Algebra and Trigonometry

# A | Proofs, Identities, and Toolkit Functions

Algebra and TrigonometryA | 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 $a>0,a≠1. a>0,a≠1.$

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

$A= s( s−a )( s−b )( s−c ) A= s( s−a )( s−b )( s−c )$

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.

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 $p+q=c. p+q=c.$

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 $q=c−p, q=c−p,$ then $q 2 = ( c−p ) 2 . q 2 = ( c−p ) 2 .$ 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 $c. c.$

$h 2 = a 2 − p 2 = (a+p)(a−p) = [ a+ ( a 2 + c 2 − b 2 ) 2c ][ a− ( a 2 + c 2 − b 2 ) 2c ] = ( 2ac+ a 2 + c 2 − b 2 )( 2ac− a 2 − c 2 + b 2 ) 4 c 2 = ( (a+c) 2 − b 2 )( b 2 − (a−c) 2 ) 4 c 2 = (a+b+c)(a+c−b)(b+a−c)(b−a+c) 4 c 2 = (a+b+c)(−a+b+c)(a−b+c)(a+b−c) 4 c 2 = 2s⋅(2s−a)⋅(2s−b)(2s−c) 4 c 2 h 2 = a 2 − p 2 = (a+p)(a−p) = [ a+ ( a 2 + c 2 − b 2 ) 2c ][ a− ( a 2 + c 2 − b 2 ) 2c ] = ( 2ac+ a 2 + c 2 − b 2 )( 2ac− a 2 − c 2 + b 2 ) 4 c 2 = ( (a+c) 2 − b 2 )( b 2 − (a−c) 2 ) 4 c 2 = (a+b+c)(a+c−b)(b+a−c)(b−a+c) 4 c 2 = (a+b+c)(−a+b+c)(a−b+c)(a+b−c) 4 c 2 = 2s⋅(2s−a)⋅(2s−b)(2s−c) 4 c 2$

Therefore,

$h 2 = 4s(s−a)(s−b)(s−c) c 2 h = 2 s(s−a)(s−b)(s−c) c h 2 = 4s(s−a)(s−b)(s−c) c 2 h = 2 s(s−a)(s−b)(s−c) c$

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

$A = 1 2 c 2 s(s−a)(s−b)(s−c) c = s(s−a)(s−b)(s−c) A = 1 2 c 2 s(s−a)(s−b)(s−c) c = s(s−a)(s−b)(s−c)$

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:

$u·u =〈 u 1 , u 2 ,... u n 〉·〈 u 1 , u 2 ,... u n 〉 = u 1 u 1 + u 2 u 2 +...+ u n u n = u 1 2 + u 2 2 +...+ u n 2 =|〈 u 1 , u 2 ,... u n 〉 | 2 =v·u u·u =〈 u 1 , u 2 ,... u n 〉·〈 u 1 , u 2 ,... u n 〉 = u 1 u 1 + u 2 u 2 +...+ u n u n = u 1 2 + u 2 2 +...+ u n 2 =|〈 u 1 , u 2 ,... u n 〉 | 2 =v·u$

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:

$( x−( −c ) ) 2 + ( y−0 ) 2 + ( x−c ) 2 + ( y−0 ) 2 =constant ( x−( −c ) ) 2 + ( y−0 ) 2 + ( x−c ) 2 + ( y−0 ) 2 =constant$

Consider a vertex.

Then, $( x−( −c ) ) 2 + ( y−0 ) 2 + ( x−c ) 2 + ( y−0 ) 2 =2a ( x−( −c ) ) 2 + ( y−0 ) 2 + ( x−c ) 2 + ( y−0 ) 2 =2a$

Consider a covertex.

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

$(x−(−c)) 2 + (y−0) 2 + (x−c) 2 + (y−0) 2 = 2a (x+c) 2 + y 2 = 2a− (x−c) 2 + y 2 (x+c) 2 + y 2 = ( 2a− (x−c) 2 + y 2 ) 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 −4a (x−c) 2 + y 2 + (x−c) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 −4a (x−c) 2 + y 2 + x 2 −2cx+ y 2 2cx = 4 a 2 −4a (x−c) 2 + y 2 −2cx 4cx−4 a 2 = 4a (x−c) 2 + y 2 − 1 4a ( 4cx−4 a 2 ) = (x−c) 2 + y 2 a− c a x = (x−c) 2 + y 2 a 2 −2xc+ c 2 a 2 x 2 = (x−c) 2 + y 2 a 2 −2xc+ c 2 a 2 x 2 = x 2 −2xc+ c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 − c 2 = x 2 − c 2 a 2 x 2 + y 2 a 2 − c 2 = x 2 ( 1− c 2 a 2 )+ y 2 (x−(−c)) 2 + (y−0) 2 + (x−c) 2 + (y−0) 2 = 2a (x+c) 2 + y 2 = 2a− (x−c) 2 + y 2 (x+c) 2 + y 2 = ( 2a− (x−c) 2 + y 2 ) 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 −4a (x−c) 2 + y 2 + (x−c) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 −4a (x−c) 2 + y 2 + x 2 −2cx+ y 2 2cx = 4 a 2 −4a (x−c) 2 + y 2 −2cx 4cx−4 a 2 = 4a (x−c) 2 + y 2 − 1 4a ( 4cx−4 a 2 ) = (x−c) 2 + y 2 a− c a x = (x−c) 2 + y 2 a 2 −2xc+ c 2 a 2 x 2 = (x−c) 2 + y 2 a 2 −2xc+ c 2 a 2 x 2 = x 2 −2xc+ c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 − c 2 = x 2 − c 2 a 2 x 2 + y 2 a 2 − c 2 = x 2 ( 1− c 2 a 2 )+ y 2$

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.

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

$(x−(−c)) 2 + (y−0) 2 − (x−c) 2 + (y−0) 2 =constant (x−(−c)) 2 + (y−0) 2 − (x−c) 2 + (y−0) 2 =constant$

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

$( x−( −c ) ) 2 + ( y−0 ) 2 − ( x−c ) 2 + ( y−0 ) 2 =2a ( x−( −c ) ) 2 + ( y−0 ) 2 − ( x−c ) 2 + ( y−0 ) 2 =2a$

$(x−(−c)) 2 + (y−0) 2 − (x−c) 2 + (y−0) 2 = 2a (x+c) 2 + y 2 − (x−c) 2 + y 2 = 2a (x+c) 2 + y 2 = 2a+ (x−c) 2 + y 2 (x+c) 2 + y 2 = ( 2a+ (x−c) 2 + y 2 ) x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (x−c) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (x−c) 2 + y 2 + x 2 −2cx+ y 2 2cx = 4 a 2 +4a (x−c) 2 + y 2 −2cx 4cx−4 a 2 = 4a (x−c) 2 + y 2 cx− a 2 = a (x−c) 2 + y 2 ( cx− a 2 ) 2 = a 2 ( (x−c) 2 + y 2 ) c 2 x 2 −2 a 2 c 2 x 2 + a 4 = a 2 x 2 −2 a 2 c 2 x 2 + a 2 c 2 + a 2 y 2 c 2 x 2 + a 4 = a 2 x 2 + a 2 c 2 + a 2 y 2 a 4 − a 2 c 2 = a 2 x 2 − c 2 x 2 + a 2 y 2 a 2 ( a 2 − c 2 ) = ( a 2 − c 2 ) x 2 + a 2 y 2 a 2 ( a 2 − c 2 ) = ( c 2 − a 2 ) x 2 − a 2 y 2 (x−(−c)) 2 + (y−0) 2 − (x−c) 2 + (y−0) 2 = 2a (x+c) 2 + y 2 − (x−c) 2 + y 2 = 2a (x+c) 2 + y 2 = 2a+ (x−c) 2 + y 2 (x+c) 2 + y 2 = ( 2a+ (x−c) 2 + y 2 ) x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (x−c) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (x−c) 2 + y 2 + x 2 −2cx+ y 2 2cx = 4 a 2 +4a (x−c) 2 + y 2 −2cx 4cx−4 a 2 = 4a (x−c) 2 + y 2 cx− a 2 = a (x−c) 2 + y 2 ( cx− a 2 ) 2 = a 2 ( (x−c) 2 + y 2 ) c 2 x 2 −2 a 2 c 2 x 2 + a 4 = a 2 x 2 −2 a 2 c 2 x 2 + a 2 c 2 + a 2 y 2 c 2 x 2 + a 4 = a 2 x 2 + a 2 c 2 + a 2 y 2 a 4 − a 2 c 2 = a 2 x 2 − c 2 x 2 + a 2 y 2 a 2 ( a 2 − c 2 ) = ( a 2 − c 2 ) x 2 + a 2 y 2 a 2 ( a 2 − c 2 ) = ( c 2 − a 2 ) x 2 − a 2 y 2$

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 $cos(α+β)=cosαcosβ−sinαsinβ cos(α−β)=cosαcosβ+sinαsinβ sin(α+β)=sinαcosβ+cosαsinβ sin(α−β)=sinαcosβ−cosαsinβ tan(α+β)= tanα+tanβ 1−tanαtanβ tan(α−β)= tanα−tanβ 1+tanαtanβ cos(α+β)=cosαcosβ−sinαsinβ cos(α−β)=cosαcosβ+sinαsinβ sin(α+β)=sinαcosβ+cosαsinβ sin(α−β)=sinαcosβ−cosαsinβ tan(α+β)= tanα+tanβ 1−tanαtanβ tan(α−β)= tanα−tanβ 1+tanαtanβ$ Double-Angle Formulas $sin(2θ)=2sinθcosθ cos(2θ)= cos 2 θ− sin 2 θ cos(2θ)=1−2 sin 2 θ cos(2θ)=2 cos 2 θ−1 tan(2θ)= 2tanθ 1− tan 2 θ sin(2θ)=2sinθcosθ cos(2θ)= cos 2 θ− sin 2 θ cos(2θ)=1−2 sin 2 θ cos(2θ)=2 cos 2 θ−1 tan(2θ)= 2tanθ 1− tan 2 θ$ Half-Angle Formulas $sin α 2 =± 1−cosα 2 cos α 2 =± 1+cosα 2 tan α 2 =± 1−cosα 1+cosα tan α 2 = sinα 1+cosα tan α 2 = 1−cosα sinα sin α 2 =± 1−cosα 2 cos α 2 =± 1+cosα 2 tan α 2 =± 1−cosα 1+cosα tan α 2 = sinα 1+cosα tan α 2 = 1−cosα sinα$ Reduction Formulas $sin 2 θ= 1−cos( 2θ ) 2 cos 2 θ= 1+cos( 2θ ) 2 tan 2 θ= 1−cos( 2θ ) 1+cos( 2θ ) sin 2 θ= 1−cos( 2θ ) 2 cos 2 θ= 1+cos( 2θ ) 2 tan 2 θ= 1−cos( 2θ ) 1+cos( 2θ )$ Product-to-Sum Formulas $cosαcosβ= 1 2 [ cos(α−β)+cos(α+β) ] sinαcosβ= 1 2 [ sin(α+β)+sin(α−β) ] sinαsinβ= 1 2 [ cos(α−β)−cos(α+β) ] cosαsinβ= 1 2 [ sin(α+β)−sin(α−β) ] cosαcosβ= 1 2 [ cos(α−β)+cos(α+β) ] sinαcosβ= 1 2 [ sin(α+β)+sin(α−β) ] sinαsinβ= 1 2 [ cos(α−β)−cos(α+β) ] cosαsinβ= 1 2 [ sin(α+β)−sin(α−β) ]$ Sum-to-Product Formulas $sinα+sinβ=2sin( α+β 2 )cos( α−β 2 ) sinα−sinβ=2sin( α−β 2 )cos( α+β 2 ) cosα−cosβ=−2sin( α+β 2 )sin( α−β 2 ) cosα+cosβ=2cos( α+β 2 )cos( α−β 2 ) sinα+sinβ=2sin( α+β 2 )cos( α−β 2 ) sinα−sinβ=2sin( α−β 2 )cos( α+β 2 ) cosα−cosβ=−2sin( α+β 2 )sin( α−β 2 ) cosα+cosβ=2cos( α+β 2 )cos( α−β 2 )$ 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

Figure A1
Figure A2
Figure A3

### Trigonometric Functions

Unit Circle

Figure A4
Angle $0 0$
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 Do you know how you learn best?
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