Chapter 1 - Worksheet 6 - TRIGONOMETRY – PART 2
NAME____________________________
1. Sketch the angle  in standard position. Then find the exact values for cos and tan  .
A.  
7
3
B.  
5
4
C.  
5
6
D.   
7

is in the 4th quadrant with reference angle .
3
3

3
sin


 1
3 
2  3
cos 
and tan 

1
3
3 2
cos
3
2
5

is in the 3rd quadrant with reference angle .
4
4


1
and tan  1
 cos  
4
4
2
5

is in the 2nd quadrant with a reference angle .
6
6

3

1
 cos  
and  tan  
6
2
6
3
 is a standard angle

cos   1 and tan   0
2. In each case, determine the quadrant(s) for angle T.
A. sin T  0 and cos T  0
quadrant 4
C. sec T  0 and csc T  0
quadrant 1
B. cos T  0 and tan T  0
quadrant 2
D. cot T  0 and cos 2 T  0
quadrants 2 and 4
3. Solve for the given angle. Express your answer in radian form and assume 0  angle  2 .
In each case there are two angles.
 3
A. sin  
B. tan   0
C. sec is undefined
2
4 5
 3

,
 ,
  0, 
3 3
2 2
4. Find the exact value for csc A .
Hint: make a right triangle and fill in the missing sides
(-7,3)
A
So the right triangle has adjacent side of 7 and opposite side of 3 and by the Pythagorean
Theorem the hypotenuse has length 58 .
csc A 
1

sin A
1
3
58

58
3
5. Solve for the variable so that 0  variable   . Express your answers in radians.
A. 2 cos 2 t  16  0 cos 2 t  8  cos t   8 which has no solution. Recall the range of the cosine
function is  1  cos t  1
B.
3
1  tan y
 0 1  tan y  0  tan y  1  y  arctan( 1) 
4
sin y
C. sin(2 x)  cos( x)  0 By double angle formula sin( 2 x)  2 cos x sin x.
2 cos x sin x  cos x  0  cos x(2 sin x  1)  0  cos x  0  x 

2
or sin x 
1
 5
x ,
2
6 6

 x 
6. Find the exact value for csc  tan 1 
  . Your answer will be in terms of x. Hint: make a right
2



triangle and fill in the missing side
x
 x 
Since tan 1 
so the right triangle has opposite side x and
   (an angle)  tan  
2
 2
adjacent side
2 so by the Pythagorean Thm the length of the hypotenuse is
1
csc  

sin 
1
x

x 2  2.
x2  2
x
x2  2
7. Express the area of this triangle in terms of A, b, and c.
C
b
a
A
c
Area =
1
a
1
ac and sin A   a  b sin A  Area  bc sin A
2
b
2
8. Simplify completely:
1  cos x  cos x  sin x  sin x
2
1  cos x 
=
cos x  cos 2 x  sin 2 x cos x  (cos 2 x  sin 2 x)
cos x  1
 (1  cos x)
1




2
2
2
2
1  cos x
(1  cos x)
(1  cos x)
(1  cos x)
(1  cos x)