Question

Find cos(β/2) and cot(β/2), if sin(β)= -(√(3)/2) and pi<β<(3pi/2)

Answer 1

answer: cos(β/2) = 1/2 and cot(β/2) = -sqrt(3,

Given sin()=-((3)/2, we can use the Pythagorean identity to calculate the value of cos():

cos(β) sqrt(1 - sin2()) sqrt(1 - sqrt(3)/2) = square (1 - 3/4) Equals square (1/4) = square 1/2

We know that is in the third quadrant, where cosine is negative, since pi (3pi/2). As a result, we have:

cos(β) = -1/2

Now we can calculate cos(/2) and cot(/2) using the half-angle formulas:

sqrt((1 + cos())/2) = cos(/2) = ±sqrt((1 - 1/2)/2) = ±sqrt(1/4) = ±1/2

We know that /2 is in the fourth quadrant, where tangent is negative, because is in the third quadrant. As a result, we have:

Tan(1/2) = -1/tan(1/2) = -1/sqrt((1 + cos())/1 - cos()) = -1/sqrt((1 - 1/2)/(1 + 1/2)) = -1/sqrt(1/3) = -sqrt (3)

The answers are therefore cos(/2) = 1/2 and cot(/2) = -sqrt (3). Nonetheless, we must establish the sign of cos(/2). We can exploit the fact that sine is negative in the third quadrant, where is located, to accomplish this. As a result, we have:

Sin(/2) is equal to sqrt((1 - cos())/2). = -sqrt((1 + 1/2)/2) = squared(3/4) = squared(3)/2

Given that cosine is positive and /2 is in the fourth quadrant, we have:

sqrt(1 - sin2(/2)) = cos(/2) 1/2 = sqrt(1 - 3/4) = sqrt(1/4)

Therefore, the solutions are cos(β/2) = 1/2 and cot(β/2) = -sqrt(3,

Answer 2

Explanation:

We know that the sine of an angle is negative in the third and fourth quadrants. So, since β is in the third quadrant (pi < β < (3pi/2)), we know that sin(β) is negative.

Given sin(β) = -√(3)/2, we can use the half-angle formulas for sine to find sin(β/2):

sin(β/2) = ±√[(1 - cos(β))/2]

Since β is in the third quadrant, we know that cos(β) is negative. We can use the Pythagorean identity to find cos(β):

cos²(β) + sin²(β) = 1

cos²(β) = 1 - sin²(β)

cos(β) = -√(1 - sin²(β))

cos(β) = -√(1 - (3/4))

cos(β) = -√(1/4)

cos(β) = -1/2

Now we can substitute cos(β) into the half-angle formula for sine:

sin(β/2) = ±√[(1 - cos(β))/2]

sin(β/2) = ±√[(1 - (-1/2))/2]

sin(β/2) = ±√(3/4)

sin(β/2) = ±(√3/2)

Since β is in the third quadrant, we know that cos(β/2) is negative. So we can use the half-angle formula for cosine to find cos(β/2):

cos(β/2) = ±√[(1 + cos(β))/2]

cos(β/2) = ±√[(1 - 1/2)/2]

cos(β/2) = ±√(1/4)

cos(β/2) = ±(1/2)

Since β is in the third quadrant, we know that cot(β/2) is negative. We can use the half-angle formulas for cosine and tangent to find cot(β/2):

cos(β/2) = ±√[(1 + cos(β))/2]

cos(β/2) = ±√[(1 - 1/2)/2]

cos(β/2) = ±√(1/4)

cos(β/2) = ±(1/2)

cot(β/2) = cos(β/2) / sin(β/2)

cot(β/2) = (±1/2) / (±(√3/2))

cot(β/2) = ±(1/√3)

cot(β/2) = ±(√3/3)

Therefore, cos(β/2) is either -1/2 or 1/2, and cot(β/2) is either -√3/3 or √3/3, depending on the choice of sign for the square roots.