Download Math 140 Lecture 24 cos(sin 5 )) = 1

Math 140
Lecture 24
Exam 4: Lectures 19-24, including this lecture.
Inverse trigonometric functions
RECALL. A function is 1-1 iff no horizontal line crosses its
graph more than once.
While not 1-1 in general, sin, cos, and tan are 1-1 on the
first quadrant angles in [0, p/2] and on the larger
heavily marked “1-1” intervals.
`arcsin(-1/2) = -p/6
arccos(-1/l2) = 3p/4
cos -1(2) = undef since 22[-1, 1]
arctan(-1) = -p/4
cos(tan-1(-1)) = cos(-p/4) = cos(p/4) = 1/ 2
Since inverses undo each other, we have
sin-1(sin q) = q if qi[-p/2, p/2] sin(sin-1 x) = x if xi[-1, 1]
cos -1(cos q) = q if qi[0, p]
sin
1
-p/2
p/2
tan-1(tan q) = q if qi(-p/2,p/2) tan(tan -1 x) = x for any x
RECALL. cosx  cos x, sin  x  sin x.
When a formula involves an angle outside the restricted
range, rewrite it in terms of an angle q in the restricted
range before applying the equations in the box.
-1
cos
p
0
`coscos 1 5  
1
tan
-p/2
NOTATION. sin -1(x), cos -1(x) and tan -1(x) are also written:
arcsin(x), arccos(x), and arctan(x).
Warning, sin -1(x) = 1/sin(x). sin -1(x) is the inverse;
(sin(x)) -1 = 1/sin(x) is the reciprocal.
cos -1(1) = 0
sin -1(1/2) = p/6
sin -1(1/l2̄) = p/4
-1
tan -1(0) = 0
cos -1(l3̄/2) = p/6 tan -1(1/l3̄) = p/6
cos -1(1/l2̄) = p/4
-1
sin (l3̄/2) = p/3
cos (1/2) = p/3
sin -1(1) = p/2
cos -1(0) = p/2
tan -1(1) = p/4
-1
tan (l3̄) = p/3
x
arccos(x)
0
1
The heavily marked half circles are the restricted ranges.
important
cossin1x  1  x2
sincos1x  1  x2
Proof of 1st. Recall: cos 2   1  sin 2 .
t cos    1  sin 2 
t
cossin 1 x   1  sinsin 1 x 2
 1  x2
The “+” was chosen over the “-” since
sin-1(x) i[-p/2, p/2] ˆ cos(sin-1(x)) > 0.
`sinsin 1  15  
1
5
cossin 1  15   1 

tansin 1  15  

1
25
sinsin 1 15 
cossin 1 15 
46
24
25
25 
6
1/5
 12
2 6 /5
In a triangle with hypotenuse 1 and side x,
sin1 x = angle opposite x,
cos 1 x = angle adjacent to x.
1
arcsin(x)
arccos(x)
-1 -p/2
important
THEOREM. For xi[-1,1],

 sec 
2  0  1/ cos 2   1/0  undef.
1 p/2
x
4
4
5  5
cos 1 cos 5   cos 1 cos 5   5

sin 1 sin 
5   5


1
sin 1 sin 4
5   sin sin 5   5
`secsin1 1  cos 1 1
The unit circle pictures are
x
arcsin(x) p
-1
1
5
cos 1 cos
p/2
DEFINITION. sin -1, cos -1, tan -1 are the inverses of the above
restrictions of sin, cos, and tan. Thus
sin -1(x) = the q i[-p/2, p/2] such that sin(q) = x,
cos -1(x) = the q i[0, p]
such that cos(q) = x,
-1
tan (x) = the q i(-p/2, p/2) such that tan(q) = x.
sin-1(x) and cos-1(x) have domain [-1, 1], for x2[-1, 1],
they are undefined. tan -1(x) has domain (-5,5).
sin -1(0) = 0
cos(cos-1 x) = x if xi[-1, 1]
x
Note that sin 1 x  cos 1 x  /2  90 o 

2 6
5