Calculus Test 2
tanO/2=
(1-cosO)/sinO
Evaluate: tan(cos-1(x/2))
(4-x^2)r/x
assuming Q1 cos(tan-1(x))
(x^2+1)r/x^2+1
cosO/2 =
+/- (1+cosO/2)r
sinO/2=
+/- (1-cosO/2)r
sin(sin-1(x)) = x if
-1<O<1
tan-1(tan-3pi/8)
-3pi/8
tan-1 (-infinity)
-pi/2
tan-1(tan(x)) = x if
-pi/2 < O < pi/2
unit circle sin-1(cos2pi/3) =
-pi/3
tan-1(cos(pi))
-pi/4
sin-1(sin7pi/6)
-pi/6
cos^2(x)+sin(x)+1 = 0
3pi/2
tan-1(tan(3pi/7))=
3pi/7
cos(sin-1(1/2))
3r/2
sin(cos-1(3/5) + tan-1(12/5))
56/65
evaluate: cos(sin-1(-2/3))
5r/3
y=3sin(-2x) period and amplitude
A: 3 P: pi (pi,0)
fd exact value cos(75)cos(15)+sin(75)sin(15)
cos(60)=1/2
tan-1(2)
defined, but not on the unit circle
cos -1(-1)
pi
sin-1(sin4)
pi-4
cos-1(cos(-5pi/3))
pi/3
sin-1(sin2pi/3)
pi/3
tan-1(tan8pi/7)
pi/7
fd exact value tan(75)-tan(15)
tan(60)=3r
tan-1(sin(3pi/2)/cos(pi/2))
tan-1 = -1/0 = undefined = tan-1(-infinity) = -pi/2
cos-1(pi)
undefined
sin(sin-1(pi/2))
undefined
sin-1(pi/2)
undefined
sin-1(sinx)
x
trig func sin(4x)=-1
x = 3pi/8, 7pi/8, 11pi/8, 15pi/8
trig func 2cosxsinx-cosx=0
x=pi/2, 3pi/2, pi/6, 5pi/6
trig func sin(4x)=1
x=pi/8, 5pi/8, 9pi/8, 13pi/8
tan(2x)=1
x=pi/8, 7pi/8, 11pi/8, 15pi/8
Definition of O=cos^-1 (y)
y=cosO y=0<O<pi
Definition of O=tan^-1(y)
y=tanO y=pi/2 < O < pi/2
cos(3x)=1/2
...
2sin(x)+tan(x)=0
0, 2pi/3, pi, 4pi/3
sin(tan-1(infinity))
1
cos-1(cos2)
2
evaluate: tan(15)
2-3r
sin(sin-1(2/3)) =
2/3
cos-1 (-1/2)
2pi/3
cos-1(sin (pi/6) + tan (3pi/4))
2pi/3
cos-1(sin(7pi/6))=
2pi/3
Sin2O =
2sinO (cosO)
tan2O=
2tanO/1-tan^2O
sin(sin-1(x)-cos-1(x)))
2x^2-1
cosO=4/5 Q1, simplify sin (O+pi/3)
3+4(3r)/10
fd and simply csc(15)
6r+2r
Fd and simply sin(15)
6r-2r/4
y=5cos(pix) period and amplitude
A: 5 P: 2 (2,5)
y=5cos(-2x) period and amplitude
A: 5 P: pi (pi, 5)
y=5tan(1/4x) period and amplitude
A: none P: 4pi (-2pi, 2pi)
cos-1(cosx)=x if
O<x<pi
Cos2O =
cos2O-sin^2O 2cos^2O -1 1-2sin^2O
tan^2(x)=sec(x)+1
pi/3,pi,5pi/3
cos -1 (2r/2)
pi/4
tan-1(1)
pi/4
cot(x) = 2cos(x)
pi/6, pi/2, 5pi/6, 3pi/2
sin-1(sin(pi/7)=
pi/7
sin(sin-1(2) =
undefined sin-1(>1) = undefined
