Solve Trig Question: cot(Arctan (-√(5)/2))

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In summary, the value of cot(Arctan (-√(5)/2)) is equal to -2/√(5). To solve this, we can use the identity cot(x) = cos(x)/sin(x) and the right triangle with sides -√(5), 2, and √(5). The relationship between cot(Arctan (-√(5)/2)) and the triangle in the question is that the value of cot(Arctan (-√(5)/2)) is equal to the tangent of the angle in the triangle. This value can be further simplified by rationalizing the denominator. In real-life situations, this solution can be applied in right triangles with sides
  • #1
majormuss
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Homework Statement


How do I solve this trig question?
cot(Arctan (-√(5)/2))

Homework Equations


The usual equation I know of which is easy o identify on the Unit circle is √3/2. I don't know how to approach this one..


The Attempt at a Solution


 
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  • #2
You do know that cot(x)= 1/tan(x) don't you? So cot(Arctan (-√(5)/2))= 1/tan(Arctan (-√(5)/2)).

And what do you think tan(Arctan (-√(5)/2)) is?
 
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Related to Solve Trig Question: cot(Arctan (-√(5)/2))

1. What is the value of cot(Arctan (-√(5)/2))?

The value of cot(Arctan (-√(5)/2)) is equal to -2/√(5).

2. How do you solve cot(Arctan (-√(5)/2))?

To solve cot(Arctan (-√(5)/2)), we can use the identity cot(x) = cos(x)/sin(x). By substituting -√(5)/2 for x in the identity, we get cos(Arctan (-√(5)/2))/sin(Arctan (-√(5)/2)). Then, we can use the right triangle with sides -√(5), 2, and √(5) to find the values of cosine and sine, giving us -2/√(5) for cot(Arctan (-√(5)/2)).

3. What is the relationship between cot(Arctan (-√(5)/2)) and the triangle in the question?

The triangle in the question has sides -√(5), 2, and √(5), and the angle in question is the angle opposite the side -√(5). The value of cot(Arctan (-√(5)/2)) is equal to the adjacent side (2) divided by the opposite side (-√(5)), which is the same as the tangent of the angle in the triangle.

4. Can the value of cot(Arctan (-√(5)/2)) be simplified further?

Yes, the value of cot(Arctan (-√(5)/2)) can be simplified further by rationalizing the denominator. By multiplying both the numerator and denominator by √(5), we get (-2√(5))/5, which is a simplified form of the original value.

5. How can we apply the solution to cot(Arctan (-√(5)/2)) in real-life situations?

The solution to cot(Arctan (-√(5)/2)) can be applied in real-life situations involving right triangles with sides -√(5), 2, and √(5). For example, if we know the length of the adjacent side and the angle opposite that side, we can use the value of cot(Arctan (-√(5)/2)) to find the length of the opposite side.

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