Find the Other Trig Values in Quadrant I sin(theta)=( square root of 3)/6

Problem

sin(θ)=√(,3)/6,0<θ<π/2

Solution

1. Identify the given information and the quadrant. We are given sin(θ)=√(,3)/6 in Quadrant I, where all trigonometric functions are positive.

2. Apply the Pythagorean identity sin2(θ)+cos2(θ)=1 to find the value of cos(θ)

3. Substitute the known value into the identity:

(√(,3)/6)2+cos2(θ)=1

4. Simplify the squared term and solve for cos2(θ)

3/36+cos2(θ)=1

1/12+cos2(θ)=1

cos2(θ)=1−1/12

cos2(θ)=11/12

5. Solve for cos(θ) by taking the square root. Since θ is in Quadrant I, we take the positive root:

cos(θ)=√(,11/12)

cos(θ)=√(,11)/(2√(,3))

cos(θ)=√(,33)/6

6. Find tan(θ) using the identity tan(θ)=sin(θ)/cos(θ)

tan(θ)=√(,3)/6/√(,33)/6

tan(θ)=√(,3)/√(,33)

tan(θ)=1/√(,11)

tan(θ)=√(,11)/11

7. Find the reciprocal functions csc(θ) sec(θ) and cot(θ)

csc(θ)=1/sin(θ)=6/√(,3)=2√(,3)

sec(θ)=1/cos(θ)=6/√(,33)=(6√(,33))/33=(2√(,33))/11

cot(θ)=1/tan(θ)=√(,11)

Final Answer

cos(θ)=√(,33)/6,tan(θ)=√(,11)/11,csc(θ)=2√(,3),sec(θ)=(2√(,33))/11,cot(θ)=√(,11)

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