Trigonometric Ratios of Angles Lesson Plan

1. Topic-

TRIGONOMETRY

2. Content-

The trigonometric ratios of angles

3. Goals: Aims/Outcomes-

1.The six trigonometric ratios of angles 0c, 30c, 45c, 60c and 90c are provided in the
following table.

2.The trigonometric ratios of the value

3.The table using the sums

4.The examples sums are developing the skill

4. Objectives-

1.The values of triangles

2.Using the trigonometric table solving the sums

3.The six trigonometric ratios of angles 0c, 30c, 45c, 60c and 90c

5. Materials and Aids-

Chart and working model

6. Procedures/Methods-

A. Introduction-

Consider Fig. 2.12 which shows a circle of radius 1 unit centered at the origin. Let P
be a point on the circle in the first quadrant with coordinates (x, y).
We drop a perpendicular PQ from P to the x-axis in
order to form the right triangle OPQ. Let +POQ = i, then
sin i = OP
PQ = y
1 = y (y coordinate of P)
cos i = Q
OP
O = x
1 = x (x coordinate of P)
tan i = Q O
PQ = x y
If OP coincides with OA, then angle i = 0c. Since the
coordinates of A are (1, 0), we have
sin 0c = 0 ( y coordinate of A) cosec0c is not defined
cos0c= 1 (x coordinate of A) sec0c = 1
tan 0c =
cos
sin
0
0
c
c =
1
0 = 0. cot0c is not defined
If OP coincides with OB, then angle i=90o . Since the coordinates of B are (0, 1), we have
sin 90c = 1 ( y coordinate of B) cosec 90c = 1
cos 90c = 0 (x coordinate of B) sec 90c is not defined
tan 90c =
cos
sin
90
90
c
c =
0
1 is not defined. cot 90c = 0
Fig. 2.12
O A(1, 0)
P(x, y)
Q
1
y
x
i
B(0, 1)
X
Y
40 Mathematics
Chapter2

B. Development-

Consider Fig. 2.12 which shows a circle of radius 1 unit centered at the orgin. Let P
be a point on the circle in the first quadrant with coordinates (x, y).
We drop a perpendicular PQ from P to the x-axis in
order to form the right triangle OPQ. Let +POQ = i, then
sin i =
OP
PQ = y
1
= y (y coordinate of P)
cos i = Q
OP
O = x
1
= x (x coordinate of P)
tan i =
Q O
PQ =
x
y
If OP coincides with OA, then angle i = 0c. Since the
coordinates of A are (1, 0), we have
sin 0c = 0 ( y coordinate of A) cosec0c is not defined
cos0c= 1 (x coordinate of A) sec0c = 1
tan 0c =
cos
sin
0
0
c
c =
1
0 = 0. cot0c is not defined
If OP coincides with OB, then angle i=90o . Since the coordinates of B are (0, 1), we have
sin 90c = 1 ( y coordinate of B) cosec 90c = 1
cos 90c = 0 (x coordinate of B) sec 90c is not defined
tan 90c =
cos
sin
90
90
c
c =
0
1 is not defined. cot 90c = 0
Fig. 2.12
O A(1, 0)
P(x, y)
Q
1
y
x
i
B(0, 1)
X
Y
40 Mathematics
Chapter2

C. Practice-

sin 0c = 0 ( y coordinate of A) cosec0c is not defined
cos0c= 1 (x coordinate of A) sec0c = 1
tan 0c =
cos
sin
0
0
c
c =
1
0 = 0. cot0c is not defined
If OP coincides with OB, then angle i=90o . Since the coordinates of B are (0, 1), we have
sin 90c = 1 ( y coordinate of B) cosec 90c = 1
cos 90c = 0 (x coordinate of B) sec 90c is not defined
tan 90c =
cos
sin
90
90
c
c =
0
1 is not defined.
cot 90c = 0

D. Independent Practice-

Example 2.8
Evaluate sin245 tan245 cos245 c + c + c.
Solution We know, sin45
2
c = 1 , tan 45c = 1 and cos45
2
c = 1
` sin245 tan245 cos245 c + c + c= (1)
2
1
2
2 2 1 2
c m + + c m
= 1
2
1
2
+ + 1 = 2

E. Accommodations (Differentiated Instruction)-

Example 2.9
Evaluate
sec
cos tan
4 45
12 30 2 60
2
2 2 -
c
c c .
Solution We know, cos 30
2
c = 3 , tan 60c = 3 and sec45c = 2
`
sec
cos tan
4 45
12 30 2 60
2
2 2 -
c
c c =
4 ( 2)
12
2
3 2 3
2
2
2
#
c #c m m- ^ #^ h h
=
4 2
12
4
3 2 3
#
` # j- ^ # h
=
8
9- 6 =
8
3
Note

F. Checking for understanding-

13. Evaluate.
(i) sin 45c + cos45c (ii) sin 60c tan 30c
(iii)
tan tan
tan
30 60
45
c + c
c (iv) cos260 sin230 tan230 cot260 c c + c c
(v) 6 cos290 3 sin290 4tan245 c + c + c (vi)
sin cos
cot sec sin
60 45
4 60 30 2 45
2 2
2 2 2
+
+ -
c c
c c c
(vii)
cosec sec cot
tan cos sec cos
30 60 30
60 4 45 3 30 5 90
2
2 2 2 2
+ -
+ + +
c c c
c c c c
(viii) 4(sin430 cos460 ) 3(cos245 sin290 ) c + c - c - c .
14. Verify the following equalities.
(i) sin230 cos230 1 c + c =
(ii) 1 tan245 sec245 + c = c
(iii) cos 60 1 2sin230 2cos230 1 c = - c = c -
(iv) cos 90 1 2sin245 2cos245 1 c = - c = c -
(v)
sin
cos
1 60 sec tan
60
60 60
1
+
=
c +
c
c c
(vi) 2 60 1
tan
tan cos
1 60
1 60
2
2
2
+
- = -
c
c c
(vii)
sec tan
sec tan
sin
sin
30 30
30 30
1 30
1 30
-
+ =
-
+
c c
c c
c

G. Closure-

7. Evaluation-

8. Teacher Reflection-

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