# EVALUATE TRIGONOMETRIC FUNCTIONS GIVEN A POINT

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.

 1st quadrant2nd quadrant3rd quadrant4th quadrant All trigonometric ratios are positivesin θ and cosec θ only positivetan θ and cot θ only positivecos 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.

r2 = x2 + y2

= (-2)2 + (-3)2

= 4 + 9

r2 = 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.

r2 = x2 + y2

= (-8)2 + (√17)2

= 64 + 17

r2 = 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.

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r2 = x2 + y2

= (6)2 + (-17)2

= 36 + 289

r2 = 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.

r2 = x2 + y2

= (17)2 + (-7)2

= 289 + 49

r2 = 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.

= -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.

= √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.

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r2 = x2 + y2

= (6)2 + (-2)2

= 36 + 4

r2 = 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.

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r2 = x2 + y2

= (-3)2 + (4)2

= 9 + 16

r= 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.

r2 = x2 + y2

= (√5)2 + (-2)2

= 5 + 4

r2 = 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.

r2 = x2 + y2

= (√7)2 + (-3)2

= 7 + 9

r2 = 16

r = 4

cosec θ = 4/-3

cosec θ = -4/3

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