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

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

Mathematics
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\[\cos^{-1} \left( \frac{ \sqrt{3} }{ 2} \right)\]
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. Answer provided by Yahoo Answers :)
The answer amber provided was for \[\sf \large-\frac{\sqrt{3}}{2}\] by the way

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Other answers:

I think your answer is wrong, but the way you explain it is right.
The easiest way to go about this sort of problem is simply to utilize the unit circle.
For cos, the range is 0
Yes, I actually based my answer on the unit circle. My answer is \[\frac{ \pi }{ 6 }\]
here is a triangle |dw:1437262431290:dw| so the cosine of which angle is \[\cos(?) = \frac{\sqrt{3}}{2}\]
It is 30 degrees.
right @campbell_st ?
that's correct...
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
@campbell_st Oh, I don't that one. It is Geometry. Thanks for the help! :)
it is a property from geometry that can be applied to any triangle.. so its handy to know for trig... seeing trig is about triangle measurements

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