [0, pi].

Saying that arcos(-sqrt(2)/2) = x is then the same as saying cos(x) = -sqrt(2)/2 because they are inverse function so their x and y coordinates flip. But note that x is in the interval
0 <= x <= pi because that's the range of arccos(x).

For which angle is cos(x) -sqrt(2)/2? (Hint: It's a special angle.)
Once you find the angle, you must then confirm that cos(x) will be negative for that angle.

Answer 10

@cohenbrian12

Answer 11

1/sqrt(2)