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Featured 7 months ago

See explanation...

Consider a right angled triangle with an internal angle

Then:

#sin theta = a/c#

#cos theta = b/c#

So:

#sin^2 theta + cos^2 theta = a^2/c^2+b^2/c^2 = (a^2+b^2)/c^2#

By Pythagoras

So given Pythagoras, that proves the identity for

For angles outside that range we can use:

#sin (theta + pi) = -sin (theta)#

#cos (theta + pi) = -cos (theta)#

#sin (- theta) = - sin(theta)#

#cos (- theta) = cos(theta)#

So for example:

#sin^2 (theta + pi) + cos^2 (theta + pi) = (-sin theta)^2 + (-cos theta)^2 = sin^2 theta + cos^2 theta = 1#

**Pythagoras theorem**

Given a right angled triangle with sides

The area of the large square is

The area of the small, tilted square is

The area of each triangle is

So we have:

#(a+b)^2 = c^2 + 4 * 1/2ab#

That is:

#a^2+2ab+b^2 = c^2+2ab#

Subtract

#a^2+b^2 = c^2#

Featured 5 months ago

Here is the graph of

Step 1: Sketch the graph of

Step 2: Sketch the line

Step 3: Reflect the graph of

Featured 3 months ago

Given:

Use this reference for Fundamental Frequency

Let

Using the fact that

The fundamental frequency is the greatest common divisor of the two frequencies:

Here is a graph:

graph{y = sin(4x) - cos(7x) [-10, 10, -5, 5]}

Please observe that it repeats every

Featured 2 months ago

A few thoughts...

It was known and studied by Euclid (approx 3rd or 4th century BCE), basically for many geometric properties...

It has many interesting properties, of which here are a few...

The Fibonacci sequence can be defined recursively as:

#F_0 = 0#

#F_1 = 1#

#F_(n+2) = F_n + F_(n+1)#

It starts:

#0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987,...#

The ratio between successive terms tends to

#lim_(n->oo) F_(n+1)/F_n = phi#

In fact the general term of the Fibonacci sequence is given by the formula:

#F_n = (phi^n - (-phi)^(-n))/sqrt(5)#

A rectangle with sides in ratio

This is related to both the limiting ratio of the Fibonacci sequence and the fact that:

#phi = [1;bar(1)] = 1+1/(1+1/(1+1/(1+1/(1+1/(1+...)))))#

which is the most slowly converging standard continued fraction.

If you place three golden rectangles symmetrically perpendicular to one another in three dimensional space, then the twelve corners form the vertices of a regular icosahedron. Hence we can calculate the surface area and volume of a regular icosahedron of given radius. See https://socratic.org/s/aFZyTQfn

An isosceles triangle with sides in ratio

Featured 2 months ago

As per problem the movement of the London Eye is periodic one and its height at

*meter* and t in *sec*.

Here *cycle* is 3o min. This means time period

Hence

So the equation[1] takes the following form

Now at

Applying this condition on [2] we get

Since at

then we can say that at

So we can write

Now adding [3] and [4] we have

Subtracting [4] from [3] we get

So finally the given equation takes the following form

The variation of height with time can be represented by following graph.

Featured 1 month ago

**Alternative**

Let

So

The given expression becomes

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