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Derivations of Key Formulas
The derivations of various key formulas presented in this book are given below.
Follow-up exercises allow you to understand the derivations better by applying
them to specific situations and/or repeating them under different conditions.
The Quadratic Formula
......................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lesson 5.6, page 291
You can derive the quadratic formula by completing the square for the general
quadratic equation ax 2 + bx + c = 0 (where a ≠ 0).
ax 2 + bx + c = 0
b
a
Standard form of general equation
c
a
x 2 + x + = 0
Divide each side by a.
b
a
c
a
c
a
x2 + x = º
b
a
Subtract }} from each side.
2ba = ºac + 2ba b º 4ac
x + 2ba = 4a
x2 + x + 2
2
2
Complete the square by adding the square of
half the coefficient of x to each side.
2
Write the left side as the square of a binomial.
Write the right side as a single fraction.
2
b2 º 4ac
4a2
b
2a
x + = ±
b
2a
x = º ±
Take square roots of each side.
b2 º 4ac
4a2
4ac
ºb ± b2º
x = 2a
b
2a
Subtract }} from each side.
Simplify.
Exercises
1. Solve the quadratic equation 3x 2 + 5x + 2 = 0 by completing the square and by
using the quadratic formula. Check to see that you get the same solutions.
2. Derive a formula for the solution of a quadratic equation of the form
x 2 + mx + n = 0. Check to see that your formula works for x 2 º 4x + 3 = 0.
3. Show that the function ƒ(x) = ax 2 + bx + c can be written in intercept form as
4ac
ºb + b2º
ƒ(x) = a x º 2a
x º
ºb º b2
º4a
c
2a
by multiplying the factors in the intercept form and simplifying.
Equation of a Parabola
......................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Lesson 10.2, page 596
Using the geometric definition of a parabola and the distance formula (page 589),
you can derive the equation of a parabola.
DEFINITION A parabola is the set of points (x, y) that are equidistant from a fixed
line, called the directrix, and a fixed point, called the focus.
Derivations of Key Formulas
895
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Equation of a Parabola (continued)
For p > 0, let the coordinates of the focus be (0, p)
and the equation of the directrix be y = ºp as in the
diagram shown. Notice that (0, 0) is p units from the
focus and also p units from the directrix. Therefore,
(0, 0) is a point on the parabola.
distance between
distance between
=
(x, y) and (0, p)
(x, y) and y = ºp
x2 + (y º p)2 = (y + p)2
2
(x, y )
x
(x, p)
Definition of a parabola
(x
º0
)2
+( yº
p
)2 = (x
ºx
)2
+( yº
(º
p))
2
2
x2
+( yº
p)2 = (y
+p)
2
focus
(0, p)
directrix
y p
For any point (x, y) on the parabola:
2
y
Distance formula
Simplify.
Square each side.
2
2
x + y º 2py + p = y + 2py + p
Multiply.
x2 = 4py
Subtract y 2 º 2py + p 2 from each side.
In the preceding derivation, p was assumed to be
positive. If p were negative, the focus would be below
the x-axis and the directrix would be above it, as
shown. There would be no change in the derivation,
however, so the equation x2 = 4py describes a
parabola that opens up when p > 0 and a parabola
that opens down when p < 0.
y
directrix
y p
(x, p)
x
(0, p)
focus
(x, y )
Exercises
1. Derive an equation for the set of all points (x, y) that are the same distance from
(0, 5) as they are from y = º5. Compare your work with each step of the
derivation shown above.
2. Show that y2 = 4px describes a parabola that opens to the right when p > 0 and a
parabola that opens to the left when p < 0.
Equation of an Ellipse
.......................................................
Lesson 10.4, page 609
Using the geometric definition of an ellipse and the distance formula (page 589),
you can derive the equation of an ellipse.
DEFINITION An ellipse is the set of
y
points (x, y) such that the sum of the
distances between (x, y) and two distinct
points, called the foci, is constant.
(0, b)
Let the foci have coordinates (ºc, 0)
and (c, 0), and let (0, b) be the point
where the ellipse intersects the positive
y-axis. Let the distance between (0, b)
and each focus be a. Then the sum of the
distances from any point (x, y) on the
ellipse to the two foci must be 2a.
896
Student Resources
a
(c, 0)
c
b
a
c
(c, 0)
x
(x, y )
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distance between
distance between
+
= 2a
(x, y) and (ºc, 0)
(x, y) and (c, 0)
Definition of an ellipse
(x
º(º
c)
)2
+( yº
0
)2 + (x
ºc
)2
+( yº
0
)2 = 2a
Distance formula
(x
+c
)2
+y2 + (x
ºc
)2
+y2 = 2a
Simplify.
Subtract (x
º
c
)2
+y2
from each side.
(x
+c
)2
+y2 = 2a º (x
ºc
)2
+y2
At this point, square each side of the equation twice to eliminate the two radicals.
2
(x + c)2 + y2 = 4a2 º 4a
(x
ºc
)2
+y
+ (x º c)2 + y2
Square each side.
Subtract (x º c)2 + y 2 from
each side, and simplify.
4cx = 4a2 º 4a(x
ºc
)2
+y2
Subtract 4a 2 from each side,
and divide each side by 4.
cx º a2 = ºa(x
ºc
)2
+y2
c2x2 º 2a2cx + a4 = a2(x º c)2 + y2
Square each side again.
c2x2 º 2a2cx + a4 = a2x2 º 2a2cx + a2c2 + a2y2
4
2 2
2 2
2 2
Multiply.
2 2
Subtract c 2x 2 º 2a 2cx from each side.
a =a x ºc x +a c +a y
a2(a2 º c2) = (a2 º c2)x2 + a2y2
Subtract a 2c 2 from each side, and factor.
Since a, b, and c are the lengths of the sides of a right triangle (see diagram on
previous page), you have a2 = b2 + c2, or a2 º c2 = b2, by the Pythagorean
theorem. Therefore:
a2b2 = b2x2 + a2y2
x2
a
y2
1 = 2 + 2
Substitute b 2 for a 2 º c 2.
Divide each side by a 2b 2.
b
You now have the equation of an ellipse whose center is (0, 0). You know that the length
of the vertical axis of the ellipse—the distance between (0, b) and (0, ºb)—is 2b. But
what is the length of the other axis?
y
Let (d, 0) be the intersection of the
ellipse with the positive x-axis as shown.
Then the distance between (d, 0) and
(ºc, 0) plus the distance between (d, 0)
and (c, 0) is 2a. That is:
[c + c + (d º c)] + (d º c) = 2a
2d = 2a
b
(c, 0)
(c, 0)
c
(d, 0)
c
b
x
dc
d=a
Therefore, the ellipse intersects the x-axis at (ºa, 0) and (a, 0), and the length of the
horizontal axis of the ellipse is 2a.
Exercises
1. Derive an equation for the set of all points (x, y) such that the sum of the
distances between (x, y) and the two points (º3, 0) and (3, 0) is 10. Compare
your work with each step of the derivation shown above.
y2
x2
2. Show that the equation 2 + 2 = 1 describes an ellipse where the two foci have
b
a
coordinates (0, ºc) and (0, c) and where (b, 0) is the point where the ellipse
intersects the positive x-axis. (Let the distance between (b, 0) and each focus be a.)
Derivations of Key Formulas
897
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Equation of a Hyperbola
................................................
Lesson 10.5, page 615
Using the geometric definition of a hyperbola and the distance formula (page 589),
you can derive the equation of a hyperbola.
DEFINITION A hyperbola is the set of points
(x, y )
y
(x, y) such that the difference of the distances
between (x, y) and two distinct points, called
the foci, is constant.
Let the two foci have coordinates (ºc, 0) and
(c, 0) where c > 0, and let the coordinates of
the points, called the vertices, where the
hyperbola intersects the x-axis have coordinates
(ºa, 0) and (a, 0) where a > 0.
(c, 0)
(a, 0)
(a, 0)
(c, 0) x
The distance between either focus and the vertex farther from it is c + a, while the
distance between either focus and the vertex closer to it is c º a. The difference in
the distances is c + a º (c º a) = 2a or c º a º (c + a) = º2a. Therefore, ±2a
is the common difference in the distances to the foci from any point on the hyperbola.
distance between
distance between
º
= ±2a
(x, y) and (ºc, 0)
(x, y) and (c, 0)
Definition of a hyperbola
(x
º(º
c)
)2
+( yº
0
)2 º (x
ºc
)2
+( yº
0
)2 = ±2a
Distance formula
(x
+c
)2
+y2 º (x
ºc
)2
+y2 = ±2a
Simplify.
(x
+c
)2
+y2 = ±2a + (x
ºc
)2
+y2
2
Add (x
º
c
)2
+y
to each side.
At this point, square each side of the equation twice to eliminate the two radicals.
ºc
)2
+y2 + (x º c)2 + y2
(x + c)2 + y2 = 4a2 ± 4a(x
4cx = 4a2 ± 4a(x
ºc
)2
+y2
cx º a2 = ±a(x
ºc
)2
+y2
Square each side.
Subtract (x º c)2 + y 2 from
each side, and simplify.
Subtract 4a 2 from each side,
and divide each side by 4.
c2x2 º 2a2cx + a4 = a2(x º c)2 + y2
Square each side again.
c2x2 º 2a2cx + a4 = a2x2 º 2a2cx + a2c2 + a2y2
Multiply.
c2x2 = a2x2 + a2c2 + a2y2 º a4
(c2 º a2)x2 º a2y2 = a2(c2 º a2)
Subtract º2a 2cx + a4 from each side.
Subtract a 2x 2 + a 2y 2 from each side, and factor.
By drawing perpendiculars to the x-axis at the vertices, you can form right triangles
each having one leg (along the x-axis) of length a and a hypotenuse of length c as
shown at the top of the next page. Let b be the length of the other leg. Then
a2 + b2 = c2, or b2 = c2 º a2, by the Pythagorean theorem. Therefore:
898
b2x2 º a2y2 = a2b2
Substitute b 2 for c 2 º a 2.
y2
x2
2 º 2 = 1
a
b
Divide each side by a 2b 2.
Student Resources
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You now have the equation of a hyperbola
whose center is (0, 0). The lines that contain
the hypotenuses of the right triangles have
b
y a x
(a, 0)
b
equations y = ±x. To see that these lines
a
(c, 0)
are asymptotes of the hyperbola, solve the
equation of the hyperbola for y:
y2
x2
2 º 2 = 1
a
b
b
y
y ax
b c
c b
a
b c
c
a
b
(a, 0)
(c, 0)
x
Equation of a hyperbola
b2x2 º a2y2 = a2b2
ºa2y2 = a2b2 º b2x2
b2
a
b2
y2 = 2 (x2 º a2)
a
b 2
y = ± x
ºa2
a
y2 = 2 x2 º b2
Multiply each side by a 2b 2.
Subtract b 2x 2 from each side.
Divide each side by ºa 2.
b2
a
Factor out }}2 .
Take square roots of each side.
As x ˘ +‡, the value of x2 becomes much greater than the value of a2 (a constant)
b
a
so that x2
ºa2 ˘ x2 = x. Therefore, the graph of y = ±
x2
ºa2 approaches
b
a
the graph of y = ±x as x ˘ +‡. You can make a similar argument for x ˘ º‡.
Exercises
1. Derive an equation for the set of all points (x, y) such that the difference of the
distances between (x, y) and the two points (º5, 0) and (5, 0) is ±8. Compare
your work with each step of the derivation shown on the previous page.
y2
x2
2. Show that the equation 2 º 2 = 1 describes a hyperbola where the two foci
a
b
have coordinates (0, ºc) and (0, c) and where the hyperbola intersects the y-axis
at (0, ºa) and (0, a).
Permutations of n Objects
Taken r at a Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lesson 12.1, page 703
Using the fundamental counting principle (page 701) and the definition of factorial
(page 681), you can derive the formula for the number of permutations of n objects
taken r at a time.
Number of
Number of
Number of
Number of
Number of
ways to choose • ways to choose • ways to choose • . . . • ways to choose
=
permutations
2nd object
3rd object
rth object
1st object
nPr
= n • (n º 1) • (n – 2) • . . . • (n º r + 1)
n • (n º 1) • (n º 2) • . . . • (n º r + 1) • (n º r) • (n º r º 1) • . . . • 2 • 1
= (n º r) • (n º r º 1) • . . . • 2 • 1
n!
(n º r)!
= Multiply and divide
by (n º r)!.
Definition of factorial
Derivations of Key Formulas
899
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Permutations of n Objects Taken r at a Time (continued)
Exercises
1. In the derivation shown on the previous page, explain why the number of ways
to choose the rth object is n º r + 1.
2. You know that order is important when arranging a group of objects. You will
learn in Lesson 12.2 that a combination is a grouping of objects where the order
of the objects is not important.
a. For any group of r objects, how many ways are there of arranging the objects?
b. Let nCr denote the number of combinations of n objects taken r at a time. Use
your answer from part (a) to write nPr in terms of nCr. (That is, complete this
? )(nCr) because for each combination of r objects there are
statement: nPr = (
?
permutations
of
those objects.)
c. Using the equation from part (b) and the formula for nPr, derive a formula for
nCr
in terms of n and r. (This is the formula given on page 708.)
Formula for Arc Length
..................................................
Lesson 13.2, page 779
By using the circumference and radian measure of a circle (page 777), you can write
a proportion to derive the formula for the length of a circular arc.
Let r be the radius of a circle, and let † be the radian measure
of a central angle that intercepts an arc of length s. Knowing
that the circumference of the circle is 2πr and that there are
2π radians in a full circle, you can write the following
proportion:
radian measure of central angle
length of arc
= circumference of circle
radian measure of full circle
s
†
= 2πr
2π
s = r†
r
†
s
Write a proportion.
Substitute.
Multiply each side by 2πr.
So, the length of the arc is just the product of the circle’s radius and the radian
measure of the central angle that intercepts the arc.
Exercises
1. Show that the arc length formula gives a correct result for a semicircular arc.
2. Derive the formula for the area of a sector formed by a central angle † (measured
in radians) in a circle of radius r. Your derivation should involve setting up and
solving a proportion, as above.
The Law of Sines
.....................................................................
C
By using the right triangle definition of sine (page 769),
you can derive the law of sines, which applies to any
triangle.
Let a, b, and c be the lengths of the sides of ¤ABC
D
having length h.
as shown. Introduce altitude C
900
Student Resources
Lesson 13.5, page 799
b
A
h
c
D
a
B
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h
b
From right ¤ACD you have sin A = , or h = b sin A, by the definition of sine.
h
a
Likewise, from right ¤BCD you have sin B = , or h = a sin B, also by the
definition of sine. Therefore:
b sin A = a sin B
Equate expressions for h.
sin A
sin B
= a
b
Divide each side by ab.
This establishes one of the three equalities from the law of sines.
Exercises
1. Introduce a different altitude in ¤ABC and derive another equality from the law
of sines. How does this result, combined with the one above, imply the third
equality from the law of sines?
2. Derive the three formulas for the area of a triangle given on page 802 using an
argument similar to the one above for the law of sines.
The Law of Cosines
..............................................................
By using the Pythagorean theorem and
the right triangle definition of cosine
(page 769), you can derive the law of
cosines, which applies to any triangle.
Lesson 13.6, page 807
C
b
x
A
c
Let a, b, and c be the lengths of the
sides of ¤ABC as shown. Introduce
D
having length h.
altitude C
a
h
D
cx
B
Let AD = x; it follows that DB = c º x. Use the Pythagorean theorem to find two
different expressions for h2:
RIGHT ¤BCD
2
RIGHT ¤ACD
2
2
x 2 + h2 = b2
h + (c º x) = a
h2 = a2 º (c º x)2
h2 = b2 º x 2
Therefore:
a2 º (c º x)2 = b2 º x2
Equate expressions for h 2.
a2 º c2 + 2cx º x2 = b2 º x2
Multiply.
a2 = b2 + c2 º 2cx
Add c 2 º 2cx + x 2 to each side.
x
b
From right ¤ACD you have cos A = , or x = b cos A, by the definition of cosine.
Therefore:
a2 = b2 + c2 º 2c(b cos A)
2
2
2
a = b + c º 2bc cos A
Substitute b cos A for x.
Commutative property of multiplication
This establishes one of the three forms of the law of cosines.
Exercises
1. Using a different altitude in ¤ABC, derive another form of the law of cosines.
2. Solve a2 = b2 + c2 º 2bc cos A for cos A.
Derivations of Key Formulas
901
Page 1 of 2
Negative Angle Identities
Lesson 14.3, page 848
............................................
By using a geometric argument, you can establish the negative angle identities.
Draw an angle † in standard position. Let P(a, b) be a
point (other than the origin) on the terminal side of †.
The angle º† has the same amount of rotation as †
but the direction of rotation is clockwise rather than
counterclockwise from the positive x-axis. The
terminal side of º† is therefore a reflection of the
terminal side of † in the x-axis. This means that the
point P§(a, ºb) is on the terminal side of º†. If r is
the distance OP = OP§, then by the definition of sine
(see page 784) you have:
ºb
r
y
P (a, b)
†
O
x
†
P ’(a, b)
b
r
sin (º†) = = º = ºsin †
Exercises
1. Continue the argument presented above to establish the identities
cos (º†) = cos † and tan (º†) = ºtan †.
π
2. Use a geometric argument to establish the cofunction identity sin º † = cos †.
2
π
(Hint: The terminal side of º † is the reflection of the terminal side of † in the
2
line y = x.)
The Difference Formula for Cosine
................
Lesson 14.6, page 869
By using the right triangle definitions of sine and cosine (page 769), the distance
formula (page 589), the law of cosines (page 807), and a Pythagorean identity
(page 848), you can derive the difference formula for cosine.
Draw two angles in standard position. Let v be the measure of the smaller angle and u
be the measure of the larger angle. Choose points P and Q on the terminal sides of the
angles so that the points are each 1 unit from the origin.
Draw perpendiculars from P and Q to the
x-axis, and let R and S be the points of
intersection of the perpendiculars with
the x-axis. Since ¤PRO and ¤QSO are
right triangles, the lengths of their legs
are sin u and cos u (for ¤PRO) and sin v
and cos v (for ¤QSO) by the right
triangle definitions of sine and cosine.
Therefore, the coordinates of P are
(cos u, sin u), and the coordinates of Q
are (cos v, sin v).
y
P (cos u, sin u)
1
sin u
uv
u
1
R
cos u
cos v
osu
ºcosv)
2+
(si
nu
ºsin
v
)2 ,
PQ = (c
or PQ2 = (cos u º cos v)2 + (sin u º sin v)2.
By the law of cosines applied to ¤PQO, you have:
PQ2 = 12 + 12 º 2(1)(1) cos (u º v)
Student Resources
sin v
v
O
By the distance formula, you have:
902
œ (cos v, sin v)
S
x
Page 1 of 2
Equating the two expressions for PQ2, using a Pythagorean identity, and solving for
cos (u º v), you have:
2 º 2 cos (u º v) = (cos u º cos v)2 + (sin u º sin v)2
2 º 2 cos (u º v) = cos2 u º 2 cos u cos v + cos2 v + sin2 u º 2 sin u sin v + sin2 v
2 º 2 cos (u º v) = (cos2 u + sin2 u) + (cos2 v + sin2 v) º 2(cos u cos v + sin u sin v)
2 º 2 cos (u º v) = 1 + 1 º 2(cos u cos v + sin u sin v)
º2 cos (u º v) = º2(cos u cos v + sin u sin v)
cos (u º v) = cos u cos v + sin u sin v
The same argument can be used for angles in any quadrants, not just Quadrant I.
You can verify the cofunction identities (page 848) using the difference formula
for cosine. And you can in turn derive the other sum and difference formulas on
page 869 using the difference formula for cosine and the cofunction and negative
angle identities (for example, see Exercises 62–64 on page 874).
Exercises
π
1. a. Verify the cofunction identity cos º † = sin † using the difference
2
formula for cosine.
π
b. Use the cofunction identity from part (a) to verify sin º † = cos †.
2
2. Use the cofunction identities from Exercise 1 to derive the sum formula for sine.
π
π
Begin by writing sin (u + v) = cos º (u + v) = cos º u º v .
2
2
3. Use the sum formula for sine (see Exercise 2) and the negative angle identities to
derive the difference formula for sine.
The Double- and Half-Angle Formulas
........
Lesson 14.7, page 875
The double-angle formulas are obtained directly from the sum formulas, and the halfangle formulas are obtained directly from the double-angle formulas.
To establish one of the double-angle formulas for cosine, simply let v = u in the sum
formula for cosine:
cos 2u = cos (u + u)
Write 2u as a sum.
= cos u cos u º sin u sin u
Use sum formula for cosine.
= cos2 u º sin2 u
Simplify.
The other variations of the double-angle formula for cosine are obtained using the
Pythagorean identity sin2 u + cos2 u = 1. For instance, replace cos2 u with 1 º sin2 u
in cos 2u = cos2 u º sin2 u to obtain:
cos 2u = (1 º sin2 u) º sin2 u
= 1 º 2 sin2 u
Substitute.
Simplify.
Likewise, replace sin2 u with 1 º cos2 u in cos 2u = cos2 u º sin2 u to obtain:
cos 2u = cos2 u º (1 º cos2 u)
= cos2 u º 1 + cos2 u
2
= 2 cos u º 1
Substitute.
Distribute.
Simplify.
Derivations of Key Formulas
903
Page 1 of 2
The Double- and Half-Angle Formulas (continued)
To establish the half-angle formula for cosine, use the double-angle formula
cos 2† = 2 cos2 † º 1:
cos 2† = 2 cos2 † º 1
u2 Double-angle formula for cosine
u
2
Substitute }} for †.
u
2
Simplify.
1 + cos u = 2 cos2 u
2
Add 1 to each side.
1 + cos u
u
= cos2 2
2
Divide each side by 2.
cos 2 = 2 cos2 º 1
cos u = 2 cos2 º 1
1 + cos u
2
u
2
± = cos u
2
Take square roots of each side.
Exercises
1. Use the sum formula for sine to establish the double-angle formula for sine.
2. Use the double-angle formula cos 2† = 1 º 2 sin2 † to establish the half-angle
formula for sine.
904
Student Resources