To evaluate the given trigonometric function given at a point, we have to follow the steps given below.
Step 1 :
Consider the terminal side, in which quadrant it lies. Draw the perpendicular from the terminal point to the x-axis.
Step 2 :
Using ASTC formula, we can fix the sign of the trigonometric ratio that we are going to evaluate.
1^{st} quadrant 2^{nd} quadrant 3^{rd} quadrant 4^{th} quadrant |
All trigonometric ratios are positive sin θ and cosec θ only positive tan θ and cot θ only positive cos θ and sec θ only positive |
Step 3 :
Evaluate hypotenuse if it is needed to evaluate the trigonometric ratio.
Use the given point on the terminal side of angle θ to find the value of the trigonometric function indicated.
Problem 1 :
csc θ
Solution :
Since the terminal side lies in the third quadrant, for the trigonometric ratios tan θ and cot θ, we will have positive.
cosec θ = hypotenuse/opposite
Let 'r' be the hypotenuse.
By using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
= (-2)^{2} + (-3)^{2}
= 4 + 9
r^{2} = 13
r = √13
cosec θ = √13/-3
cosec θ = -√13/3
Problem 2 :
csc θ
Solution :
Since the terminal side lies in the second quadrant, for the trigonometric ratios sin θ and cosec θ, we will have positive.
cosec θ = hypotenuse/opposite
Let 'r' be the hypotenuse.
By using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
= (-8)^{2} + (√17)^{2}
= 64 + 17
r^{2} = 81
r = 9
csc θ = 9/√17
Problem 3 :
cos θ
Solution :
Since the terminal side lies in the fourth quadrant, for the trigonometric ratios cos θ and sec θ, we will have positive.
cos θ = adjacent/hypotenuse
Let 'r' be the hypotenuse.
By using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
= (6)^{2} + (-17)^{2}
= 36 + 289
r^{2} = 325
r = 5√15
cos θ = 6/5√15
Problem 4 :
csc θ
Solution :
Since the terminal side lies in the fourth quadrant, for the trigonometric ratios cos θ and sec θ, we will have positive.
cosec θ = hypotenuse/opposite
Let 'r' be the hypotenuse.
By using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
= (17)^{2} + (-7)^{2}
= 289 + 49
r^{2} = 338
r = 13√2
cosec θ = 13√2/-7
cosec θ = -13√2/7
Problem 5 :
cot θ
Solution :
Since the terminal side lies in the second quadrant, for the trigonometric ratios sin θ and cosec θ, we will have positive.
cot θ = adjacent/opposite
= -3/4
cot θ = -3/4
Problem 6 :
cot θ
Solution :
Since the terminal side lies in the first quadrant, for all trigonometric ratios it is positive.
cot θ = adjacent/opposite
= √13/6
Problem 7 :
cos θ
Solution :
Since the terminal side lies in the fourth quadrant, for the trigonometric ratios cosθ and secθ it is positive.
cos θ = adjacent/hypotenuse
Let 'r' be the hypotenuse.
By using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
= (6)^{2} + (-2)^{2}
= 36 + 4
r^{2} = 40
r = √40
cos θ = 6/√40
Problem 8 :
cos θ
Solution :
Since the terminal side lies in the second quadrant, for the trigonometric ratios sin θ and cosecθ it is positive.
cos θ = adjacent/hypotenuse
Let 'r' be the hypotenuse.
By using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
= (-3)^{2} + (4)^{2}
= 9 + 16
r^{2 }= 25
r = 5
cos θ = -3/5
Problem 9 :
csc θ
Solution :
Since the terminal side lies in the fourth quadrant, for the trigonometric ratios cos θ and secθ it is positive.
cosec θ = hypotenuse/opposite
Let 'r' be the hypotenuse.
By using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
= (√5)^{2} + (-2)^{2}
= 5 + 4
r^{2} = 9
r = 3
cosec θ = 3/-2
cosec θ = -3/2
Problem 10 :
csc θ
Solution :
Since the terminal side lies in the fourth quadrant, for the trigonometric ratios cos θ and secθ it is positive.
cosec θ = hypotenuse/opposite
Let 'r' be the hypotenuse.
By using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
= (√7)^{2} + (-3)^{2}
= 7 + 9
r^{2} = 16
r = 4
cosec θ = 4/-3
cosec θ = -4/3
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM