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Algebra and Trigonometry 2e

A | Proofs, Identities, and Toolkit Functions

Algebra and Trigonometry 2eA | Proofs, Identities, and Toolkit Functions

Important Proofs and Derivations

Product Rule

log a xy= log a x+ log a y log a xy= log a x+ log a y

Proof:

Let m= log a x m= log a x and n= log a y. n= log a y.

Write in exponent form.

x= a m x= a m and y= a n . y= a n .

Multiply.

xy= a m a n = a m+n xy= a m a n = a m+n

a m+n = xy log a (xy) = m+n = log a x+ log b y a m+n = xy log a (xy) = m+n = log a x+ log b y

Change of Base Rule

log a b= log c b log c a log a b= 1 log b a log a b= log c b log c a log a b= 1 log b a

where x x and y y are positive, and a>0,a1. a>0,a1.

Proof:

Let x= log a b. x= log a b.

Write in exponent form.

a x =b a x =b

Take the log c log c of both sides.

log c a x = log c b x log c a = log c b x = log c b log c a log a b = log c b log c a log c a x = log c b x log c a = log c b x = log c b log c a log a b = log c b log c a

When c=b, c=b,

log a b= log b b log b a = 1 log b a log a b= log b b log b a = 1 log b a

Heron’s Formula

A= s( sa )( sb )( sc ) A= s( sa )( sb )( sc )

where s= a+b+c 2 s= a+b+c 2

Proof:

Let a, a, b, b, and c c be the sides of a triangle, and h h be the height.

A triangle with sides labeled: a, b and c.  A line runs through the center of the triangle, bisecting the top angle; this line is labeled: h.

So s= a+b+c 2 s= a+b+c 2 .

We can further name the parts of the base in each triangle established by the height such that p+q=c. p+q=c.

A triangle with sides labeled: a, b, and c.  A line runs through the center of the triangle bisecting the angle at the top; this line is labeled: h. The two new line segments on the base of the triangle are labeled: p and q.

Using the Pythagorean Theorem, h 2 + p 2 = a 2 h 2 + p 2 = a 2 and h 2 + q 2 = b 2 . h 2 + q 2 = b 2 .

Since q=cp, q=cp, then q 2 = ( cp ) 2 . q 2 = ( cp ) 2 . Expanding, we find that q 2 = c 2 2cp+ p 2 . q 2 = c 2 2cp+ p 2 .

We can then add h 2 h 2 to each side of the equation to get h 2 + q 2 = h 2 + c 2 2cp+ p 2 . h 2 + q 2 = h 2 + c 2 2cp+ p 2 .

Substitute this result into the equation h 2 + q 2 = b 2 h 2 + q 2 = b 2 yields b 2 = h 2 + c 2 2cp+ p 2 . b 2 = h 2 + c 2 2cp+ p 2 .

Then replacing h 2 + p 2 h 2 + p 2 with a 2 a 2 gives b 2 = a 2 2cp+ c 2 . b 2 = a 2 2cp+ c 2 .

Solve for p p to get

p= a 2 + b 2 c 2 2c p= a 2 + b 2 c 2 2c

Since h 2 = a 2 p 2 , h 2 = a 2 p 2 , we get an expression in terms of a, a, b, b, and c. c.

h 2 = a 2 p 2 = (a+p)(ap) = [ a+ ( a 2 + c 2 b 2 ) 2c ][ a ( a 2 + c 2 b 2 ) 2c ] = ( 2ac+ a 2 + c 2 b 2 )( 2ac a 2 c 2 + b 2 ) 4 c 2 = ( (a+c) 2 b 2 )( b 2 (ac) 2 ) 4 c 2 = (a+b+c)(a+cb)(b+ac)(ba+c) 4 c 2 = (a+b+c)(a+b+c)(ab+c)(a+bc) 4 c 2 = 2s(2sa)(2sb)(2sc) 4 c 2 h 2 = a 2 p 2 = (a+p)(ap) = [ a+ ( a 2 + c 2 b 2 ) 2c ][ a ( a 2 + c 2 b 2 ) 2c ] = ( 2ac+ a 2 + c 2 b 2 )( 2ac a 2 c 2 + b 2 ) 4 c 2 = ( (a+c) 2 b 2 )( b 2 (ac) 2 ) 4 c 2 = (a+b+c)(a+cb)(b+ac)(ba+c) 4 c 2 = (a+b+c)(a+b+c)(ab+c)(a+bc) 4 c 2 = 2s(2sa)(2sb)(2sc) 4 c 2

Therefore,

h 2 = 4s(sa)(sb)(sc) c 2 h = 2 s(sa)(sb)(sc) c h 2 = 4s(sa)(sb)(sc) c 2 h = 2 s(sa)(sb)(sc) c

And since A= 1 2 ch, A= 1 2 ch, then

A = 1 2 c 2 s(sa)(sb)(sc) c = s(sa)(sb)(sc) A = 1 2 c 2 s(sa)(sb)(sc) c = s(sa)(sb)(sc)

Properties of the Dot Product

u·v=v·u u·v=v·u

Proof:

u·v = u 1 , u 2 ,... u n · v 1 , v 2 ,... v n = u 1 v 1 + u 2 v 2 +...+ u n v n = v 1 u 1 + v 2 u 2 +...+ v n v n = v 1 , v 2 ,... v n · u 1 , u 2 ,... u n =v·u u·v = u 1 , u 2 ,... u n · v 1 , v 2 ,... v n = u 1 v 1 + u 2 v 2 +...+ u n v n = v 1 u 1 + v 2 u 2 +...+ v n v n = v 1 , v 2 ,... v n · u 1 , u 2 ,... u n =v·u

u·( v+w )=u·v+u·w u·( v+w )=u·v+u·w

Proof:

u·(v+w) = u 1 , u 2 ,... u n ·( v 1 , v 2 ,... v n + w 1 , w 2 ,... w n ) = u 1 , u 2 ,... u n · v 1 + w 1 , v 2 + w 2 ,... v n + w n = u 1 ( v 1 + w 1 ), u 2 ( v 2 + w 2 ),... u n ( v n + w n ) = u 1 v 1 + u 1 w 1 , u 2 v 2 + u 2 w 2 ,... u n v n + u n w n = u 1 v 1 , u 2 v 2 ,..., u n v n + u 1 w 1 , u 2 w 2 ,..., u n w n = u 1 , u 2 ,... u n · v 1 , v 2 ,... v n + u 1 , u 2 ,... u n · w 1 , w 2 ,... w n =u·v+u·w u·(v+w) = u 1 , u 2 ,... u n ·( v 1 , v 2 ,... v n + w 1 , w 2 ,... w n ) = u 1 , u 2 ,... u n · v 1 + w 1 , v 2 + w 2 ,... v n + w n = u 1 ( v 1 + w 1 ), u 2 ( v 2 + w 2 ),... u n ( v n + w n ) = u 1 v 1 + u 1 w 1 , u 2 v 2 + u 2 w 2 ,... u n v n + u n w n = u 1 v 1 , u 2 v 2 ,..., u n v n + u 1 w 1 , u 2 w 2 ,..., u n w n = u 1 , u 2 ,... u n · v 1 , v 2 ,... v n + u 1 , u 2 ,... u n · w 1 , w 2 ,... w n =u·v+u·w

u·u= | u | 2 u·u= | u | 2

Proof:

u·u = u 1 , u 2 ,... u n · u 1 , u 2 ,... u n = u 1 u 1 + u 2 u 2 +...+ u n u n = u 1 2 + u 2 2 +...+ u n 2 =| u 1 , u 2 ,... u n | 2 =u·u u·u = u 1 , u 2 ,... u n · u 1 , u 2 ,... u n = u 1 u 1 + u 2 u 2 +...+ u n u n = u 1 2 + u 2 2 +...+ u n 2 =| u 1 , u 2 ,... u n | 2 =u·u

Standard Form of the Ellipse centered at the Origin

1= x 2 a 2 + y 2 b 2 1= x 2 a 2 + y 2 b 2

Derivation

An ellipse consists of all the points for which the sum of distances from two foci is constant:

( x( c ) ) 2 + ( y0 ) 2 + ( xc ) 2 + ( y0 ) 2 =constant ( x( c ) ) 2 + ( y0 ) 2 + ( xc ) 2 + ( y0 ) 2 =constant

An ellipse centered at the origin on an x, y-coordinate plane.  Points C1 and C2 are plotted at the points (0, b) and (0, -b) respectively; these points appear on the ellipse.  Points V1 and V2 are plotted at the points (-a, 0) and (a, 0) respectively; these points appear on the ellipse.  Points F1 and F2 are plotted at the points (-c, 0) and (c, 0) respectively; these points appear on the x-axis, but not the ellipse. The point (x, y) appears on the ellipse in the first quadrant.  Dotted lines extend from F1 and F2 to the point (x, y).

Consider a vertex.

An ellipse centered at the origin.  The points C1 and C2 are plotted at the points (0, b) and (0, -b) respectively; these points are on the ellipse.  The points V1 and V2 are plotted at the points (-a, 0) and (a, 0) respectively; these points are on the ellipse.  The points F1 and F2 are plotted at the points (-c, 0) and (c, 0) respectively; these points are on the x-axis and not on the ellipse.  A line extends from the point F1 to a point (x, y) which is at the point (a, 0).  A line extends from the point F2 to the point (x, y) as well.

Then, ( x( c ) ) 2 + ( y0 ) 2 + ( xc ) 2 + ( y0 ) 2 =2a ( x( c ) ) 2 + ( y0 ) 2 + ( xc ) 2 + ( y0 ) 2 =2a

Consider a covertex.

An ellipse centered at the origin.  The points C1 and C2 are plotted at the points (0, b) and (0, -b) respectively; these points are on the ellipse.  The points V1 and V2 are plotted at the points (-a, 0) and (a, 0) respectively; these points are on the ellipse.  The points F1 and F2 are plotted at the points (-c, 0) and (c, 0) respectively; these points are on the x-axis and not on the ellipse.  There is a point (x, y) which is plotted at (0, b). A line extends from the origin to the point (c, 0), this line is labeled: c.  A line extends from the origin to the point (x, y), this line is labeled: b.  A line extends from the point (c, 0) to the point (x, y); this line is labeled: (1/2)(2a)=a.  A dotted line extends from the point (-c, 0) to the point (x, y); this line is labeled: (1/2)(2a)=a.

Then b 2 + c 2 = a 2 . b 2 + c 2 = a 2 .

(x(c)) 2 + (y0) 2 + (xc) 2 + (y0) 2 = 2a (x+c) 2 + y 2 = 2a (xc) 2 + y 2 (x+c) 2 + y 2 = ( 2a (xc) 2 + y 2 ) 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 4a (xc) 2 + y 2 + (xc) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 4a (xc) 2 + y 2 + x 2 2cx+ y 2 2cx = 4 a 2 4a (xc) 2 + y 2 2cx 4cx4 a 2 = 4a (xc) 2 + y 2 1 4a ( 4cx4 a 2 ) = (xc) 2 + y 2 a c a x = (xc) 2 + y 2 a 2 2xc+ c 2 a 2 x 2 = (xc) 2 + y 2 a 2 2xc+ c 2 a 2 x 2 = x 2 2xc+ c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 c 2 = x 2 c 2 a 2 x 2 + y 2 a 2 c 2 = x 2 ( 1 c 2 a 2 )+ y 2 (x(c)) 2 + (y0) 2 + (xc) 2 + (y0) 2 = 2a (x+c) 2 + y 2 = 2a (xc) 2 + y 2 (x+c) 2 + y 2 = ( 2a (xc) 2 + y 2 ) 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 4a (xc) 2 + y 2 + (xc) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 4a (xc) 2 + y 2 + x 2 2cx+ y 2 2cx = 4 a 2 4a (xc) 2 + y 2 2cx 4cx4 a 2 = 4a (xc) 2 + y 2 1 4a ( 4cx4 a 2 ) = (xc) 2 + y 2 a c a x = (xc) 2 + y 2 a 2 2xc+ c 2 a 2 x 2 = (xc) 2 + y 2 a 2 2xc+ c 2 a 2 x 2 = x 2 2xc+ c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 c 2 = x 2 c 2 a 2 x 2 + y 2 a 2 c 2 = x 2 ( 1 c 2 a 2 )+ y 2

Let 1= a 2 a 2 . 1= a 2 a 2 .

a 2 c 2 = x 2 ( a 2 c 2 a 2 )+ y 2 1 = x 2 a 2 + y 2 a 2 c 2 a 2 c 2 = x 2 ( a 2 c 2 a 2 )+ y 2 1 = x 2 a 2 + y 2 a 2 c 2

Because b 2 + c 2 = a 2 , b 2 + c 2 = a 2 , then b 2 = a 2 c 2 . b 2 = a 2 c 2 .

1 = x 2 a 2 + y 2 a 2 c 2 1 = x 2 a 2 + y 2 b 2 1 = x 2 a 2 + y 2 a 2 c 2 1 = x 2 a 2 + y 2 b 2

Standard Form of the Hyperbola

1= x 2 a 2 y 2 b 2 1= x 2 a 2 y 2 b 2

Derivation

A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances between two fixed points is constant.

Side-by-side graphs of hyperbole.  In Diagram 1: The foci F’ and F are labeled and can be found a little in front of the opening of the hyperbola.  A point P at (x,y) on the right curve is labeled.  A line extends from the F’ focus to the point P labeled: D1.  A line extends from the F focus to the point P labeled: D2.  In Diagram 2:  The foci F’ and F are labeled and can be found a little in front of the opening of the hyperbola.  A point V is labeled at the vertex of the right hyperbola.  A line extends from the F’ focus to the point V labeled: D1.  A line extends from the F focus to the point V labeled: D2.

Diagram 1: The difference of the distances from Point P to the foci is constant:

(x(c)) 2 + (y0) 2 (xc) 2 + (y0) 2 =constant (x(c)) 2 + (y0) 2 (xc) 2 + (y0) 2 =constant

Diagram 2: When the point is a vertex, the difference is 2a. 2a.

( x( c ) ) 2 + ( y0 ) 2 ( xc ) 2 + ( y0 ) 2 =2a ( x( c ) ) 2 + ( y0 ) 2 ( xc ) 2 + ( y0 ) 2 =2a

(x(c)) 2 + (y0) 2 (xc) 2 + (y0) 2 = 2a (x+c) 2 + y 2 (xc) 2 + y 2 = 2a (x+c) 2 + y 2 = 2a+ (xc) 2 + y 2 (x+c) 2 + y 2 = ( 2a+ (xc) 2 + y 2 ) x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (xc) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (xc) 2 + y 2 + x 2 2cx+ y 2 2cx = 4 a 2 +4a (xc) 2 + y 2 2cx 4cx4 a 2 = 4a (xc) 2 + y 2 cx a 2 = a (xc) 2 + y 2 ( cx a 2 ) 2 = a 2 ( (xc) 2 + y 2 ) c 2 x 2 2 a 2 c 2 x 2 + a 4 = a 2 x 2 2 a 2 c 2 x 2 + a 2 c 2 + a 2 y 2 c 2 x 2 + a 4 = a 2 x 2 + a 2 c 2 + a 2 y 2 a 4 a 2 c 2 = a 2 x 2 c 2 x 2 + a 2 y 2 a 2 ( a 2 c 2 ) = ( a 2 c 2 ) x 2 + a 2 y 2 a 2 ( a 2 c 2 ) = ( c 2 a 2 ) x 2 a 2 y 2 (x(c)) 2 + (y0) 2 (xc) 2 + (y0) 2 = 2a (x+c) 2 + y 2 (xc) 2 + y 2 = 2a (x+c) 2 + y 2 = 2a+ (xc) 2 + y 2 (x+c) 2 + y 2 = ( 2a+ (xc) 2 + y 2 ) x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (xc) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (xc) 2 + y 2 + x 2 2cx+ y 2 2cx = 4 a 2 +4a (xc) 2 + y 2 2cx 4cx4 a 2 = 4a (xc) 2 + y 2 cx a 2 = a (xc) 2 + y 2 ( cx a 2 ) 2 = a 2 ( (xc) 2 + y 2 ) c 2 x 2 2 a 2 c 2 x 2 + a 4 = a 2 x 2 2 a 2 c 2 x 2 + a 2 c 2 + a 2 y 2 c 2 x 2 + a 4 = a 2 x 2 + a 2 c 2 + a 2 y 2 a 4 a 2 c 2 = a 2 x 2 c 2 x 2 + a 2 y 2 a 2 ( a 2 c 2 ) = ( a 2 c 2 ) x 2 + a 2 y 2 a 2 ( a 2 c 2 ) = ( c 2 a 2 ) x 2 a 2 y 2

Define b b as a positive number such that b 2 = c 2 a 2 . b 2 = c 2 a 2 .

a 2 b 2 = b 2 x 2 a 2 y 2 a 2 b 2 a 2 b 2 = b 2 x 2 a 2 b 2 a 2 y 2 a 2 b 2 1 = x 2 a 2 y 2 b 2 a 2 b 2 = b 2 x 2 a 2 y 2 a 2 b 2 a 2 b 2 = b 2 x 2 a 2 b 2 a 2 y 2 a 2 b 2 1 = x 2 a 2 y 2 b 2

Trigonometric Identities

Pythagorean Identities cos 2 θ+ sin 2 θ=1 1+ tan 2 θ= sec 2 θ 1+ cot 2 θ= csc 2 θ cos 2 θ+ sin 2 θ=1 1+ tan 2 θ= sec 2 θ 1+ cot 2 θ= csc 2 θ
Even-Odd Identities cos(−θ)=cosθ sec(−θ)=secθ sin(−θ)=sinθ tan(−θ)=tanθ csc(−θ)=cscθ cot(−θ)=cotθ cos(−θ)=cosθ sec(−θ)=secθ sin(−θ)=sinθ tan(−θ)=tanθ csc(−θ)=cscθ cot(−θ)=cotθ
Cofunction Identities cosθ=sin( π 2 θ ) sinθ=cos( π 2 θ ) tanθ=cot( π 2 θ ) cotθ=tan( π 2 θ ) secθ=csc( π 2 θ ) cscθ=sec( π 2 θ ) cosθ=sin( π 2 θ ) sinθ=cos( π 2 θ ) tanθ=cot( π 2 θ ) cotθ=tan( π 2 θ ) secθ=csc( π 2 θ ) cscθ=sec( π 2 θ )
Fundamental Identities tanθ= sinθ cosθ secθ= 1 cosθ cscθ= 1 sinθ cotθ= 1 tanθ = cosθ sinθ tanθ= sinθ cosθ secθ= 1 cosθ cscθ= 1 sinθ cotθ= 1 tanθ = cosθ sinθ
Sum and Difference Identities cos(α+β)=cosαcosβsinαsinβ cos(αβ)=cosαcosβ+sinαsinβ sin(α+β)=sinαcosβ+cosαsinβ sin(αβ)=sinαcosβcosαsinβ tan(α+β)= tanα+tanβ 1tanαtanβ tan(αβ)= tanαtanβ 1+tanαtanβ cos(α+β)=cosαcosβsinαsinβ cos(αβ)=cosαcosβ+sinαsinβ sin(α+β)=sinαcosβ+cosαsinβ sin(αβ)=sinαcosβcosαsinβ tan(α+β)= tanα+tanβ 1tanαtanβ tan(αβ)= tanαtanβ 1+tanαtanβ
Double-Angle Formulas sin(2θ)=2sinθcosθ cos(2θ)= cos 2 θ sin 2 θ cos(2θ)=12 sin 2 θ cos(2θ)=2 cos 2 θ1 tan(2θ)= 2tanθ 1 tan 2 θ sin(2θ)=2sinθcosθ cos(2θ)= cos 2 θ sin 2 θ cos(2θ)=12 sin 2 θ cos(2θ)=2 cos 2 θ1 tan(2θ)= 2tanθ 1 tan 2 θ
Half-Angle Formulas sin α 2 =± 1cosα 2 cos α 2 =± 1+cosα 2 tan α 2 =± 1cosα 1+cosα tan α 2 = sinα 1+cosα tan α 2 = 1cosα sinα sin α 2 =± 1cosα 2 cos α 2 =± 1+cosα 2 tan α 2 =± 1cosα 1+cosα tan α 2 = sinα 1+cosα tan α 2 = 1cosα sinα
Reduction Formulas sin 2 θ= 1cos( 2θ ) 2 cos 2 θ= 1+cos( 2θ ) 2 tan 2 θ= 1cos( 2θ ) 1+cos( 2θ ) sin 2 θ= 1cos( 2θ ) 2 cos 2 θ= 1+cos( 2θ ) 2 tan 2 θ= 1cos( 2θ ) 1+cos( 2θ )
Product-to-Sum Formulas cosαcosβ= 1 2 [ cos(αβ)+cos(α+β) ] sinαcosβ= 1 2 [ sin(α+β)+sin(αβ) ] sinαsinβ= 1 2 [ cos(αβ)cos(α+β) ] cosαsinβ= 1 2 [ sin(α+β)sin(αβ) ] cosαcosβ= 1 2 [ cos(αβ)+cos(α+β) ] sinαcosβ= 1 2 [ sin(α+β)+sin(αβ) ] sinαsinβ= 1 2 [ cos(αβ)cos(α+β) ] cosαsinβ= 1 2 [ sin(α+β)sin(αβ) ]
Sum-to-Product Formulas sinα+sinβ=2sin( α+β 2 )cos( αβ 2 ) sinαsinβ=2sin( αβ 2 )cos( α+β 2 ) cosαcosβ=2sin( α+β 2 )sin( αβ 2 ) cosα+cosβ=2cos( α+β 2 )cos( αβ 2 ) sinα+sinβ=2sin( α+β 2 )cos( αβ 2 ) sinαsinβ=2sin( αβ 2 )cos( α+β 2 ) cosαcosβ=2sin( α+β 2 )sin( αβ 2 ) cosα+cosβ=2cos( α+β 2 )cos( αβ 2 )
Law of Sines sinα a = sinβ b = sinγ c a sinα = b sinβ = c sinγ sinα a = sinβ b = sinγ c a sinα = b sinβ = c sinγ
Law of Cosines a 2 = b 2 + c 2 2bccosα b 2 = a 2 + c 2 2accosβ c 2 = a 2 + b 2 2abcosγ a 2 = b 2 + c 2 2bccosα b 2 = a 2 + c 2 2accosβ c 2 = a 2 + b 2 2abcosγ
Table A1

ToolKit Functions

Three graphs side-by-side. From left to right, graph of the identify function, square function, and square root function. All three graphs extend from -4 to 4 on each axis.
Figure A1
Three graphs side-by-side. From left to right, graph of the cubic function, cube root function, and reciprocal function. All three graphs extend from -4 to 4 on each axis.
Figure A2
Three graphs side-by-side. From left to right, graph of the absolute value function, exponential function, and natural logarithm function. All three graphs extend from -4 to 4 on each axis.
Figure A3

Trigonometric Functions

Unit Circle

Graph of unit circle with angles in degrees, angles in radians, and points along the circle inscribed.
Figure A4
Angle 0 0 π 6 ,or 30° π 6 ,or 30° π 4 ,or 45° π 4 ,or 45° π 3 ,or 60° π 3 ,or 60° π 2 ,or 90° π 2 ,or 90°
Cosine 1 3 2 3 2 2 2 2 2 1 2 1 2 0
Sine 0 1 2 1 2 2 2 2 2 3 2 3 2 1
Tangent 0 3 3 3 3 1 3 3 Undefined
Secant 1 2 3 3 2 3 3 2 2 2 Undefined
Cosecant Undefined 2 2 2 2 3 3 2 3 3 1
Cotangent Undefined 3 3 1 3 3 3 3 0
Table A2
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