Problem of the DayWhich 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 onlyB) II onlyC) III onlyD) I and III onlyE) II and III only
Problem of the DayWhich 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 onlyB) II onlyC) III onlyD) I and III onlyE) 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
cos arccos
tan arctan
Domain [-π/2, π/2]Range [-1, 1] Domain [-1, 1]
Range [-π/2, π/2]
Domain [0, π]Range [-1, 1] Domain [-1, 1]
Range [0, π]
Domain [-π/2, π/2]Range [-∞, ∞]
Domain [-∞, ∞]Range [-π/2, π/2]
(sin y = x) (arcsin x = y)
(cos y = x) (arccos x = y)
(tan y = x) (arctan x = y)
sec arcsec
csc arccsc
cot arccot
Domain [0, π], x≠π/2Range |y| > 1
Domain |x| > 1Range [0, π], y≠π/2
Domain [-π/2, π/2], x≠0Range |y| > 1
Domain |x| > 1Range [-π/2, π/2], y≠0
Domain [0, π]Range [-∞, ∞]
Domain [-∞, ∞]Range [0, π]
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
Similar properties hold true for other trig functions.
Example arctan (2x - 3) = π/4 tan(arctan (2x - 3)) = tan π/4 2x - 3 = 1 x = 2
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?
Consider the triangle
u
1
dx
u
1
f(u) = sin uf '(u) = cos u dug(u) = arcsin u
Derivatives of Inverse Trig Functions
Examples
Find cos(arcsin x)sin = opp = x hyp 1
x
1θ
1.
2. a2 + b2 = c2x2 + b2 = 12 b =
3. cos = adj = hyp 1
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 liney - 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)
