To find the exact value of the trigonometric function, we follow the procedure given below.
Step 1 :
Check the given angle lies in which quadrant.
Step 2 :
Using reference angle decompose the given angle as follows
the same 180 - given angle given angle - 180 360 - given angle |
1st quadrant 2nd quadrant 3rd quadrant 4th quadrant |
Using ASTC, we can fix the sign for the answer.
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
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM