Download 1 - BrainMass

1. If sin(alpha)=1/5 where alpha is in quadrant II, find the remaining five trigonometric
functions of alpha.
sin 2   cos 2   1
 is in the 2nd quadrant, so cos  is negative.
2
2 6
1
cos    1  sin 2    1     
5
5
1
sin 
6
tan  
 5 
cos 
12
2 6

5
1
1
cot  

 2 6
tan 
6

12
1
1
5 6
sec 


cos 
12
2 6

5
1
1
cos ec 
 5
sin  1
5
2. Given that sin(alpha)= -2/3 and cos(alpha)= -root5/3>0, find the remaining four
trigonometric functions.
2
5
sin    ,cos  
, so sin   0 and cos  >0, then  is in the 4th quadrant.
3
3
2

sin 
2 5
tan  
 3 
cos 
5
5
3
1
1
5
cot  


tan 
2
2 5

5
1
1
3 5
sec 


cos 
5
5
3
1
1
3
cos ec 


sin   2
2
3
3. Sketch a graph of y=3cos(2theta+pi) using either transformations or the "5 key points"
method. Be sure to label the key points throughout the stages if you use transformations.
3
y
3
2
1
/4
0
0
/2
3/4


-1
-2
-3
-8
-3
The period
of the function
is  ,-2the amplitude
of the function
is 3.
-6
-4
0
2
The five key points are


3
(0, 3), ( ,0), ( ,3), ( ,0), ( , 3)
4
2
4
4
6
4. Sketch a graph of y= -tan(theta-(pi/4)) using transformations. Be sure to label the key
points and asymptotes throughout the stages.


y   tan(  )  tan(  )
4
4

First we graph y  tan( ) , then the shift this curve to the right by
to get y.
4
8
200
y=tan(-)
150
100
50
0
-/2
-3/2

3/2
/2
0
-50
-100
-150
-200
-5
-4
-3
-1
Now shift
this
graph-2 to right
by
200
0


4
 5 )
to get1 y   2tan( 3 )  tan(
4
4
4
y=tan(--/4)
150
100
50
0
-/4
-5/4
-3/4
7/4
3/4
0
/4

5/4
-50
-100
-150
-200
5
 3 7
4
In the-4above graph,-2 x    ,  0 ,  ,  are2 the asymptotes.
4
4 4 4
3
5
When x    ,0,  , y=0.
4
4
6
5. Sketch a graph of y=2csc(2theta+pi) using transformations or the "5 key points"
method using the sine function as a guide. Be sure to label the key points and asymptotes
throughout the stages.
150
y=2ccs(2+)
100
50
0

-/2
-3/2
0
/2
3/2
-50
-100
-150
-4
2
-3
-2
1
The period
of the function
is -1   . 0
2
3
1 1 3
In the above graph, x    ,   ,  ,  are asymptotes.
2
2 2 2
2
6. State the amplitude, range, period and phase shift of y= -4sin(2x-pi)
y  4sin(2 x  pi)  4sin(2 x)
(1)
So the amplitude is 4
1  sin 2x  1, then 4  y  4 , this is the range.
2
 .
The period T 
2
As can be found from equation (1), there is no phase shift.
8. Find the exact value of cos(inverse sin(-1/3))
 1
Assume   sin 1    , then  is in quadrant, so cos  is positive.
 3
1
We have sin   
3
2
2 2
1
cos   1  sin   1    
3
3
2
3
4

2 2
 1 
cos  sin 1      cos  
3
 3 

9. Find the exact value of sec(inverse tan(1/2))
1
Assume x  tan 1   , then x in the first quadrant, all trigonometric functions are
2
positive.
1
tan x 
2
1
sin 2 x cos 2 x
2
2
sec x  tan x 


1
cos 2 x cos 2 x cos 2 x
So
sec2 x  1  tan 2 x
2
5
1
sec x  1  tan 2 x  1    
2
2
That is

5
 1 
sec  tan 1    
 2  2

10. Solve: 2cos(theta)+root3=0
2cos  3  0
3
2
So,  could be in quadrants II and III.
5 5
5
5
,
 2 ,
 4 , ,
 2k , where k =0, 1, 2, 3, ..
When  in quadrant II,  
6 6
6
6
7 7
7
7
,
 2 ,
 4 , ,
 2k where k =0, 1, 2, 3 , ..
When  in quadrant III,  
6 6
6
6
cos  
11. Find the exact value of sin105degrees
sin105  sin(180  105 )  sin 75
Now we need to use double angle equations to find sin75.
1  cos 2 x
1  cos 2 x
cos 2 x  1  2sin 2 x  sin 2 x 
 sin x  
2
2
1  cos(2  75)
1  cos150
sin 75 


2
2
So
1
3
2  2 3
2
2
sin105  sin 75 
2 3
2
12. Find the exact value of cos(7pi/12)
7

  
cos
 cos      sin
12
12
 2 12 
It is obvious we know cos

6

problem to find sin
12
sin

12
1  cos

2

6 

3
, so we still need to use the equation in the previous
2
3
2  2 3
2
2
1
So
cos
7

2 3
  sin  
12
12
2
13. Use the fact that if cos(alpha)=1/root5, 0<alpha<pi/2 and sin(beta)= -4/5,
-pi/2<beta<0
1
cos  
5
0  

2
2
2
 1 
sin   1  cos   1  
 
5
 5
4
sin    ,
5
2


2
 0
2
3
 4
cos   1  sin   1     
5
 5
2
A) Find the exact value of sin(alpha-beta)
2 3 1  4
6
4
2 5
sin(   )  sin  cos   cos  sin  
 
  


5
5 5
5  5 5 5
5
B) Find the exact value of sin(2alpha)
sin 2  2sin  cos   2 
2
1
4


5
5 5
C) Find the exact value of cos(beta/2)
 
cos   2cos 2    1
2
   1  cos 
 cos 2   
2
2

1  cos 
cos 

2
2
3
5  42 5
2
5
5
1