anonymous
  • anonymous
PRE-CALCULUS: Find the exact value of cos-1 the quantity square root of three divided by two.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\cos^{-1} \left( \frac{ \sqrt{3} }{ 2} \right)\]
anonymous
  • 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 :)
anonymous
  • anonymous
The answer amber provided was for \[\sf \large-\frac{\sqrt{3}}{2}\] by the way

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

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