EVALUATING TRIGONOMETRIC FUNCTIONS GIVEN A POINT ON THE TERMINAL SIDE

The coordinates of a point P on the terminal arm of each angle are shown. Write the exact trigonometric ratios sin θ, cos θ, and tan θ for each.

Problem 1 :

terminalpointq1

Solution :

By drawing a perpendicular line from P to the x-axis, we will get a right triangle. Then OP be the hypotenuse.

Use the Pythagorean Theorem,

OP = √(3² + 4²)

OP = √(9 + 16)

OP = √25

OP = 5

Opposite side = 4

Adjacent side = 3

Hypotenuse = 5

sin θ = Opposite side / Hypotenuse

sin θ = 4/5

cos θ = Adjacent side / Hypotenuse

cos θ = 3/5

tan θ = Opposite side / Adjacent side

tan θ = 4/3

Problem 2 :

terminalpointq2

Solution :

By drawing a perpendicular line from P to the x-axis, we will get a right triangle. Then OP be the hypotenuse.

OP = √(-12)² + (-5)²

OP = √(144 + 25)

OP = √169

OP = 13

Opposite side = -5

Adjacent side = -12

Hypotenuse = 13

sin θ = Opposite side / Hypotenuse

sin θ = -5/13

cos θ = Adjacent side / Hypotenuse

cos θ = -12/13

tan θ = Opposite side / Adjacent side

tan θ = -5/-12

tan θ = 5/12

Problem 3 :

terminalpointq3

Solution :

By drawing a perpendicular line from P to the x-axis, we will get a right triangle. Then OP be the hypotenuse.

OP = √8² + (-15)²

OP = √(64 + 225)

OP = √289

OP = 17

Opposite side = -15

Adjacent side = 8

Hypotenuse = 17

sin θ = Opposite side / Hypotenuse

sin θ = -15/17

cos θ = Adjacent side / Hypotenuse

cos θ = 8/17

tan θ = Opposite side / Adjacent side

tan θ = -15/8

Problem 4 :

terminalpointq4

Solution :

OP = √ (1)² + (-1)²

OP = √ (1 + 1)

OP = √2

Opposite side = -1

Adjacent side = 1

Hypotenuse = √2

sin θ = Opposite side / Hypotenuse

sin θ = -1/√2

sin θ = -√2/2

cos θ = Adjacent side / Hypotenuse

cos θ = 1/√2

cos θ = √2/2

tan θ = Opposite side / Adjacent side

tan θ = -1/1

tan θ = -1

Determine the exact of sin θ, cos θ, and tan θ if the terminal arm of an angle in standard position passes through the given point.

Problem 5 :

P (-5, 12)

Solution :

Since the given point is in the form of (-x, y), it will be in second quadrant.

Accordingly ASTC, for the trigonometric ratios sin θ and cosec θ alone we have positive.

Finding hypotenuse :

√(-5)² + (12)²

√(25 + 144)

√169

13

Opposite side = 12

Adjacent side = -5

Hypotenuse = 13

sin θ = Opposite side / Hypotenuse

sin θ = 12/13

cos θ = Adjacent side / Hypotenuse side

cos θ = -5/13

tan θ = Opposite side / Adjacent side

tan θ = -12/5

Problem 6 :

P (5, -3)

Solution :

Since the given point is in the form of (x, -y), it will be in fourth quadrant.

Accordingly ASTC, for the trigonometric ratios cos θ and sec θ alone we have positive.

Finding hypotenuse :

√5² + (-3)²

√(25 + 9)

√34

Opposite side = -3

Adjacent side = 5

Hypotenuse = √34

sin θ = Opposite side / Hypotenuse side

sin θ = -3/√34

cos θ = Adjacent side / Hypotenuse side

cos θ = 5/√34

tan θ = Opposite side / Adjacent side

tan θ = -3/5

Problem 7 :

P (6, 3)

Solution :

Since the given point is in the form of (x, y), it will be in fourth quadrant.

Accordingly ASTC, for all trigonometric ratios we have to use positive sign.

Finding hypotenuse :

√6² + (3)²

√(36 + 9)

√45

3√5

Opposite side = 3

Adjacent side = 6

Hypotenuse side = 3√5

sin θ = Opposite side / Hypotenuse

sin θ = 3√5

= 1/√5

sin θ = 1/√5

cos θ = Adjacent side / Hypotenuse

cos θ = 6/ 3√5

cos θ = 2/√5

tan θ = Opposite side / Adjacent side

tan θ = 3/6

tan θ = 1/2

Problem 8 :

P (-24, -10)

Solution :

Since the given point is in the form of (-x, -y), it will be in third quadrant.

Accordingly ASTC, for the trigonometric ratios tan θ and cot  θ alone we have positive.

Finding hypotenuse :

√(-24)² + (-10)²

√(576 + 100)

√676

26

Opposite side = -10

Adjacent side = -24

Hypotenuse = 26

sin θ = Opposite side / Hypotenuse

sin θ = -10/26

sin θ = -5/13

cos θ = Adjacent side / Hypotenuse

cos θ = -24/26

cos θ = -12/13

tan θ = Opposite side / Adjacent side

tan θ = -10/-24

tan θ = 5/12

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