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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

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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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