let theta be an angle in quadrant II

Question

Let theta be an angle in quadrant II such that cos(theta) = -4/7

 

Find the exact values of csc(theta) and tan(theta)

Answer 1

Let    x=theta

 

 

First, we use the identity      sin²x + cos²x = 1        to find sin(x).

 

We know that cos²x = 16/49

 

 

sin(x) = √(1 - (16/49))

          = √((49/49) - (16/49))

          = √(33/49)

          = √(33) / 7

 

 

Then,

 

csc(x) = 1 / sin(x)

 

          = 1 / (√(33) / 7)

 

 

 

tan(x) = sin(x) / cos(x)

 

          = (√(33) / 7) / (-4 / 7)

Answer 2

csc(theta) = 1/sin(theta)

 

sin(theta) = sqrt[1-cos^2(theta)] = sqrt[1-(4/7)^2] =sqrt(33)/7

 

csc(theta) = 1/sin(theta) = 1/[sqrt(33)/7] = 7/sqrt(33) = 7(√33)/33

 

tan(theta) = sin(theta)/cos(theta) = [sqrt(33)/7](-7/4) = -(√33)/4

Answer 3

• Since theta is in QII and equal to -4/7, find sine of theta by using the Pythagorean Theorem.
• x^2 + y^2 = r^2,       where x = -4 and r = 7.
• (-4)^2 + y^2 = (7)^2
• 16 + y^2 = 49
• y^2 = 49 - 16 = 33
• y = ±√33.           Because theta is in QII,         y = +√33.
• sin(theta) = y/r = √33/7.
• To solve for csc(theta) realize that csc(theta) = 1/sin(theta).
• csc(theta) = 1/(√33/7) = 7/√33. You must radicalize the denominator.
• 7/√33 * √33/√33 = 7*√33/33.
• The tan(theta) = sin(theta)/cos(theta) = (√33/7)/(-4/7) = √33/7 * 7/-4 =-√33/4.

Answer 4

SOHCAHTOA

 

Cos = adjacent/hypotenuse

Since we are in quadrant II, adjacent is -4

 

csc = hypotenuse/opposite

 

(-4)² + b² = 7²

16 + b² = 49

b² = 33

b = √33

 

csc = 7/√33 = 7√33/33

 

tan = opposite/adjacent = -√33/4