Download Take out a sheet of paper and break it into 6 equal parts My Own

11/17/2011
Chapter 5 Review Material
Take out a sheet of paper and
break it into 6 equal parts
My Own Mental Checklist
• I can use the fundamental identities.
• Given 1 trigonometric function of an angle, I can find
the remaining 5 trigonometric function exact values.
• I understand the meaning of “verify the identity.”
• I can verify identities using the fundamental
identities.
• I can verify identities using the sum and difference,
and double angle identities.
• I know how to identify positive and negative values
of the trigonometric functions.
1
11/17/2011
Lottery Question #1
• Given csc x = -5 find the five remaining exact
trigonometric function values. X terminates in
QIV.
1
1
sin x 
csc x
 sin x  
5
1
2 6
sin 2 x  cos 2 x  1  (  ) 2  cos 2 x  1  cos x 
5
5
1
5 6
sec x 
 sec x 
cos x
12
sin x
6
 tan x  
cos x
12
1
cot x 
 cot x  2 6
tan x
tan x 
Lottery Question #2
• Find cos (x – y) given sin x = ¼ and cos y = 3/5
both x and y terminate in quadrant I.
cos( x  y )  cos x cos y  sin x sin y
sin 2 x  cos 2 x  1
1
( ) 2  cos 2 x  1  cos x 
4
cos 2 y  sin 2 y  1
15
4
3
4
( ) 2  sin 2 y  1  sin y 
5
5
15 3
1 4
)( )  ( )( )
4
5
4 5
3 15  4
cos( x  y ) 
20
cos( x  y )  (
2
11/17/2011
Lottery Question #3
3
• Find sin (z + q) given tan z = 4 and sin q = -2/3
both z and q terminate in QIII.
sin( z  q )  sin z cos q  cos z sin q
sin 2 q  cos 2 q  1
2
5
(  ) 2  cos 2 q  1  cos q  
3
3
1
1  tan 2 z  sec 2 z;cos z 
sec z
3
5
1  ( ) 2  sec 2 z  sec z  
4
4
4
4
3
cos z   ; cos 2 z  sin 2 z  1  (  ) 2  sin 2 z  1  sin z  
5
5
5
3
5
4
2
sin( z  q )  ( )(
)  ( )( )
5
3
5
3
3 5 8
sin( z  q ) 
15
Lottery Question #4
• Find sin 2x, cos 2x and tan 2x given sin x = 2/9
and x terminates in QI.
sin 2 x  2sin x cos x
2
77
cos2 x  sin 2 x  1  cos 2 x  ( ) 2  1  cos x 
9
9
2
77
4 77
sin 2 x  2( )(
)  sin 2 x 
9 9
81
2
cos 2 x  2 cos x  1
77 2
73
) 1  
9
81
sin 2 x
4 77
tan 2 x 

cos 2 x
81
cos 2 x  2(
3
11/17/2011
Lottery Question #5
• Simplify to a single trigonometric function value,
or number:
1
1 2 0
2
0
cos 15  sin 15
2
2
2 1
1
( cos 2 15  sin 2 15)
2 2
2
1 2 1
1
( ( cos 2 15  sin 2 15))
2 1 2
2
1
2
2
(cos 15  sin 15)
2
1
1
1 3
3
cos(2 *15)  cos 30  ( ) 
2
2
2 2
4
Lottery Question #7
• Simplify to a single trigonometric function value
or number:
2 tan 22.50
1  tan 2 22.50
 tan(222.5)  tan 450  1
4
11/17/2011
Just for fun…
fun…
Verify : sin 3 x  sin x  cos 2 x sin x
 sin x(1  cos 2 x)
 sin x(sin 2 x)
 sin 3 x
5