0.1 Proofs, identities, and toolkit functions

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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.

$x y = a^{m} a^{n} = a^{m + n}$

$\begin{matrix} a^{m + n} & = & x y \\ \log_{a} (x y) & = & m + n \\ = & \log_{a} x + \log_{b} y \end{matrix}$

Change of Base Rule

$\begin{array}{l} \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \\ \log_{a} b = \frac{1}{\log_{b} a} \end{array}$

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

Proof:

Let $x = \log_{a} b .$

Write in exponent form.

$a^{x} = b$

Take the $\log_{c}$ of both sides.

$\begin{matrix} \log_{c} a^{x} & = & \log_{c} b \\ x \log_{c} a & = & \log_{c} b \\ x & = & \frac{\log_{c} b}{\log_{c} a} \\ \log_{a} b & = & \frac{\log_{c} b}{\log_{a} b} \end{matrix}$

When $c = b,$

$\log_{a} b = \frac{\log_{b} b}{\log_{b} a} = \frac{1}{\log_{b} a}$

Heron’s Formula

$A = \sqrt{s (s - a) (s - b) (s - c)}$

where $s = \frac{a + b + c}{2}$

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 $s = \frac{a + b + c}{2}$ .

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

$p = \frac{a^{2} + b^{2} - c^{2}}{2 c}$

Since $h^{2} = a^{2} - p^{2},$ we get an expression in terms of $a,$ $b,$ and $c .$

$\begin{matrix} h^{2} & = & a^{2} - p^{2} \\ = & (a + p) (a - p) \\ = & [a + \frac{(a^{2} + c^{2} - b^{2})}{2 c}] [a - \frac{(a^{2} + c^{2} - b^{2})}{2 c}] \\ = & \frac{(2 a c + a^{2} + c^{2} - b^{2}) (2 a c - a^{2} - c^{2} + b^{2})}{4 c^{2}} \\ = & \frac{({(a + c)}^{2} - b^{2}) (b^{2} - {(a - c)}^{2})}{4 c^{2}} \\ = & \frac{(a + b + c) (a + c - b) (b + a - c) (b - a + c)}{4 c^{2}} \\ = & \frac{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}{4 c^{2}} \\ = & \frac{2 s \cdot (2 s - a) \cdot (2 s - b) (2 s - c)}{4 c^{2}} \end{matrix}$

Therefore,

$\begin{matrix} h^{2} & = & \frac{4 s (s - a) (s - b) (s - c)}{c^{2}} \\ h & = & \frac{2 \sqrt{s (s - a) (s - b) (s - c)}}{c} \end{matrix}$

And since $A = \frac{1}{2} c h,$ then

$\begin{matrix} A & = & \frac{1}{2} c \frac{2 \sqrt{s (s - a) (s - b) (s - c)}}{c} \\ = & \sqrt{s (s - a) (s - b) (s - c)} \end{matrix}$

Properties of the Dot Product

$u \cdot v = v \cdot u$

Proof:

$\begin{matrix} u \cdot v & = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ v_{1}, v_{2}, ... v_{n} ⟩ \\ = u_{1} v_{1} + u_{2} v_{2} + ... + u_{n} v_{n} \\ = v_{1} u_{1} + v_{2} u_{2} + ... + v_{n} v_{n} \\ = ⟨ v_{1}, v_{2}, ... v_{n} ⟩ \cdot ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \\ = v \cdot u \end{matrix}$

$u \cdot (v + w) = u \cdot v + u \cdot w$

Proof:

$\begin{matrix} u \cdot (v + w) & = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot (⟨ v_{1}, v_{2}, ... v_{n} ⟩ + ⟨ w_{1}, w_{2}, ... w_{n} ⟩) \\ = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ v_{1} + w_{1}, v_{2} + w_{2}, ... v_{n} + w_{n} ⟩ \\ = ⟨ u_{1} (v_{1} + w_{1}), u_{2} (v_{2} + w_{2}), ... u_{n} (v_{n} + w_{n}) ⟩ \\ = ⟨ u_{1} v_{1} + u_{1} w_{1}, u_{2} v_{2} + u_{2} w_{2}, ... u_{n} v_{n} + u_{n} w_{n} ⟩ \\ = ⟨ u_{1} v_{1}, u_{2} v_{2}, ..., u_{n} v_{n} ⟩ + ⟨ u_{1} w_{1}, u_{2} w_{2}, ..., u_{n} w_{n} ⟩ \\ = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ v_{1}, v_{2}, ... v_{n} ⟩ + ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ w_{1}, w_{2}, ... w_{n} ⟩ \\ = u \cdot v + u \cdot w \end{matrix}$

$u \cdot u = {| u |}^{2}$

Proof:

$\begin{matrix} u \cdot u & = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \\ = u_{1} u_{1} + u_{2} u_{2} + ... + u_{n} u_{n} \\ = u_{1}^{2} + u_{2}^{2} + ... + u_{n}^{2} \\ = | ⟨ u_{1}, u_{2}, ... u_{n} ⟩ |^{2} \\ = v \cdot u \end{matrix}$

Standard Form of the Ellipse centered at the Origin

$1 = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$

Derivation

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

$\sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} + \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = constant$

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, $\sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} + \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = 2 a$

Consider a covertex.

Then $b^{2} + c^{2} = a^{2} .$

$\begin{matrix} \sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} + \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} & = & 2 a \\ \sqrt{{(x + c)}^{2} + y^{2}} & = & 2 a - \sqrt{{(x - c)}^{2} + y^{2}} \\ {(x + c)}^{2} + y^{2} & = & {(2 a - \sqrt{{(x - c)}^{2} + y^{2}})}^{2} \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} - 4 a \sqrt{{(x - c)}^{2} + y^{2}} + {(x - c)}^{2} + y^{2} \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} - 4 a \sqrt{{(x - c)}^{2} + y^{2}} + x^{2} - 2 c x + y^{2} \\ 2 c x & = & 4 a^{2} - 4 a \sqrt{{(x - c)}^{2} + y^{2}} - 2 c x \\ 4 c x - 4 a^{2} & = & 4 a \sqrt{{(x - c)}^{2} + y^{2}} \\ - \frac{1}{4 a} (4 c x - 4 a^{2}) & = & \sqrt{{(x - c)}^{2} + y^{2}} \\ a - \frac{c}{a} x & = & \sqrt{{(x - c)}^{2} + y^{2}} \\ a^{2} - 2 x c + \frac{c^{2}}{a^{2}} x^{2} & = & {(x - c)}^{2} + y^{2} \\ a^{2} - 2 x c + \frac{c^{2}}{a^{2}} x^{2} & = & x^{2} - 2 x c + c^{2} + y^{2} \\ a^{2} + \frac{c^{2}}{a^{2}} x^{2} & = & x^{2} + c^{2} + y^{2} \\ a^{2} + \frac{c^{2}}{a^{2}} x^{2} & = & x^{2} + c^{2} + y^{2} \\ a^{2} - c^{2} & = & x^{2} - \frac{c^{2}}{a^{2}} x^{2} + y^{2} \\ a^{2} - c^{2} & = & x^{2} (1 - \frac{c^{2}}{a^{2}}) + y^{2} \end{matrix}$

Let $1 = \frac{a^{2}}{a^{2}} .$

\begin{matrix} a^{2} - c^{2} & = & x^{2} (\frac{a^{2} - c^{2}}{a^{2}}) + y^{2} \\ 1 & = & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} \end{matrix}

Because $b^{2} + c^{2} = a^{2},$ then $b^{2} = a^{2} - c^{2} .$

\begin{matrix} 1 & = & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} \\ 1 & = & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \end{matrix}

Standard Form of the Hyperbola

$1 = \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$

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:

$\sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = constant$

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

$\sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = 2 a$

$\begin{matrix} \sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} & = & 2 a \\ \sqrt{{(x + c)}^{2} + y^{2}} - \sqrt{{(x - c)}^{2} + y^{2}} & = & 2 a \\ \sqrt{{(x + c)}^{2} + y^{2}} & = & 2 a + \sqrt{{(x - c)}^{2} + y^{2}} \\ {(x + c)}^{2} + y^{2} & = & (2 a + \sqrt{{(x - c)}^{2} + y^{2}}) \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} + 4 a \sqrt{{(x - c)}^{2}} + y^{2} \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} + 4 a \sqrt{{(x - c)}^{2} + y^{2}} + x^{2} - 2 c x + y^{2} \\ 2 c x & = & 4 a^{2} + 4 a \sqrt{{(x - c)}^{2} + y^{2}} - 2 c x \\ 4 c x - 4 a^{2} & = & 4 a \sqrt{{(x - c)}^{2} + y^{2}} \\ c x - a^{2} & = & a \sqrt{{(x - c)}^{2} + y^{2}} \\ {(c x - a^{2})}^{2} & = & a^{2} ({(x - c)}^{2} + y^{2}) \\ c^{2} x^{2} - 2 a^{2} c^{2} x^{2} + a^{4} & = & a^{2} x^{2} - 2 a^{2} c^{2} x^{2} + a^{2} c^{2} + a^{2} y^{2} \\ c^{2} x^{2} + a^{4} & = & a^{2} x^{2} + a^{2} c^{2} + a^{2} y^{2} \\ a^{4} - a^{2} c^{2} & = & a^{2} x^{2} - c^{2} x^{2} + a^{2} y^{2} \\ a^{2} (a^{2} - c^{2}) & = & (a^{2} - c^{2}) x^{2} + a^{2} y^{2} \\ a^{2} (a^{2} - c^{2}) & = & (c^{2} - a^{2}) x^{2} - a^{2} y^{2} \end{matrix}$

Define $b$ as a positive number such that $b^{2} = c^{2} - a^{2} .$

\begin{matrix} a^{2} b^{2} & = & b^{2} x^{2} - a^{2} y^{2} \\ \frac{a^{2} b^{2}}{a^{2} b^{2}} & = & \frac{b^{2} x^{2}}{a^{2} b^{2}} - \frac{a^{2} y^{2}}{a^{2} b^{2}} \\ 1 & = & \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} \end{matrix}

Trigonometric identities

Pythagorean Identity	$\begin{array}{l} \cos^{2} t + \sin^{2} t = 1 \\ 1 + \tan^{2} t = \sec^{2} t \\ 1 + \cot^{2} t = \csc^{2} t \end{array}$
Even-Odd Identities	$\begin{array}{l} \cos (- t) = c o s t \\ \sec (- t) = \sec t \\ \sin (- t) = - \sin t \\ \tan (- t) = - \tan t \\ \csc (- t) = - \csc t \\ \cot (- t) = - \cot t \end{array}$
Cofunction Identities	$\begin{array}{l} \cos t = \sin (\frac{π}{2} - t) \\ \sin t = \cos (\frac{π}{2} - t) \\ \tan t = \cot (\frac{π}{2} - t) \\ \cot t = \tan (\frac{π}{2} - t) \\ \sec t = \csc (\frac{π}{2} - t) \\ \csc t = \sec (\frac{π}{2} - t) \end{array}$
Fundamental Identities	$\begin{array}{l} \tan t = \frac{\sin t}{\cos t} \\ \sec t = \frac{1}{\cos t} \\ \csc t = \frac{1}{\sin t} \\ c o t t = \frac{1}{tan t} = \frac{cos t}{sin t} \end{array}$
Sum and Difference Identities	$\begin{array}{l} \cos (α + β) = \cos α \cos β - \sin α \sin β \\ \cos (α - β) = \cos α \cos β + \sin α \sin β \\ \sin (α + β) = \sin α \cos β + \cos α \sin β \\ \sin (α - β) = \sin α \cos β - \cos α \sin β \\ \tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β} \\ \tan (α - β) = \frac{\tan α - \tan β}{1 + \tan α \tan β} \end{array}$
Double-Angle Formulas	$\begin{array}{l} \sin (2 θ) = 2 \sin θ \cos θ \\ \cos (2 θ) = \cos^{2} θ - \sin^{2} θ \\ \cos (2 θ) = 1 - 2 \sin^{2} θ \\ \cos (2 θ) = 2 \cos^{2} θ - 1 \\ \tan (2 θ) = \frac{2 \tan θ}{1 - \tan^{2} θ} \end{array}$
Half-Angle Formulas	$\begin{array}{l} \sin \frac{α}{2} = \pm \sqrt{\frac{1 - \cos α}{2}} \\ \cos \frac{α}{2} = \pm \sqrt{\frac{1 + \cos α}{2}} \\ \tan \frac{α}{2} = \pm \sqrt{\frac{1 - \cos α}{1 + \cos α}} \\ \tan \frac{α}{2} = \frac{\sin α}{1 + \cos α} \\ \tan \frac{α}{2} = \frac{1 - \cos α}{\sin α} \end{array}$
Reduction Formulas	$\begin{array}{l} \sin^{2} θ = \frac{1 - \cos (2 θ)}{2} \\ \cos^{2} θ = \frac{1 + \cos (2 θ)}{2} \\ \tan^{2} θ = \frac{1 - \cos (2 θ)}{1 + \cos (2 θ)} \end{array}$
Product-to-Sum Formulas	$\begin{array}{l} \cos α \cos β = \frac{1}{2} [\cos (α - β) + \cos (α + β)] \\ \sin α \cos β = \frac{1}{2} [\sin (α + β) + \sin (α - β)] \\ \sin α \sin β = \frac{1}{2} [\cos (α - β) - \cos (α + β)] \\ \cos α \sin β = \frac{1}{2} [\sin (α + β) - \sin (α - β)] \end{array}$
Sum-to-Product Formulas	$\begin{array}{l} \sin α + \sin β = 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \\ \sin α - \sin β = 2 \sin (\frac{α - β}{2}) \cos (\frac{α + β}{2}) \\ \cos α - \cos β = - 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2}) \\ \cos α + \cos β = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \end{array}$
Law of Sines	$\begin{array}{l} \frac{\sin α}{a} = \frac{\sin β}{b} = \frac{\sin γ}{c} \\ \frac{a}{\sin α} = \frac{b}{\sin β} = \frac{c}{\sin γ} \end{array}$
Law of Cosines	$\begin{matrix} a^{2} = b^{2} + c^{2} - 2 b c \cos α \\ b^{2} = a^{2} + c^{2} - 2 a c \cos β \\ c^{2} = a^{2} + b^{2} - 2 a b cos γ \end{matrix}$

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

Angle	$0$	$\frac{π}{6}, or 30 °$	$\frac{π}{4}, or 45 °$	$\frac{π}{3}, or 60 °$	$\frac{π}{2}, or 90 °$
Cosine	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
Sine	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1
Tangent	0	$\frac{\sqrt{3}}{3}$	1	$\sqrt{3}$	Undefined
Secant	1	$\frac{2 \sqrt{3}}{3}$	$\sqrt{2}$	2	Undefined
Cosecant	Undefined	2	$\sqrt{2}$	$\frac{2 \sqrt{3}}{3}$	1
Cotangent	Undefined	$\sqrt{3}$	1	$\frac{\sqrt{3}}{3}$	0

Questions & Answers

how to study physic and understand

Ewa Reply

what is conservative force with examples

Moses

what is work

Fredrick Reply

the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the direction of the displacement and the magnitude of the displacement

AI-Robot

why is it from light to gravity

Esther Reply

difference between model and theory

Esther

Is the ship moving at a constant velocity?

Kamogelo Reply

The full note of modern physics

aluet Reply

introduction to applications of nuclear physics

aluet Reply

the explanation is not in full details

Moses Reply

I need more explanation or all about kinematics

Moses

yes

zephaniah

I need more explanation or all about nuclear physics

aluet

Show that the equal masses particles emarge from collision at right angle by making explicit used of fact that momentum is a vector quantity

Muhammad Reply

Isaac

A wave is described by the function D(x,t)=(1.6cm) sin[(1.2cm^-1(x+6.8cm/st] what are:a.Amplitude b. wavelength c. wave number d. frequency e. period f. velocity of speed.

Majok Reply

what is frontier of physics

Somto Reply

A body is projected upward at an angle 45° 18minutes with the horizontal with an initial speed of 40km per second. In hoe many seconds will the body reach the ground then how far from the point of projection will it strike. At what angle will the horizontal will strike

Gufraan Reply

Suppose hydrogen and oxygen are diffusing through air. A small amount of each is released simultaneously. How much time passes before the hydrogen is 1.00 s ahead of the oxygen? Such differences in arrival times are used as an analytical tool in gas chromatography.

Ezekiel Reply

please explain

Samuel

what's the definition of physics

Mobolaji Reply

what is physics

Nangun Reply

the science concerned with describing the interactions of energy, matter, space, and time; it is especially interested in what fundamental mechanisms underlie every phenomenon

AI-Robot

what is isotopes

Nangun Reply

nuclei having the same Z and different N s

AI-Robot

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