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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

ToolKit Functions

Figure A1
Figure A2
Figure A3

Trigonometric Functions

Unit Circle

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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