Wie bewerten Sie `cos ^ -1 (-1 / 2)`?

`color(brown)("Consider the vertex A as being at the origin of an x y graph plane")`
`color(brown)("In which case the length of triangle side AB is always positive.")`
`color(brown)("Also the only way a trig ratio of the triangles vertex A")`
`color(brown)("can be negative is for either x or y to by negative.")`

Lassen Sie den unbekannten Winkel sein `theta`

`cos(/_A) =cos(60^0)= x/("hypotenuse")=x/c = 1/2" "`

Also, wenn dies die Bedingung war (ist es nicht!), Dann `cos^(-1)(1/2)=60^0`

Aber wir haben `cos(theta)= ("adjacent")/("hypotenuse")=x/("hypotenuse")=-1/2->(-1)/2`

Da die Hypotenuse dann positiv ist `x` muss negativ sein

So `cos(120^0)=(-x)/c=-cos(180-120)=-cos(60)=-1/2`

So `color(blue)(theta= 120^0)`

so `cos^(-1)(-1/2)=theta=120^0`

Für Bogenmaß`-> 120/180xxpi = 2/3 pi" radians"`