A sliced shape has the same total area as the cumulative area of its pieces.
Similarly, a volume sliced by planes has the same total volume as the volume of the individual pieces.
The area of a parallelogram, or that of a rectangle, is calculated with the lenght of the base b, times the length of the height h, perpendicular to the base. $$A=\b\h$$
A rectangle split in half by a straight line crossing two of it's opposing vertices is a right triangle. By Cavalieri's principle, all triangles with the same base and height have the same area: Half a parallelogram. $$A={1/2}\b\h$$
The constant $π$ is defined as the ratio between the perimeter P and the diameter D.
$π=\P/\D$ $\D=2\r$ $\P=2π\r$
The perimeter of a regular polygon is the lenght of its base b, times the number of sides n: $$\P=\b⋅{n}$$
Its area is the area of one of its triangle part times the number of sides n. $$A={1/2}\b\h⋅{n}$$ The perimeter previously defined can be factored out. $$A={1/2}\h⋅\P$$ Letting n reach infinity to get a circle, sides s and h tend to a radius r. $$A={1/2}\r⋅\P$$ Using the formula for the perimeter of a circle, we can write the area as
$A={1/2}\r⋅2π\r$ which simplifies as $A=π\r^2$.
The volume of an n-sided polygonal prism is: $V=\text"Base Area"⋅\h$,
its surface area is: $A=(\text"Base Perimeter"⋅\h) + (2⋅\text"Base Area")$.
A circular prism, (a cylinder) has volume $V=π\r^2⋅\h$ and a surface area of $A=(2π\r⋅\h) + (2⋅π\r^2)$.
Note that if a cylinder has the same height as its width, $(\text"Height"=\text"Diameter"=2\r)$, then the volume could be written as: $V=π\r^2⋅2\r= 2π\r^3$, its surface area: $A=(4π\r^2) + (2⋅π\r^2) = 6π\r^2$.
An open pyramid, with a regular n-sided polygon base has an area $A={1/2}\b{s}⋅n$
The perimeter of the base can be written as $\P=\b{n}$ so the area is also $A={1/2}\P{s}$
Letting n reach infinity to get a cone, the perimeter of the base is then $\P=2π\r$
So the surface area of an open cone is then $A={1/2}⋅2π\r⋅\s$, or $A=π\r\s$
If it becomes flat like a disk, s then equals r, so its area is $A=π\r⋅\r$, which is $A = π\r^2$
A triangular prism can be sliced in three equal volumes by planes which cross its vertices: Each pair of sub-pyramid share a base and a height perpendicular to this base. By cavalieri's principle, all pyramids are then 1/3 the volume of prisms of same height and base.
We can also find the volume of pyramids with $n$ polygonal prisms.
With rectangular prisms, each slice x has volume $V=x\L/{n}⋅x\W/{n}⋅{\H/n}$
The sum is then $V=1^2{LW}/n^2{H/n}+2^2{LW}/n^2{H/n}+...+n^2{LW}/n^2{H/n}$
Factor out ${LWH}/n^3$ to get $V={LWH}/n^3⋅(1^2+2^2+...+n^2)$
Recognize the sum of power series $∑↙{i=1}↖n i^2 = {n(2n+1)(n+1)}/6$
So that $V={LWH}/n^3⋅{n(2n+1)(n+1)}/6$, cancel an $n$ to get
$V={LWH}/n^2⋅{(2n+1)(n+1)}/6$, multiply out the binomes,
$V={LWH}/n^2⋅{2n^2+3n+1}/6$, separate as $V={LWH}/n^2⋅({n^2}/3+n/2+1/6)$, and cancel $1/{n^2}$ to get
$V={LWH}⋅(1/3+1/{2n}+1/{6n^2})$. By increasing the number of slices, n, reach infinity, $V={LWH}⋅(1/3+1/{2∞}+1/{6∞^2})$ which is $V={LWH}⋅1/3$ where $LW$ is the area of the base.
$V={LWH}⋅1/3$ being the formula for the volume of a rectangular pyramid, where $LW$ is the area of the base, we can deduce that $V={(\text"Base Area")⋅H}⋅1/3$ is the formula for the volume of any pyramid.
To get a hold of the properties of spheres, it is necessary to prove that the shapes of horizontal sections of an inverted cone have the same area, at all heights, of that of horizontal sections of the outside of a hemisphere inscribed in a cylinder:
First, we acknowledge that the area of a horizontal section of an inverted cone, at a given height h, where the height of the cone is equal to its radius, is computed as $A=π\h^2$.
Next, we subtract from the area of a section of a cylinder, $π\r^2$, the area of a section of a hemisphere at height h, $π(√{\r^2-\h^2})^2$ to get the area of the ring outside the hemisphere:
$$A=π\r^2 - π(√{\r^2-\h^2})^2$$
It can be simplified to $A=π\r^2 - π(\r^2-\h^2)$, and further to $A=π\r^2 - π\r^2 + π\h^2$, which is $A=π\h^2$.
Having shown that those areas are equal at all heights, $π\h^2$, it follows that the volume of a cone and that of the outside of a hemisphere inscribed in a cylinder are also equal.
Then from the volume of a cylinder, $\h⋅π\r^2$, if we subtract the volume of a cone of same height and radius, $1/3⋅\h⋅π\r^2$, we obtain the formula for the volume of a hemisphere: $2/3⋅h⋅π\r^2$, or if the height equals the radius, $2/3⋅π\r^3$. Doubling this value gives the formula for a whole sphere: $V=4/3π\r^3$.
If we consider a sphere to be made up of infinitely thin pyramids of height r having their apex at the center and their bases facing outward as the outside surface. We can write a formula for the volume $V=4/3π\r^3$ as: ${1/3b_1\r}+{1/3b_2\r}+{1/3b_3\r}+...$
By factoring out common parts of the terms we get: ${1/3\r}⋅(b_1+b_2+b_3 + ... )$
The sum of the bases, $(b_1+b_2+b_3+...)$ is the outside area of the sphere, which can be isolated by dividing the volume's formula $V=4/3πr^3$, by ${1/3\r}$, so that, the surface area of a sphere can be found: $$A=4π\r^2$$
Starting with $ax^2+bx+c=0$, where $a≠0$, subtract $c$ from both sides: $ax^2+bx=-c$.
Factor out the $a$ like so: $a(x^2+{bx}/a)= -c$ then divide by $a$: $x^2+{bx}/a= -c/a$ .
Completing the square on the left yields the extra term $b^2/{4a^2}$ which is added to the right side. $$ (x+b/{2a})^2 = b^2/{4a^2}-c/a$$ Multiply the $-c/a$ term by ${4a}/{4a}$ to have a common denominator. $$(x+b/{2a})^2 = {b^2-4ac}/{4a^2}$$ Take the square root on both sides: $$ x+b/{2a} = ±√{{b^2-4ac}/{4a^2}}$$ The denominator on the right side can be taken out of the square root. $$x+b/{2a} = ±√{{b^2-4ac}}/{2a}$$ Lastly, the $b/{2a}$ term can be sent to the right side, which already has a term with the same denominator. $$x={-b±√{b^2-4ac}}/{2a}$$
Starting with $ax^2+bx+c=0$, where $a≠0$, subtract $c$ from both sides: $ax^2+bx=-c$
Now complete the square on the left side to get an extra ${b^2}/{4a}$. $$ (√ax + b/{2√a})^2 = {b^2}/{4a} -c $$ Have the terms on the right side share a common denominator. $$ (√ax + b/{2√a})^2 ={b^2-4ac}/{4a} $$ Take the square root of both sides. $$ √ax + b/{2√a} = √{{b^2-4ac}/{4a}} $$ Take the denominator on the right side out of the radical sign. $$ √ax + b/{2√a} = √{{b^2-4ac}}/{2√a} $$ Finally divide both sides by $√a$ $$x={-b±√{b^2-4ac}}/{2a}$$
The silver ratio $δ_s$ is a proportion between a longer length 'a' and a smaller length 'b' where that proportion is the same as twice the longer length 'a' plus the smaller length 'b' over the longer length 'a'.
Starting with $δ_s=a/b={2a+b}/a$, or $a/b=({2a}/a+b/a)$ which is ${δ_s}/1=2+1/{δ_s}$
Square both sides by multiplying by $δ_s$ gives ${{δ_s}^2}/1= {2δ_s}/1+{δ_s}/{δ_s}$ which is ${δ_s}^2=2δ_s+1$, a quadratic.
Using the quadratic formula $x={-b±√{b^2-4ac}}/{2a}$ where $a=1$, $b=-2$ and $c=-1$.
We get for the value of ${δ_s}$ : ${δ_s}= {2±√{(-2)^2-4(-1)}}/{2}= {2±√{8}}/{2} = {1+√{2}}$
A silver rectangle has proportions such that when cut in half it has again the same proportions. Taking $a/b=b/{a\text"/2"}$ and bringing the divisor of 1/2 on top gives: $a/b = {2b}/a$. Squaring both sides by multiplying by $a/b$ gives ${a^2}/{b^2} = {2ba}/{ab}$ which is ${a^2}/{b^2}= 2$. So $a/b = √{2}$.
The golden ratio $φ$ is a proportion between a longer length 'a' and a smaller length 'b' where the proportion is the same ratio for the sum of both 'a' and 'b' in relation to the longer length 'a'.
Starting with $φ= a/b={a+b}/a$, expanding the right side gives $a/a+b/a$ which is $1+1/φ$ Squaring both sides of $φ=1+1/φ$ by multiplying by $φ$ gives $φ^2=φ+1$. This is a quadratic $φ^2-φ-1=0$ where $a=1$, $b=-1$ and $c=-1$.
Using the quadratic formula $x={-b±√{b^2-4ac}}/{2a}$, we get $φ= {1±√{(-1)^2-4(-1)}}/{2}= {1+√{5}}/{2}$
Golden rectangle.
The exponent states how many times the multiplicative identity, 1, is factored by the base. Correspondingly, the logarithm is the exponent that would be needed to raise the base to get a given number.
$a^{-2}=1/{a⋅a}$ $a^{-1}=1/{a}$ $a^0=1$ $a^1=a$ $a^2=a⋅a$
$log_a(1/{a⋅a})= -2$ $log_a(1/a)= -1$ $log_a(1)= 0$ $log_a(a)= 1$ $log_a(a⋅a)= 2$
Factorisation and exponentiation of powers of a common base
${a^x}⋅{a^y} = a^{x+y}$ ${a^x}/{a^y} = a^{x-y}$ $(a^x)^y = a^{x⋅y}$
$log_a(x⋅y)= log(x)+log(y)$ $log_a(x/y)= log(x)-log(y)$ $log_a(x^y)= y⋅log(x)$
Negative and rational exponents along with logarithms with inverted and changed base
$a^{-x}=1/{a^x}$ $a^{1/x}=√^x{a}$ $a^{x/y}=√^y{a^x}$ $a^{-{x/y}}=1/{√^y{a^x}}$
$log_{1/a}(b) = -log_a(b)$ $log_a(1/x) = -log_a(x)$ $log_a(b) = 1/{log_b(a)}$ $log_a(b) = {log_c(b)}/{log_c(a)}$
Solve for x:
$\ln(x+1) + \ln(x-1) = 2y + \ln(y)$
Combine logarithms on the left side, $$\ln(x^2-1)=2y+\ln(y)$$ Exponentiate both sides, $$x^2-1 = e^{2y+\ln(y)}$$ Separate the exponent on the right side, $$x^2-1 = e^{2y}⋅e^{\ln(y)}$$ Simplify the right side, $$x^2-1 = ye^{2y}$$ Add 1 to both sides, $$x^2 = ye^{2y}+1$$ Finally, take the square root on both sides, $$x = √{ye^{2y}+1}$$
$\log(x+1) = y^2 + \log(x-1)$
Exponentiate both sides using base 10, $$x+1 = 10^{y^2+\log(x-1)}$$ Separate the exponent on the right side, $$x+1 = 10^{y^2}⋅10^{\log(x-1)}$$ Simplify the right side, $$x+1 = 10^{y^2}⋅(x-1)$$ Distribute on the right side, $$x+1 = x10^{y^2}-10^{y^2}$$ Separate terms to have all x on the left, $$x - x10^{y^2} = -10^{y^2}-1$$ Factor on the left side, $$x(1-10^{y^2}) = -10^{y^2}-1$$ Divide both sides by $1-10^{y^2}$, $$x = {-10^{y^2}-1}/{1-10^{y^2}}$$ Multiply both numerator and denominator by -1, $$x = {10^{y^2}+1}/{10^{y^2}-1}$$
$2\ln(x) = \ln(x+1) + y$
Bring the factor of 2 in the log on the left side, $$\ln(x^2) = \ln(x+1) + y$$ Exponentiate and simplify both sides, $$x^2 = (x+1)⋅e^y$$ Distribute on the right side, $$x^2 = xe^y+e^y$$ Bring every term on the right side, $$x^2 - xe^y - e^y = 0$$ Use the quadratic equation. $$x= { e^y+√{e^{2y}+4e^y} }/2$$ Note that the original question implied the log of $x$ which cannot be negative hence the solution only uses the positive version of the quadratic equation.
${e^x + e^{-x}} / {e^x - e^{-x}} = y$
Substitute $u$ for $e^x$, $${u^1 + u^{-1}} / {u^1 - u^{-1}} = y$$ Multiply both numerator and denominator by $u$, $${u^2 + 1} / {u^2 - 1} = y$$ Multiply both sides by $(u^2 - 1)$, $$u^2 + 1 = yu^2 - y$$ Gather all $u$ on the left, other terms on the right, $$u^2 - yu^2 = - y - 1$$ Factor out $u^2$ on the right side, $$u^2(1-y) = -y -1$$ Divide both sides by $(1-y)$, $$u^2 = {-y -1} / {1-y}$$ Multiply both numerator and denominator by -1 on the right
then take the square root on both sides, $$u=√{{y+1}/{y-1}}$$ Substitute back from $u$ to $e^x$ and take the natural log on both sides, $$x=\ln(√{{y+1}/{y-1}})$$ This solution can be rewritten or simplified in many other ways such as: $$x=1/2\ln({y+1}/{y-1})$$
$y = e^x + e^{-x}$
Substitute $u$ for $e^x$, $$y = u^1 + u^{-1}$$ Multiply both sides by $u$, $$yu = u^2 + 1$$ Bring every term on one side, $$0 = u^2 - yu +1$$ Solve for $u$ with the quadratic equation, $$u = {y ± √{ y^2-4 }} /2$$ Remember that $u$ is $e^x$, so taking the natural log on both sides gives: $$x = \ln({{y ± √{ y^2-4 }} /2})$$ This solution can be rewritten or simplified in many other ways such as: $$x = \ln({{y ± 1/2√{{y^2}/4-1}}})-\ln(2)$$
Trigonometric functions are defined in relation to an angle θ in a right triangle: It is the proportion that the opposite side, or the adjacent side to θ makes with the hypotenuse: $\sin(θ)=\opp/\hyp$, $\cos(θ)=\adj/\hyp$, and $\tan(θ)=\opp/\adj$.
Sine is an odd function that is also periodic, so negating the angle negates the result: $$\sin(-θ)=-\sin(θ)$$ It is repeating at every 2π and is a copy of cosine if the angle is shifted by π/2 or -3π/2.
Cosine is an even function which is also periodic, so negating the angle does not change the result: $$\cos(-θ)=\cos(θ)$$ It is repeating at every 2π and is a copy of sine if the angle is shifted by -π/2 or 3π/2.
The theorem of pythagore says that for a right triangle, the length of the hypotenuse squared is the sum of the squares of the other two sides. It follows from the definition of sine and cosine that
$\cos^2(θ)+\sin^2(θ)=1$.
Dividing both sides by either $\cos^2(θ)$ or $\sin^2(θ)$ yield respectively:
$1+\tan^2(θ)=\sec^2(θ)$ $\cot^2(θ)+1=\csc^2(θ)$.
Sine and cosine values from the pythagorean theorem:
Multiples of $π/2$ angles (90 degrees) yield trigonometric values of either 1 or 0. An angle of π/4, (45 degrees), has equal sine and cosine values for being exactly in the middle range of both axis. To find this value 'x', we start with the hypotenuse squared which is equal to twice that value squared: $1^2=x^2+x^2$.
This is equivalent to $1=√{x^2+x^2}$ or $1=√{2x^2}$.
Squaring both sides gives: $1=2x^2$. Dividing both sides by two: $1/2=x^2$.
Taking the square root gives: $1/{√{2}}=x$. Rationalizing the denominator yields ${√{2}/2$.
Multiples of $π/6$ angles have sine and cosine values that can be deduced from the pythagorean theorem by starting with an equilateral triange of side length 1. Then cut that triangle in half with a vertical line to keep only one of the right triangles.
Its base is then exactly half the hypotenuse: $1/2$. Its remaining side x is then $x=√{1^2-{1/2}^2}$, which is also $x=√{1-{1/4}}$, the equivalent of $x=√{3/4}$ or $√{3}/2$.
$\sin(α+β)$ is directly ${\BC}/{\AB}$.
BC can also be measured from the segments on the right ${\EF+\BD}/{\AB}$.
Separating this fraction in two terms gives: ${\EF}/{\AB}+{\BD}/{\AB}$.
Since EF and BD dont make triangles with AB, nor relate directly to it,
we make EF relate to AE and multiply it with ${\AE}/{\AB}$ to keep the same value for this term.
Both those ratiosnow constitute meaningfull triangles in the diagram.
The same is done with BD, in relation with BE, and multiplied by ${\BE}/{\AB}$ to keep its value. $${\EF}/{\AE}⋅{\AE}/{\AB}+{\BD}/{\BE}⋅{\BE}/{\AB}$$ $$\sin(α)\cos(β)+\cos(α)\sin(β)$$
From the same diagram, we see that $\cos(α+β)$ is ${\AC}/{\AB}$.
AC can also be measured from segments of other triangles: ${\AF-\DE}/{\AB}$.
Separating this fraction in two terms gives: ${\AF}/{\AB}-{\DE}/{\AB}$.
Since AF and DE dont make triangles with AB, nor relate directly to it,
we make AF relate to AE and multiply it with ${\AE}/{\AB}$ to keep the same value for this term.
Those ratios then constitute meaningfull triangles in the diagram.
The same is done with DE, in relation with BE, and multiplied by ${\BE}/{\AB}$ to keep its value. $${\AF}/{\AE}⋅{\AE}/{\AB}-{\DE}/{\BE}⋅{\BE}/{\AB}$$ $$\cos(α)\cos(β)-\sin(α)\sin(β)$$
Starting with a double angle as a sum: $\sin(2α)=\sin(α+α)$.
From the sum of angles identity we know that:
$\sin(α+α)=\cos(α)\sin(α)+\sin(α)\cos(α)$, which is finally: $\sin(2α)=2\sin(α)\cos(α)$.
For the double angle with cosine we start with: $\cos(2α)=\cos(α+α)$.
Again from the sum of angles identity we know that:
$\cos(α+α)=\cos(α)\cos(α)-\sin(α)\sin(α)$, which gives: $\cos(2α)=\cos^2(α)-\sin^2(α)$.
From the pythagorean identities, since $\sin^2(θ)=1-\cos^2(θ)$, we also find: $\cos(2α)=\cos^2(α)- (1-\cos^2(θ))$, which means that $\cos(2α)=2\cos^2(α)-1$.
From the double angle formula of the cosine, its angle formula can be found. Adding one and multiplying by 2 on both sides gives ${1+\cos(2α)}/2=\cos^2(α)$ which is used to reduce the powers of a cosine.
Taking the square root: $√{{1+\cos(2α)}/2}=\cos(α)$ then reverses the double angle formula.
To find the double and half angle formulas for the sine, we start with the cosine double angle formula: $\cos(2α)=\cos^2(α)-\sin^2(α)$. With a pythagorean identity we can rewrite: $\cos(2α)=(1-\sin^2(θ))-\sin^2(α)$ which is the same as: $\cos(2α)=(1-2\sin^2(θ))$ Adding one and multiplying by 2 on both sides gives ${\cos(2α)-1}/2=-\sin^2(α)$, or after inverting, ${1-\cos(2α)}/2=\sin^2(α)$ which is used to reduce the powers of a sine.
Taking the square root: $√{{1-\cos(2α)}/2}=\sin(α)$ then reverses the double angle formula.
To compute the lenght of side $A$, if the opposite angle $α$ is known as well as the two other sides,
first recognise that the vertical $h$ is measurable with trigonometry : $h=C⋅{sin(α)}$.
Also, that $e=C⋅{cos(α)}$ and because $f=B-e$ it follows that $f=B-C⋅{cos(α)}$.
Using the pythagorian theorem in the form $A^2=h^2+f^2$ you can write :
$A^2= (Csinα)^2+(B-Ccosα)^2 $
Expanding $h$ and $f$ gives :
$A^2= C^2sin^2α + B^2 +2BC⋅cos(α) + C^2{cos^2(α)}$
Group the terms with a $C^2$ and factor
$A^2= C^2(sin^2(α)+cos^2(α)) + B^2 +2BC⋅cos(α) $
Recognize the trigonometric identity $sin^2+cos^2=1$ in the term with $C^2$ to get
$A^2= C^2 + B^2 +2BC⋅cos(α) $ Often rearanged as : $A = √{B^2 + C^2 +2BC⋅cos(α)} $
