Find the exact value of each expression.

Question

1. arcsin (−1/2)

2. csc^−1 (−2)

3. arccot(1)

4. sec^−1 (−2)

5. arccos (1/2)

6. tan^−1 (−1)

Answer 1

(1) Recall that the range of arcsin(x) is [-pi/2, pi/2]. If we let arcsin(-1/2)=A, then sin(A)=-1/2. This means we are looking for an angle A in [-pi/2, pi/2] whose sine value is -1/2. This angle A is clearly -pi/6. Therefore, arcsin(-1/2) = -pi/6.

(2) I assume that csc^-1(x) stands for the inverse of csc(x). Let A=csc^-1(-2); then csc(A)=-2. Since csc(A)=1/sin(A), then 1/sin(A)=-2 or sin(A)=-1/2. Since the range of csc^-1(x) is [-pi/2, 0)U(0, pi/2], then A must be in [-pi/2, 0)U(0, pi/2]. Therefore, sin(A)=-1/2 gives A=-pi/6. Now we conclude that csc^-1(-2)=-pi/6.

(3) Let A=arccot(1). So cot(A)=1. As the range of arccot(x) is (0, pi), then we are looking for an angle in(0,pi) whose cotangent value is1. This angle is clearly pi/4. So arccot(1) =pi/4.

(4) Let A=sec^-1(-2). Then sec(A)=-2. As sec(A)=1/cos(A), then 1/cos(A)=-2 or cos(A)=-1/2. Recall that the range of sec^-1(x) is [0, pi/2)U(pi/2, pi]. So we are looking for angle A in [0, pi/2)U(pi/2, pi] whose cosine value is -1/2. This angle A is clearly 2pi/3. Therefore, sec^-1(-2)=2pi/3.

(5) Let A=arccos(1/2). Then cos(A)=1/2. Since the range of arccos(x) is [0, pi], then we are looking for an angle A in [0, pi] whose cosine value is 1/2; this angle A is clearly pi/3. Therefore, arccos(1/2)=pi/3.

(6) Let A=tan^-1(-1). Then tan(A)=-1. Since the range of tan^-1(x) is (-pi/2, pi/2), then we are looking for an angle A in (-pi/2, pi/2) whose tangent value is -1; this angle A is clearly -pi/4. Therefore, tan^-1(-1)=-pi/4.

Answer 2

(1) arcsin(-1/2)

theta = arcsin(-1/2)

theta = the angle whose sin is (-1/2)

theta = (7/6)pi OR (11/6)pi

theta = (7/6)pi + 2k*pi OR (11/6) + 2k*pi

(2) [ csc{-2) ]^(-1)

= sin(-2) Note: sin(-2) ~ sin(-114 degrees) in Q3

= - sin(2)

(3) ArcCot(1)

theta = ArcCot(1)

theta = the angle whose cot is 1

theta = (1/4)pi OR (5/4)pi

theta = (1/4)pi + 2k*pi OR (5/4)pi + 2k*pi

(4) [ sec(-2) ]^(-1)

= cos(-2) Note: cos(-2) ~ cos(-115 degrees) in Q3

= cos(2)

(5) ArcCos(1/2)

theta = the angle whose cos is 1/2

theta = (1/3)pi OR (5/3)pi

theta = (1/3)pi + 2k*pi OR (5/3)pi + 2k*pi

(6) [ tan(-1) ]^(-1)

= cot(-1) Note; cot(-1) ~ cot(-57 degrees) in Q4

= - cot(1)

NOTES:

When working with trig functions, with standard angles in radians,

the quadrant in which the angle resides is easy to determine,

For example, we know the cos[ (1/3 pi ) is in Q1

When working with trig functions, with non-standard angles in radians,

it may be difficult to determine the quadrant in which the angle resides.

To help determine where the angle is located (i.e. which quadrant)

it may be helpful to convert the radians to degrees.

For example, 2 radians = (2 rad)*(180 deg/ pi rad) = (2*180) / (pi) = 115 deg.