Simplify cot(arcsin(-( square root of 6)/4))

Problem

cot(arcsin(−√(,6)/4))

Solution

1. Identify the angle θ such that θ=arcsin(−√(,6)/4) By the definition of the inverse sine function, sin(θ)=−√(,6)/4 where −π/2≤θ≤π/2

2. Determine the quadrant of θ Since the sine value is negative, θ must be in the fourth quadrant (Q*I*V, where cos(θ) is positive and cot(θ) is negative.

3. Use the Pythagorean identity cos2(θ)+sin2(θ)=1 to find cos(θ)

cos2(θ)+(−√(,6)/4)2=1

cos2(θ)+6/16=1

cos2(θ)=1−3/8

cos2(θ)=5/8

4. Solve for cos(θ) by taking the square root. Since θ is in Q*I*V cos(θ) is positive.

cos(θ)=√(,5/8)

cos(θ)=√(,5)/(2√(,2))

cos(θ)=√(,10)/4

5. Apply the definition of the cotangent function, cot(θ)=cos(θ)/sin(θ)

cot(θ)=√(,10)/4/(−√(,6)/4)

cot(θ)=−√(,10)/√(,6)

6. Simplify the radical expression by dividing the terms inside the square roots and rationalizing the denominator.

cot(θ)=−√(,10/6)

cot(θ)=−√(,5/3)

cot(θ)=−√(,15)/3

Final Answer

cot(arcsin(−√(,6)/4))=−√(,15)/3

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