Question

PRE-CALCULUS: Find the exact value of cos-1 the quantity square root of three divided by two.

Answer 1

\[\cos^{-1} \left( \frac{ \sqrt{3} }{ 2} \right)\]

Answer 2

we know that the range arccos(x) is 0 ≤ x ≤ π/2 ==> 0° ≤ x ≤ 180°. So we need to find the solution to cos(x) = -√3/2 this in the range of [0°, 180°]. If we construct a 30-60-90 triangle, we see that the 30° angle corresponds to this and so the angle required is x = 180° - 30° = 150°.

Thus, arccos(-√3/2) = 150° = 5π/6.
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Answer 3

The answer amber provided was for \[\sf \large-\frac{\sqrt{3}}{2}\] by the way

Answer 4

I think your answer is wrong, but the way you explain it is right.

Answer 5

I know @LegendarySadist :)

Answer 6

The easiest way to go about this sort of problem is simply to utilize the unit circle.

Answer 7

For cos, the range is 0<y<pi, right?

Answer 8

Yes, I actually based my answer on the unit circle. My answer is \[\frac{ \pi }{ 6 }\]

Answer 9

here is a triangle



so the cosine of which angle is

\[\cos(?) = \frac{\sqrt{3}}{2}\]

Answer 10

It is 30 degrees.

Answer 11

right @campbell_st ?

Answer 12

that's correct...

Answer 13

you just need to remember the exact value triangles...

I always think about the longest side is opposite the largest angle

2 = 90

1 = 30 shortest and smallest angle

Answer 14

@campbell_st Oh, I don't that one. It is Geometry. Thanks for the help! :)