FIND THE EXACT VALUE OF TRIGONOMETRIC FUNCTIONS

Problem 1 :

cot (5π/6)

Solution:

Given angle lies in second quadrant, then the reference angle will be (π - θ)

cot (5π/6) = cot (π - π/6)

For the trigonometric ratios which lies lies in the second quadrant, sin θ and its reciprocal csc θ it is positive. The rest of the trigonometric ratios will have negative sign.

= -cot (π/6)

= -1/tan(π/6)

= -1/(1/√3)

= -√3

Problem 2 :

csc (5π/4)

Solution:

Given angle lies in third quadrant, then the reference angle will be (θ - π)

csc (5π/4) = csc (5π/4 - π)

Lies in third quadrant, tan θ and its reciprocal cot θ it is positive. The rest of the trigonometric ratios will have negative sign.

= csc (π/4)

= -1/sin (π/4)

= -1/(√2/2)

= -2/√2

Doing rationalization, 

= -2√2/2

= -√2

Problem 3 :

cos (π/4)

Solution:

= cos (π/4)

The given angle is less than 90, then we dont have to decompose into further and this is special angle.

= cos (π/4)

= √2/2

Problem 4 :

csc π

Solution:

= csc π

= 1/sin π

= 1/0

= infinity

Problem 5 :

csc (3π/2)

Solution:

= csc (3π/2)

= 1/sin (3π/2)

= 1/(-1)

= -1

Problem 6 :

tan (2π/3)

Solution:

= tan (2π/3)

Given angle lies in second quadrant, then the reference angle will be (π - θ)

= tan (π - 2π/3)

For the trigonometric ratio for which the angle measure lies in the second quadrant, for sin θ and its reciprocal csc θ it is positive. Since the given trigonometric ratio is tan, we use negative sign.

= -tan (π/3)

= -√3

Problem 7 :

sin (5π/6)

Solution:

= sin (5π/6)

Given angle lies in second quadrant, then the reference angle will be (π - θ)

= sin(π - 5π/6)

For the trigonometric ratio for which the angle measure lies in the second quadrant, for sin θ and its reciprocal csc θ it is positive. Since the given trigonometric ratio is sin, the we use positive sign.

= sin(π/6)

= 1/2

Problem 8 :

cos (4π/3)

Solution:

= cos (4π/3)

Given angle lies in third quadrant, then the reference angle will be (θ - π)

= cos (4π/3 - π)

In third quadrant for the trigonometric ratios, tan and cot we have positive sign. The rest of the trigonometric ratios will have negative sign. So,

= -cos (π/3)

= -1/2

Problem 9 :

sec (5π/3)

Solution:

= sec (5π/3)

Given angle lies in fourth quadrant, then the reference angle will be (360 - θ)

= sec (2π - 5π/3)

In fourth quadrant for the trigonometric ratios, cos and sec we will have positive sign. The rest of the trigonometric ratios will have negative sign. So,

= sec (π/3)

= 1/cos (π/3)

= 1/(1/2)

= 2

Problem 10 :

tan (7π/6)

Solution:

= tan (7π/6)

Given angle lies in third quadrant, then the reference angle will be (θ - 180)

= tan (7π/6 - π)

In third quadrant for the trigonometric ratios, tan and cot we have positive sign. The rest of the trigonometric ratios will have negative sign. Then

= tan (π/6)

= 1/√3

Problem 11 :

cot (π/2)

Solution:

= cot (π/2)

= 1/tan(π/2)

= 1/infinity

= 0

Problem 12 :

sec (11π/6)

Solution:

= sec (11π/6)

The given angle lies in fourth quadrant, then the reference angle will be 360 - θ

= sec (2π - 11π/6)

In fourth quadrant for the trigonometric ratios, cos and sec we will have positive sign. The rest of the trigonometric ratios will have negative sign. So,

= sec (π/6)

= 1/cos (π/6)

= 1/√3/2

= 2/√3