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Republic of the Philippines
DEPARTMENT OF EDUCATION
Region I
Division of Ilocos Sur
CIRCULAR
FUNCTIONS
OBJECTIVES
 illustrate the six circular functions,
 find its exact values using reference
angles,
 determine its domain and range and
 graph the six circular functions and
identify their (a) amplitude, (b) period,
and (c) phase shift.
Have you ever played
the Super Mario game?
Have you ever observed
Mario glide so smoothly
over game obstacles?
SOH-CAH-TOA is a mnemonic device in trigonometry that you
must know. It stands for sine-opposite-hypotenuse (SOH), cosine-
adjacent-hypotenuse (CAH), tangent-opposite-adjacent (TOA).
Studying the right triangle below (Figure 1) and looking at the sides
with respect to the angle θ, y is the opposite side, x is the adjacent
side, and r is the hypotenuse. Then
Figure 1
cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
sec 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑛𝑐𝑒𝑛𝑡
sin 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
csc 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
cot 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
Circular functions which is commonly known as the trigonometric
functions because the radian measures of the angles are calculated
by the length and coordinates of the terminal point of the arc on the
unit circle.
Circular functions are function whose domain and range
correspond to the measures of angles with respect to the
trigonometric functions. The basic circular functions are sine, cosine
and tangent and their reciprocal functions are cosecant, secant and
cotangent respectively.
The difference is that the domain of the trigonometric functions is the
set of angles in standard position while the range of the circular
functions is real numbers.
Figure 2
The angle is 45°. Since it is a unit
circle, the intercepted arc is
𝜋
4
.
In Figure 2, a unit circle and an angle
in standard position were drawn on
the rectangular coordinate plane
whose center and vertex are at the
origin
Figure 3
Notice that if a vertical line is
drawn from point P to the initial
side, a right triangle is formed.
(See Figure 3)
Line segment OP (radius) has
a length of 1 unit since circle O
is a unit circle. Considering the
central angle
𝜋
4
, x is the length of
the adjacent side and y is the
length of the opposite side.
Let 𝜃 be an angle in the standard position and P(x,y) be a point on
the terminal side of the angle. Then we have the six-circular function as
follows:
sin 𝜃 =
𝑦
𝑟
= 𝑦 csc 𝜃 =
1
sin 𝜃
=
1
𝑦
cos 𝜃 =
𝑥
𝑟
= 𝑥 sec 𝜃 =
1
cos 𝜃
=
1
𝑥
𝑡𝑎𝑛 𝜃 =
𝑦
𝑥
; 𝑥 ≠ 0 cot 𝜃 =
1
tan 𝜃
=
𝑥
𝑦
; 𝑦 ≠ 0
Therefore, if the
coordinates of point P on
the unit circle are known,
finding the values of the 6
circular functions would
not be difficult. Revisit the
unit circle and the
coordinates of point P on
the special angles
Figure 4. (etc.usf.edu)
ILLUSTRATIVE EXAMPLES
1. Find the exact values of sin
3𝜋
2
,
cos
3𝜋
2
, and tan
3𝜋
2
Solution:
Let P
3𝜋
2
be the point on the unit circle and on the
terminal side of the angle in the standard position with
measure
3𝜋
2
rad. Then P
3𝜋
2
= (0, -1), and so
sin
3𝜋
2
= −1, cos
3𝜋
2
= 0 ,
but tan
3𝜋
2
is undefined
1
0
.
ILLUSTRATIVE EXAMPLES
2. Find the values of cos 60°
and tan 60°
Solution:
We may also approach the problem
using the properties of 30° - 60° right
triangles. Review the figure on the right.
Thus, the coordinates of point P at
θ=60° are
1
2
,
3
2
.
Therefore, we get
cos 60° =
1
2
and tan 60° =
3
2
∙ 2 = 3
Find the values of the six circular functions of 𝜃 given that the terminal point is
P −
2
2
,
2
2
.
Solution:
Given: x = −
2
2
and y =
2
2
, Therefore,
sin 𝜃 = 𝑦 =
2
2
csc 𝜃 =
1
2
2
=
2
2
=
2
2
⋅
2
2
=
2 2
2
= 2
cos 𝜃 = 𝑥 = −
2
2
sec 𝜃 =
1
−
2
2
=−
2
2
= −
2
2
⋅
2
2
=
2 2
2
= − 2
𝑡𝑎𝑛 𝜃 =
𝑦
𝑥
=
2
2
−
2
2
=
2
2
⋅ −
2
2
= −1 cot 𝜃 =
1
−1
= −1
Give the six circular functions of the angle 𝜃 generated by an arc
whose length is
5𝜋
6
.
Thus the six circular functions are,
sin 𝜃 = 𝑦 =
1
2
csc 𝜃 =
1
1
2
= 2
cos 𝜃 = 𝑥 = −
3
2
sec 𝜃 =
1
−
3
2
=−
2
3
= −
2
3
⋅
3
3
= −
2 3
3
𝑡𝑎𝑛 𝜃 =
𝑦
𝑥
=
1
2
−
3
2
=
1
2
⋅ −
2
3
= −
1
3
⋅
3
3
= −
3
3
cot 𝜃 =
1
−
3
3
= −
3
3
⋅
3
3
= −
3 3
3
=− 3
ACTIVITY 1
Directions: Find all the
exact values of the six
circular functions at
𝑃
𝜋
6
.
Simplify and
rationalize (if needed)
your final answers.

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circular-functions.pptx

  • 1. Republic of the Philippines DEPARTMENT OF EDUCATION Region I Division of Ilocos Sur
  • 3. OBJECTIVES  illustrate the six circular functions,  find its exact values using reference angles,  determine its domain and range and  graph the six circular functions and identify their (a) amplitude, (b) period, and (c) phase shift.
  • 4. Have you ever played the Super Mario game? Have you ever observed Mario glide so smoothly over game obstacles?
  • 5. SOH-CAH-TOA is a mnemonic device in trigonometry that you must know. It stands for sine-opposite-hypotenuse (SOH), cosine- adjacent-hypotenuse (CAH), tangent-opposite-adjacent (TOA). Studying the right triangle below (Figure 1) and looking at the sides with respect to the angle θ, y is the opposite side, x is the adjacent side, and r is the hypotenuse. Then Figure 1 cos 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 sec 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑎𝑑𝑗𝑎𝑛𝑐𝑒𝑛𝑡 sin 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 csc 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 tan 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 cot 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
  • 6. Circular functions which is commonly known as the trigonometric functions because the radian measures of the angles are calculated by the length and coordinates of the terminal point of the arc on the unit circle. Circular functions are function whose domain and range correspond to the measures of angles with respect to the trigonometric functions. The basic circular functions are sine, cosine and tangent and their reciprocal functions are cosecant, secant and cotangent respectively.
  • 7. The difference is that the domain of the trigonometric functions is the set of angles in standard position while the range of the circular functions is real numbers. Figure 2 The angle is 45°. Since it is a unit circle, the intercepted arc is 𝜋 4 . In Figure 2, a unit circle and an angle in standard position were drawn on the rectangular coordinate plane whose center and vertex are at the origin
  • 8. Figure 3 Notice that if a vertical line is drawn from point P to the initial side, a right triangle is formed. (See Figure 3) Line segment OP (radius) has a length of 1 unit since circle O is a unit circle. Considering the central angle 𝜋 4 , x is the length of the adjacent side and y is the length of the opposite side.
  • 9. Let 𝜃 be an angle in the standard position and P(x,y) be a point on the terminal side of the angle. Then we have the six-circular function as follows: sin 𝜃 = 𝑦 𝑟 = 𝑦 csc 𝜃 = 1 sin 𝜃 = 1 𝑦 cos 𝜃 = 𝑥 𝑟 = 𝑥 sec 𝜃 = 1 cos 𝜃 = 1 𝑥 𝑡𝑎𝑛 𝜃 = 𝑦 𝑥 ; 𝑥 ≠ 0 cot 𝜃 = 1 tan 𝜃 = 𝑥 𝑦 ; 𝑦 ≠ 0
  • 10. Therefore, if the coordinates of point P on the unit circle are known, finding the values of the 6 circular functions would not be difficult. Revisit the unit circle and the coordinates of point P on the special angles Figure 4. (etc.usf.edu)
  • 11. ILLUSTRATIVE EXAMPLES 1. Find the exact values of sin 3𝜋 2 , cos 3𝜋 2 , and tan 3𝜋 2 Solution: Let P 3𝜋 2 be the point on the unit circle and on the terminal side of the angle in the standard position with measure 3𝜋 2 rad. Then P 3𝜋 2 = (0, -1), and so sin 3𝜋 2 = −1, cos 3𝜋 2 = 0 , but tan 3𝜋 2 is undefined 1 0 .
  • 12. ILLUSTRATIVE EXAMPLES 2. Find the values of cos 60° and tan 60° Solution: We may also approach the problem using the properties of 30° - 60° right triangles. Review the figure on the right. Thus, the coordinates of point P at θ=60° are 1 2 , 3 2 . Therefore, we get cos 60° = 1 2 and tan 60° = 3 2 ∙ 2 = 3
  • 13.
  • 14. Find the values of the six circular functions of 𝜃 given that the terminal point is P − 2 2 , 2 2 . Solution: Given: x = − 2 2 and y = 2 2 , Therefore, sin 𝜃 = 𝑦 = 2 2 csc 𝜃 = 1 2 2 = 2 2 = 2 2 ⋅ 2 2 = 2 2 2 = 2 cos 𝜃 = 𝑥 = − 2 2 sec 𝜃 = 1 − 2 2 =− 2 2 = − 2 2 ⋅ 2 2 = 2 2 2 = − 2 𝑡𝑎𝑛 𝜃 = 𝑦 𝑥 = 2 2 − 2 2 = 2 2 ⋅ − 2 2 = −1 cot 𝜃 = 1 −1 = −1
  • 15. Give the six circular functions of the angle 𝜃 generated by an arc whose length is 5𝜋 6 . Thus the six circular functions are, sin 𝜃 = 𝑦 = 1 2 csc 𝜃 = 1 1 2 = 2 cos 𝜃 = 𝑥 = − 3 2 sec 𝜃 = 1 − 3 2 =− 2 3 = − 2 3 ⋅ 3 3 = − 2 3 3 𝑡𝑎𝑛 𝜃 = 𝑦 𝑥 = 1 2 − 3 2 = 1 2 ⋅ − 2 3 = − 1 3 ⋅ 3 3 = − 3 3 cot 𝜃 = 1 − 3 3 = − 3 3 ⋅ 3 3 = − 3 3 3 =− 3
  • 16. ACTIVITY 1 Directions: Find all the exact values of the six circular functions at 𝑃 𝜋 6 . Simplify and rationalize (if needed) your final answers.

Editor's Notes

  1. Well Mario wasn’t really jumping along the horizontal axis straightly, but he was jumping slightly on a curved path or a parabolic path to avoid the obstacles on his way. And calculating Mario’s jump over these obstacles were circular functions comes in
  2. We define the six trigonometric function in such a way that the domain of each function is the set of angles in standard position. In this lesson, we will modify these trigonometric functions so that the domain will be real numbers rather than set of angles.
  3. Remember that in trigonometry, particularly in the circular functions, 𝜋 4 is often used as the measure of the central angle instead of the 45° Also, point P on the unit circle intersected by the radius has coordinates (x,y).
  4. Can you figure out how many degrees are there in each division? These are easy to memorize since they all have the same value with different signs depending on the quadrant. The signs of the coordinates of 𝑃(𝜃) depends on the quadrant or axis where it terminates. It is important to know the sign of each circular function in each quadrant. In QI, 𝑥 & 𝑦 are positive, therefore all circular functions are positive. In QII, 𝑥 is negative & 𝑦 is positive, therefore sine & cosecant functions are positive. In QIII, 𝑥 & 𝑦 are negative, therefore tangent and cotangent functions are positive. In the 4th Quadrant, 𝑥 is positive & 𝑦 is negative, therefore the only positive are cosine and secant functions.
  5. The circle is 2𝜋 all the way around so half way is 𝜋. The circle is divided into 8 pieces, so each piece is 𝜋/4.
  6. Since the length of the arc generated is 5𝜋 6 units then the measure of the angle is also equal to 5𝜋 6 or equivalent to 1500. Hence, the angle is in the second quadrant so the coordinates of the terminal point would be − 3 2 , 1 2 .