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Inverse trigonometric functions
ƒ If we restrict the sine function to the interval
[ − π2 , π2 ] , it becomes one-to-one;
the inverse of
this function is denoted by arcsin or sin −1 :
(1, p2 )
domain:
p
2
p
2
y = sin x
– derivative:
y = arcsin x
range:
[− 1, 1]
[ − π2 , π2 ]
( 1, p2 )
d
1
arcsin x =
dx
1 − x2
– properties: arcsin(sin x ) = x if x ∈ [ − π2 , π2 ]
sin(arcsin x ) = x if x ∈ [ −1,1]
ƒ If we restrict the cosine function to the interval
[0, π] , it becomes one-to-one;
the inverse of
this function is denoted by arccos or cos −1 :
y = cos x
( 1,p)
y = arccos x
p
domain:
range:
[− 1, 1]
[0, π]
(1, 0)
– derivative:
d
−1
arccos x =
dx
1 − x2
– properties: arccos(cos x ) = x if x ∈ [ 0, π]
cos(arccos x ) = x if x ∈ [ −1,1]
ƒ If we restrict the tangent function to the interval
( − π2 , π2 ) , it becomes one-to-one;
the inverse
of this function is denoted by arctan or tan −1 :
y = tan x
p
2
domain:
p
2
p
2
p
2
– derivative:
1
d
arctan x =
1+ x2
dx
– properties: arctan(tan x ) = x if x ∈ ( − π2 , π2 )
tan(arctan x ) = x
y = arctan x
range:
(− ∞, ∞ )
( − π2 , π2 )
ƒ Simplifying trig(trig
−1
x)
– First, draw a figure answering the question posed by the inverse function. Be sure to
draw the triangle in the correct region, depending on the range of the inverse-trig
function:
– Determine the third side via the Pythagorean theorem (it will be positive).
– Read off whatever trig function you need from the figure.
For example:
3
5
q
sin(arccos( −54 )) =
3
5
-4
−1
ƒ Simplifying trig (trig x ) : first evaluate the inside function, then answer the question posed
by the inverse-trig function.
ƒ Solving problems via trigonometric and inverse trigonometric functions: in general, you’re
looking to find right triangles, from which you can read off trig relations.
l’Hôpital’s rule
f ( x)
gives either 0 0 or ∞ ∞ ,
x →… g ( x )
f ′( x )
and if (2) lim
exists or is infinite,
x →… g ′( x )
then the original limit does the same thing.
ƒ l’Hôpital’s rule states that if:
(1) lim
– Note that both hypotheses must be true for the rule to apply, and that the logic only
works right-to-left; don’t just carelessly write “=”!
1
∞
ƒ Other applications of l’Hôpital’s rule: 0 ⋅ ∞ ; ∞ − ∞ ; 0 , 1 , ∞
0
– 0 ⋅ ∞ : first turn the product into a quotient by inverting one factor and moving it to
the denominator.
– ∞ − ∞ : first use algebra to turn this into a single quotient.
0
– 0 0 , 1∞ , ∞ :turn this into a natural exponential via a b = eb⋅ln a ,
take the limit of the exponent,
and then use the limit of the exponent to find the original limit.