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Appendix

Important proofs and derivations

Product Rule

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

Proof:

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

Write in exponent form.

x = a m and y = a n .

Multiply.

x y = a m a n = a m + n

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

where x and y are positive, and a > 0 , a 1.

Proof:

Let x = log a b .

Write in exponent form.

a x = b

Take the 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 a b

When c = b ,

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

Heron’s Formula

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

where s = a + b + c 2

Proof:

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

We can further name the parts of the base in each triangle established by the height such that 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 and h 2 + q 2 = b 2 .

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

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

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

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

Solve for p to get

p = a 2 + b 2 c 2 2 c

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

h 2 = a 2 p 2 = ( a + p ) ( a p ) = [ a + ( a 2 + c 2 b 2 ) 2 c ] [ a ( a 2 + c 2 b 2 ) 2 c ] = ( 2 a c + a 2 + c 2 b 2 ) ( 2 a c 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 = 2 s ( 2 s a ) ( 2 s b ) ( 2 s c ) 4 c 2

Therefore,

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

And since A = 1 2 c h , then

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

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 + 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 · 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

Standard Form of the Ellipse centered at the Origin

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

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

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 .

( x ( c ) ) 2 + ( y 0 ) 2 + ( x c ) 2 + ( y 0 ) 2 = 2 a ( x + c ) 2 + y 2 = 2 a ( x c ) 2 + y 2 ( x + c ) 2 + y 2 = ( 2 a ( x c ) 2 + y 2 ) 2 x 2 + 2 c x + c 2 + y 2 = 4 a 2 4 a ( x c ) 2 + y 2 + ( x c ) 2 + y 2 x 2 + 2 c x + c 2 + y 2 = 4 a 2 4 a ( x c ) 2 + y 2 + x 2 2 c x + y 2 2 c x = 4 a 2 4 a ( x c ) 2 + y 2 2 c x 4 c x 4 a 2 = 4 a ( x c ) 2 + y 2 1 4 a ( 4 c x 4 a 2 ) = ( x c ) 2 + y 2 a c a x = ( x c ) 2 + y 2 a 2 2 x c + c 2 a 2 x 2 = ( x c ) 2 + y 2 a 2 2 x c + c 2 a 2 x 2 = x 2 2 x c + 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 .

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 , then 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

Standard Form of the Hyperbola

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

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

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

( x ( c ) ) 2 + ( y 0 ) 2 ( x c ) 2 + ( y 0 ) 2 = 2 a ( x + c ) 2 + y 2 ( x c ) 2 + y 2 = 2 a ( x + c ) 2 + y 2 = 2 a + ( x c ) 2 + y 2 ( x + c ) 2 + y 2 = ( 2 a + ( x c ) 2 + y 2 ) x 2 + 2 c x + c 2 + y 2 = 4 a 2 + 4 a ( x c ) 2 + y 2 x 2 + 2 c x + c 2 + y 2 = 4 a 2 + 4 a ( x c ) 2 + y 2 + x 2 2 c x + y 2 2 c x = 4 a 2 + 4 a ( x c ) 2 + y 2 2 c x 4 c x 4 a 2 = 4 a ( x c ) 2 + y 2 c x a 2 = a ( x c ) 2 + y 2 ( c x 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 as a positive number such that 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

Trigonometric identities

Pythagorean Identity cos 2 t + sin 2 t = 1 1 + tan 2 t = sec 2 t 1 + cot 2 t = csc 2 t
Even-Odd Identities cos ( t ) = c o s t sec ( t ) = sec t sin ( t ) = sin t tan ( t ) = tan t csc ( t ) = csc t cot ( t ) = cot t
Cofunction Identities cos t = sin ( π 2 t ) sin t = cos ( π 2 t ) tan t = cot ( π 2 t ) cot t = tan ( π 2 t ) sec t = csc ( π 2 t ) csc t = sec ( π 2 t )
Fundamental Identities tan t = sin t cos t sec t = 1 cos t csc t = 1 sin t c o t t = 1 tan t = cos t sin t
Sum and Difference Identities cos ( α + β ) = cos α cos β sin α sin β cos ( α β ) = cos α cos β + sin α sin β sin ( α + β ) = sin α cos β + cos α sin β sin ( α β ) = sin α cos β cos α sin β tan ( α + β ) = tan α + tan β 1 tan α tan β tan ( α β ) = tan α tan β 1 + tan α tan β
Double-Angle Formulas sin ( 2 θ ) = 2 sin θ cos θ cos ( 2 θ ) = cos 2 θ sin 2 θ cos ( 2 θ ) = 1 2 sin 2 θ cos ( 2 θ ) = 2 cos 2 θ 1 tan ( 2 θ ) = 2 tan θ 1 tan 2 θ
Half-Angle Formulas sin α 2 = ± 1 cos α 2 cos α 2 = ± 1 + cos α 2 tan α 2 = ± 1 cos α 1 + cos α tan α 2 = sin α 1 + cos α tan α 2 = 1 cos α sin α
Reduction Formulas sin 2 θ = 1 cos ( 2 θ ) 2 cos 2 θ = 1 + cos ( 2 θ ) 2 tan 2 θ = 1 cos ( 2 θ ) 1 + cos ( 2 θ )
Product-to-Sum Formulas cos α cos β = 1 2 [ cos ( α β ) + cos ( α + β ) ] sin α cos β = 1 2 [ sin ( α + β ) + sin ( α β ) ] sin α sin β = 1 2 [ cos ( α β ) cos ( α + β ) ] cos α sin β = 1 2 [ sin ( α + β ) sin ( α β ) ]
Sum-to-Product Formulas sin α + sin β = 2 sin ( α + β 2 ) cos ( α β 2 ) sin α sin β = 2 sin ( α β 2 ) cos ( α + β 2 ) cos α cos β = 2 sin ( α + β 2 ) sin ( α β 2 ) cos α + cos β = 2 cos ( α + β 2 ) cos ( α β 2 )
Law of Sines sin α a = sin β b = sin γ c a sin α = b sin β = c sin γ
Law of Cosines a 2 = b 2 + c 2 2 b c cos α b 2 = a 2 + c 2 2 a c cos β c 2 = a 2 + b 2 2 a b cos γ

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.
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.
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.

Trigonometric functions

Unit Circle

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

Questions & Answers

x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
HERVE Reply
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
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0.75
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0.75
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depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
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I want to learn the calculations
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I need matrices
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Cromwell
a
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Cromwell
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Solve the x? x=18+(24-3)=72
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x-39=72 x=111
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Solve the formula for the indicated variable P=b+4a+2c, for b
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b=-4ac-2c+P
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b=p-4a-2c
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b= p - 4a - 2c
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p=2(2a+C)+b
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b=p-2(2a+c)
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P=4a+b+2C
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b=P-4a-2c
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A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
Kaitlyn Reply
The sequence is {1,-1,1-1.....} has
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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