Problem of the Day
Which of the following are antiderivatives of
f(x) = sin x cos x?
I. F(x) = ½ sin2x
II. F(x) = ½ cos2x
III. F(x) = -¼ cos(2x)
A) I only
B) II only
C) III only
D) I and III only
E) II and III only
Problem of the Day
Which of the following are antiderivatives of
f(x) = sin x cos x?
I. F(x) = ½ sin2x
II. F(x) = ½ cos2x
III. F(x) = -¼ cos(2x)
A) I only
B) II only
C) III only
D) I and III only
E) II and III only
Inverse Trig Functions?
How? None of the 6 basic trig functions has an inverse
because they are not 1 to 1.
Inverse Trig Functions?
How? None of the 6 basic trig functions has an inverse
because they are not 1 to 1.
Restrict the domain and they do.
(see page 373)
sin
arcsin
(sin y = x)
(arcsin x = y)
Domain [-π/2, π/2]
Range [-1, 1]
cos
Domain [-1, 1]
Range [-π/2, π/2]
arccos
(cos y = x)
(arccos x = y)
Domain [0, π]
Range [-1, 1]
tan
arctan
(tan y = x)
Domain [-1, 1]
Range [0, π]
(arctan x = y)
Domain [-π/2, π/2]
Range [-∞, ∞]
Domain [-∞, ∞]
Range [-π/2, π/2]
sec
arcsec
Domain |x| > 1
Range [0, π], y≠π/2
Domain [0, π], x≠π/2
Range |y| > 1
csc
arccsc
Domain |x| > 1
Range [-π/2, π/2], y≠0
Domain [-π/2, π/2], x≠0
Range |y| > 1
cot
arccot
Domain [-∞, ∞]
Range [0, π]
Domain [0, π]
Range [-∞, ∞]
Evaluate - arcsin(-½)
infers that sin y = -½
in the interval [-π/2, π/2],
the angle that gives -½ as its sin is -π/6
Evaluate - arcsin(0.3) using a calculator in radian
mode
y ≈ .3047
Inverse Properties (see page 373)
If you are in the restricted interval then
tan(arctan x) = x and arctan (tan y) = y
Example
arctan (2x - 3) = π/4
tan(arctan (2x - 3)) = tan π/4
2x - 3 = 1
x=2
Similar properties hold true for other trig functions.
Remember the relationship between the derivative
of a function and it's inverse?
Remember the relationship between the derivative
of a function and it's inverse?
dx
Consider the triangle
1
u
1
f(u) = sin u
f '(u) = cos u du
g(u) = arcsin u
u
Derivatives of Inverse Trig Functions
Examples
Find cos(arcsin x)
1.
2.
3.
sin = opp = x
hyp 1
a2 + b2 = c2
x2 + b2 = 12
b=
cos = adj =
hyp
1
θ
1
x
Find an equation for the line tangent to the graph of
y = cot-1x at x = -1
Find an equation for the line tangent to the graph of
y = cot-1x at x = -1
evaluated at x = -1 gives -½ which
gives you the slope of the tangent line
find y when x = -1
y = cot-1(-1) = π/2 - tan-1)(-1)
= π/2 - (-π/4)
= 3π/4
find equation of tangent line
y - 3π/4 = -½(x + 1)
Conversions
sec-1 x = cos-1(1/x)
csc-1 x = sin-1(1/x)
cot-1 x = π/2 - tan-1(x)
Page 378 summarizes all rules learned so far
(See page 376)