sin−1 x or arcsin x
h
i
sin x: D (sin) = − π2 , π2 , R (sin) = [−1, 1]
³
´
³
´
h
sin−1 x: D sin−1 = [−1, 1], R sin−1 = − π2 , π2
i
Cancellation equalities
³
´
sin sin−1 x = x,
−1 ≤ x ≤ 1,
sin−1 (sin x) = x,
−
π
π
≤x≤
2
2
How to express trigonometric functions through each other
·
¸
q
π π
x∈ − ,
,
2 2
cos x ≥ 0,
cos x = + 1 − sin2 x
sin x
sin x
=√
cos x
1 − sin2 x
sin2 x + cos2 x
1
1
tan2 x + 1 =
=
,
cos x = √
2
2
cos x
cos x
1 + tan2 x
tan x
sin x = tan x cos x = √
1 + tan2 x
tan x =
combinations of direct and inverse functions
³
cos sin
−1
³
r
´
x =
´
tan sin−1 x =
MATH115 T2 Winter 04, A.Potapov
³
´
1 − sin2 sin−1 x =
³
sin sin−1 x
³
cos sin
— 1 —
−1
√
1 − x2
´
x
´ = √
x
1 − x2
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Where do we need this? Substitution x = sin t, dx = cos t dt,
Z √
Z
1−
x2 dx
=
1Z
t sin 2t
+C =
cos t dt =
(1 + cos 2t) dt = +
2
2
4
2
1
1
1
1 √
= t + sin t cos t + C = sin−1 x + x 1 − x2 + C.
2
2
2
2
Derivative
y = sin−1 x,
0
0
x = sin y
0
q
(x) = 1 = (sin y) = cos y y =
³
sin−1 x
´0
=√
1 − sin2 y y 0 =
Z
1
,
1 − x2
√
√
1 − x2 y 0
dx
= sin−1 x + C
2
1−x
cos−1 x or arccos x
cos x: D (cos) = [0, π], R (cos) = [−1, 1]
cos−1 x: D (cos−1 ) = [−1, 1], R (cos−1 ) = [0, π]
Cancellation equalities
³
´
cos cos−1 x = x,
cos−1 (cosx) = x,
MATH115 T2 Winter 04, A.Potapov
— 2 —
−1 ≤ x ≤ 1,
0≤x≤π
www.math.ualberta.ca/∼apotapov/math115.htm
How to express trigonometric functions through each other
sin x ≥ 0,
cot x =
cot2 x + 1 =
√
sin x = + 1 − cos2 x
x ∈ [0, π] ,
cos x
cos x
=√
sin x
1 − cos2 x
cos2 x + sin2 x
1
=
,
2
sin x
sin2 x
cos x = cot x sin x = √
sin x = √
1
1 + cot2 x
cot x
1 + cot2 x
combinations of direct and inverse functions
³
´
sin cos−1 x =
√
³
sin−1 x + cos−1 x =
´
cot cos−1 x = √
1 − x2 ,
π
,
2
cos−1 x =
x
1 − x2
π
− sin−1 x
2
Derivative
³
cos−1 x
´0
³
= − sin−1 x
´0
= −√
1
,
1 − x2
tan−1 x or arctan x
³
´
tan x: D (tan) = − π2 , π2 , R (tan) = (−∞, ∞)
MATH115 T2 Winter 04, A.Potapov
— 3 —
www.math.ualberta.ca/∼apotapov/math115.htm
³
tan−1 x: D (tan−1 ) = (−∞, ∞), R (tan−1 ) = − π2 , π2
lim tan−1 x =
lim
tan x = ∞,
π−
x→∞
x→ 2
π
2
lim tan−1 x = −
limπ + tan x = −∞,
x→−∞
x→− 2
´
π
2
Cancellation equalities
³
´
tan tan−1 x = x,
−1 < x < 1,
tan−1 (tan x) = x,
−
π
π
<x<
2
2
How to express trigonometric functions through each other
µ
¶
q
π π
x∈ − ,
,
2 2
cos x > 0,
sin x = √
tan x
,
1 + tan2 x
cos x = + 1 − sin2 x
cos x = √
1
1 + tan2 x
combinations of direct and inverse functions
³
´
sin tan−1 x = √
x
,
1 + x2
³
´
cos tan−1 x = √
1
.
1 + x2
Derivative
y = tan−1 x,
(x)0 = 1 = (tan y)0 =
³
−1
tan
x
´0
x = tan y
³
´
1
1
0
0
2
y
=
y
=
1
+
x
y0
cos2 y
cos2 (tan−1 x)
Z
1
,
=
1 + x2
dx
= tan−1 x + C
1 + x2
Example. Using implicit differentiation, find derivative of y = sec−1 x, x ∈ [1, ∞), y ∈ [0, π2 )
x = sec y = 1/ cos y, cos y = 1/x,
MATH115 T2 Winter 04, A.Potapov
— 4 —
www.math.ualberta.ca/∼apotapov/math115.htm
d
d
x=1=
dx
dx
y0 = √
Ã
1
cos y
!
sin y 0
=
y =
cos2 y
cos2 y
1
= q
2
1 − cos y
x2 1 −
1
x2
√
1 − cos2 y 0
y,
cos2 y
1
= √ 2
.
x x −1
Example. Find
lim tan−1 (ex )
x→∞
Denote u = ex , u → ∞ as x → ∞, so
lim tan−1 (ex ) = u→∞
lim tan−1 u =
x→∞
π
.
2
Example. Evaluate integral
Z
sin−1 x
√
dx
1 − x2
√
Substitution u = sin−1 x, du = dx/ 1 − x2 ,
Z
³
Z
sin−1 x
sin−1 x
1 2
√
dx
=
udu
=
u
+
C
=
2
2
1 − x2
´2
+ C.
Example. Evaluate integral
Z
dx
x2 + a2
Substitution u = x/a, x = au, dx = adu,
Z
Z
adu
1 Z du
dx
=
=
=
x2 + a 2
a2 + a2 u2
a 1 + u2
µ
1
1
x
= tan−1 u + C = tan−1
) + C.
a
a
a
MATH115 T2 Winter 04, A.Potapov
— 5 —
www.math.ualberta.ca/∼apotapov/math115.htm