## anonymous one year ago PRE-CALCULUS: Find the exact value of cos-1 the quantity square root of three divided by two.

1. anonymous

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

2. anonymous

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 :)

3. anonymous

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

4. anonymous

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

5. anonymous

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. campbell_st

here is a triangle |dw:1437262431290:dw| so the cosine of which angle is $\cos(?) = \frac{\sqrt{3}}{2}$

10. anonymous

It is 30 degrees.

11. anonymous

right @campbell_st ?

12. campbell_st

that's correct...

13. campbell_st

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

14. anonymous

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

15. campbell_st

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