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Algebra II Items to Support Formative Assessment
Unit 7: Trigonometric Functions
Prove and apply trigonometric identities.
F.TF.C.8 Prove the Pythagorean identity sin2( q ) + cos2( q ) = 1 and use it to find sin( q ), cos( q ),
or tan( q ), given sin( q ), cos( q ), or tan( q ), and the quadrant of the angle.
F.TF.C.8 Task
Given the rectangle below, identify the following values.
sin 
cos 
tan 
sin  cos 
sin2   cos2  
Solutions:
11
61
60
cos 
61
11
tan 
60
sin 
71
61
2
2
sin   cos   1
sin  cos 
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F.TF.C.8 Item 1
Given the triangle below, determine if the triangle is a right triangle. If so, show that
sin2 cos2 1 .
Solution:
Yes, the triangle is a right triangle.
202  212  292
400 441  841
841  841
sin 
20
29
2
 20   21 
 29    29 
cos 
21
29
2
400 441

841 841
841
1
841
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F.TF.C.8 Item 2
3
If a triangle is located in quadrant III and cos x   , what else can you conclude? Justify your
5
reasoning using mathematics.
Solution:
3
Knowing sin2 cos2 1 and the value of cos x   , we can find the value of sin .
5
sin 2   cos2   1
2
 3
sin       1
 5
2
9
1
25
16
sin 2  
25
sin 2  
sin 2  
sin  
16
25
4
5
Since we know the triangle is located in quadrant III, we know the value of sin must equal
4
 .
5
Using the information we can then determine the value of tan .
sin
cos
4

tan  5
3

5
4
tan 
3
tan 
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F.TF.C.8 Item 3
Find the values of cosine, sine and tangent. Verify your solutions for sine and cosine using your
knowledge of Pythagorean Identities.
Solution
In order to find the values of sine, cosine and tangent, the measure of the missing side must be
found.
a2  b 2  c 2
82  42  c 2
64 16  c 2
80  c 2
80  c
4 5c
The three dimension ratios can be used to determine the values of sine  , cosine  , and
tangent  .
sin  
4
4 5

5
5
cos 
8
4 5

2 5
5
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tan 
4
1
tan    
8
2
or
sin
cos
5
1
tan  5  
2
2 5
5

Verify your solutions using your knowledge of Pythagorean Identities. The Pythagorean Identity
states that sin2 cos2 1
sin2   cos2   1
sin2   cos2   1
2
2
 5  2 5
  
 1
 5   5 
5 20
 1
25 25
25
1
25
11
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F.TF.C.8 Item 4
Given the angle measure complete the table
Angle
cos
sin
cos2 
sin2
sin2 cos2
48
139
240
357
Do you notice a pattern about sin2 cos2 from the table?
Solution
Angle
cos
sin
cos2 
sin2
sin2 cos2
48
0.669
0.743
0.448
0.552
1
139
-0.755
0.656
0.570
0.430
1
240
-0.407
-0.914
0.165
0.835
1
357
0.999
-0.052
0.997
0.003
1
Based on the table, sin2 cos2 1 for all angles.
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under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.