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## Can anyone Prove (sin theta)^2+(cos theta)^2=1 ?

George C.
Featured 7 months ago

See explanation...

#### Explanation:

Consider a right angled triangle with an internal angle $\theta$:

Then:

$\sin \theta = \frac{a}{c}$

$\cos \theta = \frac{b}{c}$

So:

${\sin}^{2} \theta + {\cos}^{2} \theta = {a}^{2} / {c}^{2} + {b}^{2} / {c}^{2} = \frac{{a}^{2} + {b}^{2}}{c} ^ 2$

By Pythagoras ${a}^{2} + {b}^{2} = {c}^{2}$, so $\frac{{a}^{2} + {b}^{2}}{c} ^ 2 = 1$

So given Pythagoras, that proves the identity for $\theta \in \left(0 , \frac{\pi}{2}\right)$

For angles outside that range we can use:

$\sin \left(\theta + \pi\right) = - \sin \left(\theta\right)$

$\cos \left(\theta + \pi\right) = - \cos \left(\theta\right)$

$\sin \left(- \theta\right) = - \sin \left(\theta\right)$

$\cos \left(- \theta\right) = \cos \left(\theta\right)$

So for example:

${\sin}^{2} \left(\theta + \pi\right) + {\cos}^{2} \left(\theta + \pi\right) = {\left(- \sin \theta\right)}^{2} + {\left(- \cos \theta\right)}^{2} = {\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\textcolor{w h i t e}{}$
Pythagoras theorem

Given a right angled triangle with sides $a$, $b$ and $c$ consider the following diagram:

The area of the large square is ${\left(a + b\right)}^{2}$

The area of the small, tilted square is ${c}^{2}$

The area of each triangle is $\frac{1}{2} a b$

So we have:

${\left(a + b\right)}^{2} = {c}^{2} + 4 \cdot \frac{1}{2} a b$

That is:

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

Subtract $2 a b$ from both sides to get:

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

## How do you graph y=sin^-1x over the interval -1<=x<=1?

Wataru
Featured 5 months ago

Here is the graph of $y = {\sin}^{- 1} x$.

#### Explanation:

Step 1: Sketch the graph of $y = \sin x$ on $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

Step 2: Sketch the line $y = x$.

Step 3: Reflect the graph of $y = \sin x$ about the line $y = x$.

## What is the frequency of f(t)= sin(4t) - cos(7t)?

Douglas K.
Featured 3 months ago

${f}_{0} = \frac{1}{2 \pi} \text{ Hz}$

#### Explanation:

Given: $f \left(t\right) = \sin \left(4 t\right) - \cos \left(7 t\right)$ where t is seconds.

Use this reference for Fundamental Frequency

Let ${f}_{0}$ be the fundamental frequency of the combined sinusoids, in Hz (or ${\text{s}}^{-} 1$).

${\omega}_{1} = 4 \text{ rad/s}$
${\omega}_{2} = 7 \text{ rad/s}$

Using the fact that $\omega = 2 \pi f$

${f}_{1} = \frac{4}{2 \pi} = \frac{2}{\pi} \text{ Hz}$ and ${f}_{2} = \frac{7}{2 \pi} \text{ Hz}$

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

${f}_{0} = \gcd \left(\frac{2}{\pi} \text{ Hz", 7/(2pi)" Hz}\right)$

${f}_{0} = \frac{1}{2 \pi} \text{ Hz}$

Here is a graph:

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

Please observe that it repeats every $2 \pi$

## What is phi, how was it discovered and are its uses?

George C.
Featured 2 months ago

A few thoughts...

#### Explanation:

$\phi = \frac{1}{2} + \frac{\sqrt{5}}{2} \approx 1.6180339887$ is known as the Golden Ratio.

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:

$0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987 , \ldots$

The ratio between successive terms tends to $\phi$. That is:

${\lim}_{n \to \infty} {F}_{n + 1} / {F}_{n} = \phi$

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

${F}_{n} = \frac{{\phi}^{n} - {\left(- \phi\right)}^{- n}}{\sqrt{5}}$

A rectangle with sides in ratio $\phi : 1$ is called a Golden Rectangle. If a square of maximal size is removed from one end of a golden rectangle then the remaining rectangle is a golden rectangle.

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 $\phi : \phi : 1$ has base angles $\frac{2 \pi}{5}$ and apex angle $\frac{\pi}{5}$. This allows us to calculate exact algebraic formulae for $\sin \left(\frac{\pi}{10}\right)$, $\cos \left(\frac{\pi}{10}\right)$ and ultimately for any multiple of $\frac{\pi}{60}$ (${3}^{\circ}$). See https://socratic.org/s/aFZztx8s

dk_ch
Featured 2 months ago

As per problem the movement of the London Eye is periodic one and its height at $t$ th instant $h \left(t\right)$ is given by the following equation

$h \left(t\right) = a + b \cos \left(c t\right) \ldots \ldots \left[1\right]$, where $h \left(t\right)$ in meter and t in sec.

Here $c$ should be the angular velocity associated with the periodic motion. Again it is also given that time taken by the London Eye to complete one cycle is 3o min. This means time period $T = 30 \min$

Hence $\text{Angular velocity} = c = \omega = \frac{2 \pi}{T} = \frac{2 \pi}{30} = \frac{\pi}{15}$ rad/min.

So the equation[1] takes the following form

$h \left(t\right) = a + b \cos \left(\frac{2 \pi}{30} t\right) \ldots \ldots \left[2\right]$

Now at $t = 0$ the London Eye is at the bottom i.e.$h \left(0\right) = 0$

Applying this condition on [2] we get

$h \left(0\right) = a + b \cos \left(\frac{2 \pi}{30} \times 0\right)$

$0 = a + b \implies a + b = 0. \ldots . \left[3\right]$

Since at $t = 0$ it is at the bottom and it completes the cycle returning at the bottom at $t = 30$ min,
then we can say that at $t = 15$ min it will reach at maximum height 135 m i.e. $h \left(15\right) = 135$m
So we can write

$h \left(15\right) = 135 = a + b \cos \left(\frac{2 \pi}{30} \times 15\right)$

$\implies a + b \cos \left(\pi\right) = 135$

$\implies a - b = 135. \ldots . \left[4\right]$

Now adding [3] and [4] we have $2 a = 135 \implies a = 67.5$ m

Subtracting [4] from [3] we get $2 b = - 135 \implies b = - 67.5$m

So finally the given equation takes the following form

$\textcolor{red}{h \left(t\right) = 67.5 - 67.5 \cos \left(\frac{2 \pi}{30} t\right)}$

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

## Using addition formula how to solve: arctan(x) + arctan(1/2) + arctan(sqrt3/2) = pi/2?

dk_ch
Featured 1 month ago

$\arctan \left(x\right) + \arctan \left(\frac{1}{2}\right) + \arctan \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{2}$

$\implies \arctan \left(\frac{1}{2}\right) + \arctan \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{2} - \arctan \left(x\right)$

$\implies a r c \cot \left(2\right) + a r c \cot \left(\frac{2}{\sqrt{3}}\right) = a r c \cot \left(x\right)$

$\left[\textcolor{red}{\text{since } \frac{\pi}{2} - a r c \tan x = a r c \cot x}\right]$

$\implies a r c \cot \left(\frac{2 \times \frac{2}{\sqrt{3}} - 1}{\frac{2}{\sqrt{3}} + 2}\right) = a r c \cot \left(x\right)$

$\implies x = \left(\frac{2 \times \frac{2}{\sqrt{3}} - 1}{\frac{2}{\sqrt{3}} + 2}\right)$

$\implies x = \left(\frac{4 - \sqrt{3}}{2 + 2 \sqrt{3}}\right)$

Alternative

Let
arctan(x)=alpha ; arctan(1/2) =beta and arctan(sqrt3/2) = gamma

So x=tanalpha;1/2= tanbeta andsqrt3/2= tangamma

The given expression becomes

$\arctan \left(x\right) + \arctan \left(\frac{1}{2}\right) + \arctan \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{2}$

$\implies \alpha + \beta + \gamma = \frac{\pi}{2}$

$\implies \beta + \gamma = \frac{\pi}{2} - \alpha$

$\implies \tan \left(\beta + \gamma\right) = \tan \left(\frac{\pi}{2} - \alpha\right)$

$\implies \frac{\tan \beta + \tan \gamma}{1 - \tan \beta \tan \gamma} = \cot \alpha$

$\implies \frac{\frac{1}{2} + \frac{\sqrt{3}}{2}}{1 - \frac{1}{2} \times \frac{\sqrt{3}}{2}} = \cot \alpha$

$\implies \frac{4 \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right)}{4 \left(1 - \frac{1}{2} \times \frac{\sqrt{3}}{2}\right)} = \frac{1}{\tan} \alpha$

$\implies \frac{2 + 2 \sqrt{3}}{4 - \sqrt{3}} = \frac{1}{x}$

$\implies x = \left(\frac{4 - \sqrt{3}}{2 + 2 \sqrt{3}}\right)$

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