Download Simplify.

Honors Geometry
Advanced Trigonometry
Review Worksheet for Lessons 9-12 answers
Evaluate.
1. sec 45 = 1  cos45
2
= 1 
2
2
=
2
= 2
2. csc 120 = 1  sin120
3
= 1
2
2
=
3
2 3
=
3
3. cot 60 = 1  tan60
= 1 3
1
=
3
3
=
3
4. sec 90 = 1  cos90
= 1  (0)
undefined
9. cot-1(3) = tan-1(1/3)
= 18.4
10. sec-1(-1) = cos-1(1/-1) = cos-1(-1)
= 180

11. csc-1( 2 ) = sin-1 



= sin-1 

1 

2
2

2 
= 45
12. cot-1(0) = tan-1(1/0) = 90
(because tangent is undefined at 90)
Solve. 0 <  < 360
13. sec = 2
cos = ½
ref  : 60
I:  = 60
IV :  = 360 – 60 = 300
Solution : { 60, 300 }
14. csc = -½
sin = 1/-½ = -2
undefined
5 csc(-40) = 1  sin(-40)
= -1.5557
solution : 
6. cot 70 = 1  tan70
= 0.3640
15. cot = 1
tan = 1/1 = 1
7. sec-1(4) = cos-1(1/4)
= 75.5
ref  : 45
I:  = 45
III :  = 180 + 45 = 225
8. csc-1(-2) = sin-1(1/-2)
= -30
Solution : { 45, 225 }
Honors Geometry
Advanced Trigonometry
Review Worksheet for Lessons 9-12 answers
16. sec2 - 3sec + 2 = 0
(sec -1)(sec - 2) = 0
sec - 1 = 0
sec = 1
cos = 1
 = 0
sec - 2 = 0
sec = 2
cos = ½
ref  : 60
I:  = 60
IV:  = 360-60 = 300
=
sin 
1

1
cos 
=
sin 
cos 
= tan
20. tan2 + sin2 + cos2
Solution : { 0, 60, 300 }
= tan2 + (sin2 + cos2)
17. 3cot2 - 1 = 0
3cot2 = 1
cot2 = 1/3
tan2 = 3
tan =  3
= tan2 + 1
tan = 3
tan = - 3
ref  : 60
ref  : 60
I:  = 60
II:  = 180-60
III:  = 180+60
= 120
= 240
IV:  = 360-60
= 300
Solution : { 60, 120, 240, 300 }
18. csc2 = csc + 2
csc2 - csc - 2 = 0
(csc + 1)(csc - 2) = 0
csc + 1 = 0 or csc - 2 = 0
csc = -1
csc = 2
sin = -1
sin = ½
 = 270
 = 30, 150
Solution : { 30, 150, 270 }
= sec2
21. cos(-)
sin(-)
= cos
-sin
= - cos
sin
= -cot
22. sec ∙sin(90-)
1
 cos 
=
cos 
= 1
23. (cos - 1)(cos + 1)
(1 + sin)(1 - sin)
= cos2 - 1
1 – sin2
Simplify.
= -(1 – cos2)
cos2
19. sin∙sec
=
-sin2
cos2
Honors Geometry
Advanced Trigonometry
Review Worksheet for Lessons 9-12 answers
= -tan2
III:  + 10 = 180 + 68.2
 + 10 = 248.2
 = 238.2
24. (sin + cos)2 – 1
cos∙cos(90-)
= (sin2 + 2sin∙cos + cos2) - 1
cos∙sin
= 2sin∙cos + (sin2+cos2) – 1
sin∙cos
= 2sin∙cos +
1
sin∙cos
= 2sin∙cos
sin∙cos
= 2
Solve. 0 <  < 360
25. sin(-) = 1
-sin = 1
sin = -1
 = 270
solution : { 270 }
26. sec(-) = - 2
sec() = - 2
1
 2
cos =

2
 2
ref  : 45
II :  = 135 III:  = 225
solution : { 135, 225 }
27. tan( + 10) = 2.5
ref  : 68.2
I:  + 10 = 68.2
 = 58.2
-1
solution : { 58.2, 238.2 }
28. csc( + 20) = -2
sin( + 20) = - ½
ref  : 30
III:  + 20 = 180 + 30 = 210
 = 190
IV:  + 20 = 360 – 30 = 330
 = 310
solution : { 190, 310 }
29. cos(2) = ½
Since 0 <  < 360,
then 0 < 2 < 720
ref  : 60
I: 2 = 60
or 2 = 60 + 360 = 420
IV: 2 = 360 – 60 = 300
or 2 = 300 + 360 = 660
So 2 = 60, 300, 420, or 660
If 2 = 60, then  = 30
If 2 = 300, then  = 150
If 2 = 420, then  = 210
If 2 = 660, then  = 330
solution : { 30, 150, 210, 330 }
30. cot( ½) = -1
Since 0 <  < 360, then 0 < ½ < 180
tan( ½) = 1/-1 = -1
ref  : 45
II: ½ = 180 – 45 = 135
IV: ½ = 360 – 45 = 315
But 0 < ½ < 180 !!!
So, ½ = 135
Honors Geometry
Advanced Trigonometry
Review Worksheet for Lessons 9-12 answers
If ½ = 135, then  = 270
solution : { 270 }
31. sin(-3) = 0
-sin(3) = 0
sin(3) = 0
Since 0 <  < 360,
then 0 < 3 < 1080
Quadrant angle.
 = 0
or  = 0 + 360 = 360
or  = 360 + 360 = 720
 = 180
or  = 180 + 360 = 540
or  = 540 + 360 = 900
So, 3 = 0, 180, 360, 540, 720, or
900
If 3 = 0, then  = 0
If 3 = 180, then  = 60
If 3 = 360, then  = 120
If 3 = 540, then  = 180
If 3 = 720, then  = 240
If 3 = 900, then  = 300
solution : {0,60,120,180,240,300}
32. tan(-5) = 1
-tan(5) = 1
tan(5) = -1
Since 0 <  < 360,
then 0 < 5 < 1800
Round 3
5 = 495 + 360
= 855
5 = 675 + 360
= 1035
Round 4
5 = 855 + 360
= 1215
5 = 1035 + 360
= 1395
Round 5
5 = 1215 + 360
= 1575
5 = 1395 + 360
= 1755
If 5 = 135, then  =
If 5 = 315, then  =
If 5 = 495, then  =
If 5 = 675, then  =
If 5 = 855, then  =
If 5 = 1035, then  =
If 5 = 1215, then  =
If 5 = 1395, then  =
If 5 = 1575, then  =
If 5 = 1755, then  =
27
63
99
135
171
207
243
279
315
351
solution : {27, 63, 99, 135, 171,
207, 243, 279, 315, 351}
33. cos(100 - ) = 0.5
cos[-( -100)] = 0.5
cos( -100) = 0.5
ref  : 60
I:  -100 = 60
 = 160
Since 0 < 5 < 1800,
we can go around the circle five times
IV:  -100 = 300
 = 400
Round 1
5 = 135
5 = 315
Since 400 > 360,
 = 400 – 360 = 40
Round 2
5 = 135 + 360
= 495
5 = 315 + 360
= 675
solution : {40, 160}