Circular Trigonometric FunctionsY

X

rθ

• circle…center at (0,0) •radius r…vector with length/direction

• angle θ… determines direction

Quadrant IQuadrant II

Quadrant III Quadrant IV

Y-axis

X-axis

90º

0º180º

270º

Initial side

Terminal side

θ

r r

360º

Quadrant IQuadrant II

Quadrant III Quadrant IV

Y-axis

X-axis

-270º

-360º-180º

-90º

Initial side

θr0ºTerminal

side

• angle θ…measured from positive x-axis,or initial side, to terminal

side

counterclockwise: positive directionclockwise: negative direction

• four quadrants…numbered I, II, III, IV counterclockwise

• six trigonometric functions for angle θwhose terminal side passes thru point(x, y) on circle of radius r

sin θ = y / r csc θ = r / y

cos θ = x / r sec θ = r / x

tan θ = y / x cot θ = x / y

These apply to any angle in any quadrant.

For any angle in any quadrant x2 + y2 = r2 …So, r is positive by Pythagorean theorem.

rθ

xy

(x,y)

Y

X

rθ

NOTE: right-triangle definitions arespecial case of circularfunctionswhen θ is inquadrant I

x

y

(x,y)

sin θ = y / r and csc θ = r / y

cos θ = x / r and sec θ = r / x

tan θ = y / x and cot θ = x / y

*Reciprocal Identities

sinθ cosθtanθ = cot θ =cosθ sinθ

*Ratio Identities

*Both sets of identities are useful to determine trigonometric

functions of any angle.

Y

X

(-, +) (+, +)

(-, -) (+, -)

Positive trig values in each quadrant:

All all six positive

Students

Take Classes

sin positive (csc)

tan positive (cot)

cos positive (sec)

I

III

II

IV

REMEMBER:In the ordered pair (x, y),x represents cosine and

y represents sine.

Y

X

(-, +) (+, +)

(-, -) (+, -)

I

III

II

IV

#1 Draw each angle whose terminal sidepasses through the given point, and findall trigonometric functions of each angle.

θ1: (4, 3)θ2: (- 4, 3)θ3: (- 4, -3)θ4: (4, -3)

SOLUTION

Ix = y = r = (4,3)

sin θ =cos θ =tan θ =csc θ =sec θ =cot θ =

θ1

SOLUTION

II x = y = r =

(-4,3)

sin θ =cos θ =tan θ =csc θ =sec θ =cot θ =

θ2

SOLUTION

III

x = y = r =

(-4,-3)

sin θ =cos θ =tan θ =csc θ =sec θ =cot θ =

θ3

SOLUTION

IV

x = y = r =

(4,-3)

sin θ =cos θ =tan θ =csc θ =sec θ =cot θ =

θ4

SOLUTION

Y

Xθ1

III

III IV

ref θ2ref θ3 ref θ4

Perpendicular line from point on circle always drawn to the x-axis forming a reference triangle

Y

Xθ1

III

III IV

ref θ2ref θ3 ref θ4

Value of trigfunction of angle inany quadrantis equal to trig function of its

reference angle,or it differsonly in sign.

#2 Given: tan θ = -1 and cos θ is positive:• Draw θ. Show the values for x, y, and r.

SOLUTION

Given: tan θ = -1 and cos θ is positive:• Find the six trigonometric functions of θ.

SOLUTION

# 1 Find the value of sin 110º. (First determine the reference angle.)

SOLUTION

#2 Find the value of tan 315º. (First determine the reference angle.)

SOLUTION

#3 Find the value of cos 230º. (First determine the reference angle.)

SOLUTION

#1 Draw the angle whose terminal side passes through the given point . 1, 3

SOLUTION

Find all trigonometric functions for angle whose terminal side passes thru . 1, 3

SOLUTION

#2 Draw angle: sin θ = 0.6, cos θ is negative.

SOLUTION

Find all six trigonometric functions: sin θ = 0.6, cos θ is negative.

SOLUTION

#3 Find remaining trigonometric functions:sin θ = - 0.7071, tan θ = 1.000

SOLUTION

Find remaining trigonometric functions:sin θ = - 0.7071, tan θ = 1.000

SOLUTION

#1 Express as a function of a reference angle and find the value: cot 306º .

SOLUTION

#2 Express as a function of a reference angle and find the value: sec (-153º) .

SOLUTION

#3 Find each value on your calculator.(Key in exact angle measure.)

sin 260.5º tan 150º 10’

SOLUTION

csc 450ºcot (-240º)

SOLUTION

sec (7/4) cos 5.41

SOLUTION

π/2 = 1.57

π = 3.14

3π/2 = 4.71

02π = 6.28

# 1 The refraction of a certain prism is

Calculate the value of n.

sin 100°n =

sin 47°

SOLUTION

#2 A force vector F has components Fx

= - 4.5 lb and Fy = 8.5 lb. Find sin θ and cos θ.

Fy = 8.5 lb

Fx=-4.5 lb

θ

SOLUTION

Fy = 8.5 lb

Fx=-4.5 lb

θ

SOLUTION