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Trigonometric Functions Evaluation in Quadrant III: y = x, Assignments of Trigonometry

The definitions and steps to evaluate the six trigonometric functions (sin, cos, tan, cot, sec, csc) for an angle θ in standard position with a point (x, y) on the terminal side, given that r = √(x² + y²) > 0. The example in this document calculates the values of these functions for θ in quadrant iii with y = x.

Typology: Assignments

Pre 2010

Uploaded on 07/30/2009

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Evaluating Trigonometric Functions
If we have any angle, θ, in standard position with a point (x, y) on the terminal side of θ
and r=px2+y2>0, then use the following definitions to evaluate the six trigonometric
functions:
sin θ=y
rcos θ=x
r
tan θ=y
x, x 6= 0 cot θ=x
y, y 6= 0
sec θ=r
x, x 6= 0 csc θ=r
y, y 6= 0
The following figure shows us the quadrants and will also help us to evaluate the
functions:
Quadrant II Quadrant I
sin θ: + sin θ: +
cos θ:cos θ: +
tan θ:tan θ: +
Quadrant III Quadrant IV
sin θ:sin θ:
cos θ:cos θ: +
tan θ: + tan θ:
Problem 7.
Determine the value of the six trigonometric functions of θ.
y=x;θ lies in quadrant III
Solution Step 1:
First, find a point on the line. Since y=x, any value will work as long
as we use the same value for both xand y. Let’s use (1,1) because we
are in quadrant III.
Solution Step 2:
pf2

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Evaluating Trigonometric Functions

If we have any angle, θ, in standard position with a point (x, y) on the terminal side of θ

and r =

x^2 + y^2 > 0, then use the following definitions to evaluate the six trigonometric

functions:

sin θ=

y r

cos θ=

x r

tan θ=

y x

, x 6 = 0 cot θ=

x y

, y 6 = 0

sec θ=

r

x

, x 6 = 0 csc θ=

r

y

, y 6 = 0

The following figure shows us the quadrants and will also help us to evaluate the functions:

Quadrant II Quadrant I sin θ : + sin θ : + cos θ : − cos θ : + tan θ : − tan θ : +

Quadrant III Quadrant IV sin θ : − sin θ : − cos θ : − cos θ : + tan θ : + tan θ : −

Problem 7.

Determine the value of the six trigonometric functions of θ.

y = x; θ lies in quadrant III

Solution Step 1:

First, find a point on the line. Since y = x, any value will work as long

as we use the same value for both x and y. Let’s use (− 1 , −1) because we

are in quadrant III.

Solution Step 2:

Next, solve for r.

r =

x^2 + y^2

r =

(−1)^2 + (−1)^2

r =