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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