Trigonometric identities - Wilkes University
Formula Sheet
Trigonometric identities:
sin2 x + cos2 x = 1 tan2 x + 1 = sec2 x sin 2x = 2 sin x cos x
cos 2x = cos2 x - sin2 x sin2 x = (1 - cos 2x)/2 cos2 x = (1 + cos 2x)/2
The line through the point (x0, y0, z0) parallel to the vector v = v1i + v2j + v3k is given by
r(t) = (x0 + tv1)i + (y0 + tv2)j + (z0 + tv3)k
The plane through the point (x0, y0, z0) with normal vector n = Ai + Bj + Ck is given by the equation
A(x - x0) + B(y - y0) + C(z - z0) = 0.
u ? v = |u||v| cos and |u ? v| = |u||v| sin
The (arc) length of a curve C given by r(t) for a t b is
b
L = ds = |v(t)|dt,
C
a
where
v(t)
=
dr dt
.
Distance from a point S to a line passing through point P , parallel to
v:
d
=
|PS?v| |v|
.
Given a point S in space, and a plane with normal n and point P on
the
plane,
the
distance
from
S
to
the
plane
is
d
=
|PS?n| |n|
.
N = (dT/dt)/|dT/dt| (principal unit normal vector, here T is the unit tangent vector).
= (1/|v|)|dT/dt| (curve curvature).
(acceleration) a = aT T+aN N, where aT
=
d(|v|) dt
and aN
=
|a|2 - a2T .
The area enclosed by a polar curve r = f () between two angles 1 and 2 is given by
1
2 r2d = 1
2
(f ())2 d.
2 1
2 1
Directional Derivative: The derivative of the function f in the direction of a unit vector u at the point P0 is Duf (P0) = f (P0) ? u.
The linearization of f (x, y) at a point P0 = (x0, y0) is L(x, y) = f (P0) + fx(P0)(x - x0) + fy(P0)(y - y0).
The Second Derivative Test for Functions of Two Variables: Let f (x, y) be a twice differentiable function, and assume its second
partial derivatives are continuous. Let (a, b) be a critical point for f ,
and define the Hessian of f at (a, b) to be H(a, b) = fxx(a, b)fyy(a, b) - [fxy(a, b)]2.
? If H(a, b) < 0, then (a, b) is a saddle point. ? If H(a, b) > 0 and fxx(a, b) > 0, then (a, b) is a local min. ? If H(a, b) > 0 and fxx(a, b) < 0, then (a, b) is a local max. ? If H(a, b) = 0, the test is inconclusive.
Lagrange Multipliers (problems with equality constraints): To find min/max of a function f , subject to a constraint g = 0, solve the system of equations f = g
g=0
for x, y, and . The max and min values of f are attained at the points among these solutions.
Polar Coordinates: ? x = r cos , y = r sin , r = ? dA = r drd
x2 + y2,
tan
=
y x
Cylindrical Coordinates: ? x = r cos , y = r sin , z = z, r = ? dV = dz r drd
x2 + y2,
tan
=
y x
Spherical Coordinates:
? x = sin cos , y = sin sin , z = cos
? = x2 + y2 + z2 = r2 + z2, r = sin ? dV = 2 sin ddd
The center of mass (x?, y?, z?) of a three-dimensional object D with
density (x, y, z): x? =
D xdV D dV
,
y?
=
( D dV is the mass of the object.)
D ydV D dV
,
z?
=
D zdV D dV
.
Change of Variables in Double Integrals: Suppose a transformation T given by x = g(u, v) and y = h(u, v) is one-to-one (except perhaps on the boundary of G), maps a region G in the uv-plane to a region R in the xy-plane, and g and h have continuous partial derivatives in G. If f is a continuous function over the region R, then
f (x, y)dx dy = f (g(u, v), h(u, v)) |J(u, v)|du dv,
R
G
where the Jacobian of the transformation is given by
x x
u v
J(u, v) =
.
y y
u v
Given a vector field F and a smooth curve C, parameterized by r(t), a t b:
? flow/work/circulation along/around C is given by:
C F ? dr =
b a
F(r(t))
?
dr dt
dt.
? flux of F = M i + N j across C in the plane is given by:
C M dy - N dx =
ab(M
dy dt
-
N
dx dt
)dt.
Component Test for Conservative Vector Fields: A vector field F = M, N, P is conservative if and only if M N M P N P = , =, =. y x z x z y Also: there is a potential function f so that f = F and line integrals
are independent of path between any two points A and B:
C F ? dr = f (B) - f (A).
Green's Theorem (R is a region enclosed by a loop C):
C F ? dr = C M dx + N dy =
R
(
N x
-
M y
)
dxdy
C F ? n ds = C M dy - N dx =
R(
M x
+
N y
)
dxdy
Surface integral of f (x, y, z) over surface S, parameterized by r(u, v) over the region R in the uv-plane:
f (x, y, z) d = f (r(u, v))|ru ? rv| dudv.
S
R
For surface area, use f (x, y, z) = 1.
Flux of a vector field F through surface S parameterized by r(u, v) is
F ? n d = F(r(u, v)) ? (ru ? rv) dudv
S
S
.
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