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## How would one solve this trigonometric model?

A08
Featured 6 months ago

Alternate solution for part (c)

#### Explanation:

For Solution posted by @dk_ch
The modelling equation becomes

$f \left(t\right) = 2.5 t + 20 + 10 \sin \left(\frac{\pi t}{6}\right)$

Using inbuilt graphics tool the plotted equation looks like

The maximum and minimum of the curve are located for $t = 3.951$ and $8.049$ respectively. These values correspond to months of May and September.
As such the stock actually lost value from May to August.

--.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-..-

For Solution posted by @Cesareo R

$f \left(t\right) = 2.25 t + 18.49 + 7.29 \sin \left(\frac{\pi t}{6}\right)$, values rounded to two decimal places

$m = {a}_{1} = 2.25439$, $b = {a}_{2} = 18.49$, $A = {a}_{3} = 7.29285$

Using inbuilt graphics tool the plotted equation looks like

The maximum and minimum of the curve are located for $t = 4.204$ and $7.796$ respectively. These values correspond to months of May and September.
As such the stock actually lost value from May to August.

## How do you find the exact value cos(x+y) if tanx=5/3,siny=1/3?

mason m
Featured 5 months ago

$\cos \left(x + y\right) = \frac{6 \sqrt{2} - 5}{3 \sqrt{34}}$

#### Explanation:

Use the cosine angle addition formula:

$\cos \left(x + y\right) = \cos x \cos y - \sin x \sin y$

We need to determine $\sin x$, $\cos x$, and $\cos y$ from whatever we have.

Determining $m a t h b f \left(\sin x , \cos x :\right)$

${\tan}^{2} x + 1 = {\sec}^{2} x \text{ "" "" }$use $\tan x = \frac{5}{3}$

$\frac{25}{9} + 1 = {\sec}^{2} x$

$\sec x = \sqrt{\frac{34}{9}} = \frac{\sqrt{34}}{3}$

Then:

color(blue)(cosx=1/secx=3/sqrt34

Now using:

${\sin}^{2} x + {\cos}^{2} x = 1 \text{ "" "" }$where $\cos x = \frac{3}{\sqrt{34}}$

${\sin}^{2} x + \frac{9}{34} = 1$

color(blue)(sinx=sqrt(25/34)=5/sqrt34

Determining $m a t h b f \left(\cos y :\right)$

${\sin}^{2} y + {\cos}^{2} y = 1 \text{ "" "" }$use $\sin y = \frac{1}{3}$

$\frac{1}{9} + {\cos}^{2} y = 1$

color(blue)(cosy=sqrt(8/9)=(2sqrt2)/3

Returning to $m a t h b f \left(\cos \left(x + y\right) :\right)$

$\cos \left(x + y\right) = \cos x \cos y - \sin x \sin y$

$\textcolor{w h i t e}{\cos \left(x + y\right)} = \frac{3}{\sqrt{34}} \left(\frac{2 \sqrt{2}}{3}\right) - \frac{5}{\sqrt{34}} \left(\frac{1}{3}\right)$

$\textcolor{w h i t e}{\cos \left(x + y\right)} = \frac{6 \sqrt{2} - 5}{3 \sqrt{34}}$

Note these are all assuming that $x$ and $y$ are in the first quadrant (everything is positive).

## How to find amplitude and cycle for y = 1+4 sin (2x)?

Love K. Soother
Featured 3 months ago

Period of $y = 1 + 4 \setminus \sin \left(2 x\right)$ is $n \setminus \pi$ and amplitude is $4$

### GENERAL WAVE EQUATION

$f \left(x\right) = a \setminus \sin \left(b x + c\right) + d$,

where $\setminus \pm c$, $\setminus \pm d$ are wave right (left) or up (down) shifts along $x \text{-axis}$ and $y \text{-axis}$ respectively.

### SOLUTION (for period)

Ignoring the shifts along axes i.e. assuming $c = d = 0$ we can have our period and amplitude without any sort of error.

$f \left(x\right) = y = 1 + 4 \setminus \sin \left(2 x\right)$

$\setminus R i g h t a r r o w y - 1 = 4 \setminus \sin \left(2 x\right)$

$\setminus R i g h t a r r o w g \left(x\right) = 4 \setminus \sin \left(2 x\right)$

We know from fundamental trigonometric ideas that the period of sine function is $2 n \setminus \pi$ i.e.

$\setminus \sin \left(x\right) = \setminus \sin \left(2 n \setminus \pi + x\right)$,

Let,

$\setminus \sin \left(u\right) = \setminus \sin \left(2 n \setminus \pi + u\right)$

Then for $u = 2 x$, we have;

$\setminus \sin \left(2 x\right) = \setminus \sin \left(2 n \setminus \pi + 2 x\right)$

Multiplying both sides by $4$,

$4 \setminus \sin \left(2 x\right) = 4 \setminus \sin 2 \left(n \setminus \pi + x\right)$

$\setminus \equiv g \left(x\right) = g \left(n \setminus \pi + x\right)$

Hence period is $n \setminus \pi$

### SOLUTION (amplitude)

In general the amplitude of sine function is given as;

If $y = \setminus \sin \left(x\right)$ then amplitude is $1$ unit long. as followed by graph

graph{y=\sin(x)[-3,3,-1.5,1.5]}

Our equation is a sort of $y = a \sin \left(b x\right)$

Which becomes;

$y = a \left[\setminus \sin \left(b x\right)\right]$

And that expands the graph of ordinary sine function i.e. $\sin \left(x\right)$ up to $| a |$ times along $y \text{-axis}$

graph{y=4\sin(2x)}

Therefore our amplitude is $4$

The graph of our real function $y = 1 + 4 \setminus \sin \left(2 x\right)$ is shifted upwards $1$ unit

graph{y=1+4 \sin(2x)}

But amplitude remains same.

## Expressing in different form and finding a maximum? See picture below (7 marks)

dk_ch
Featured 3 months ago

(a)

$\sqrt{15} \sin \left(2 x\right) + \sqrt{5} \cos \left(2 x\right)$

$= \sqrt{20} \left(\frac{\sqrt{15}}{\sqrt{20}} \sin \left(2 x\right) + \frac{\sqrt{5}}{\sqrt{20}} \cos \left(2 x\right)\right)$

[Taking $\frac{\sqrt{15}}{\sqrt{20}} = \cos \alpha \mathmr{and} \frac{\sqrt{5}}{\sqrt{20}} = \sin \alpha$]

$= \sqrt{20} \left(\cos \alpha \sin \left(2 x\right) + \sin \alpha \cos \left(2 x\right)\right)$

$= 2 \sqrt{5} \sin \left(2 x + \alpha\right)$

Where $\alpha = {\tan}^{-} 1 \left(\frac{\sqrt{5}}{\sqrt{15}}\right) = {\tan}^{-} 1 \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

(b)

(i)
$f \left(x\right) = \frac{2}{5 + \sqrt{15} \sin \left(2 x\right) + \sqrt{\cos} \left(2 x\right)}$

$\implies f \left(x\right) = \frac{2}{5 + 2 \sqrt{5} \sin \left(2 x + \alpha\right)}$

The value of $f \left(x\right)$ will be maximum when $\sin \left(2 x + \alpha\right)$ is minimum i.e. $\sin \left(2 x + \alpha\right) = - 1$

So

$f {\left(x\right)}_{\text{max}} = \frac{2}{5 - 2 \sqrt{5}}$

$\implies f {\left(x\right)}_{\text{max}} = \frac{2 \left(5 + 2 \sqrt{5}\right)}{\left(5 - 2 \sqrt{5}\right) \left(5 + 2 \sqrt{5}\right)}$

$\implies f {\left(x\right)}_{\text{max}} = \frac{10 + 4 \sqrt{5}}{25 - 20} = 2 + \frac{4}{5} \sqrt{5}$

(ii)Here $f \left(x\right)$ is maximum when

$\sin \left(2 x + \alpha\right) = - 1$

$\implies \sin \left(2 x + \frac{\pi}{6}\right) = \sin \left(\frac{3 \pi}{2}\right)$

$\implies 2 x + \frac{\pi}{6} = \frac{3 \pi}{2}$

$\implies 2 x = \frac{3 \pi}{2} - \frac{\pi}{6} = \frac{4 \pi}{3}$

$\implies x = \frac{2 \pi}{3}$

## Explain how to solve an equation in the form sin x=A ?

Shwetank Mauria
Featured 1 month ago

#### Explanation:

It isassumed that in the equation $\sin x = A$, $A$ is a number so that $- 1 \le A \le 1$ i.e. $A$ lies between $1$ and $- 1$, including these values. This is because sine ratio of an angle always lies between $- 1$
and $1$.

Also cosider two equations $y = \sin x$ and $y = A$. When we draw their graph, they intersect where the solution of $\sin x = A$ exists. Assume two values of $A$, say $A = 0.75$ or $A = - 0.5$. The two graphs (shown separately) appear as shown below:

graph{(y-sinx)(y-0.75)=0 [-1.25, 8.75, -2.48, 2.52]}

graph{(y-sinx)(y+0.5)=0 [-1.25, 8.75, -2.48, 2.52]}

Observe that each has two solutions. In first case when $A = 0.75$, we have solution in first quadrant, as $A$ is positive, which is approximately ${48.6}^{\circ}$ or $0.848$ radians. As $\sin x = \sin \left(\pi - x\right)$, wealso have a solution $3.1416 - 0.848 = 2.2936$ i.e. in $Q 1$ and $Q 2$.

Similarly, solution for $\sin x = - 0.5$ is $x = {210}^{\circ}$ or $x = 3.6652$ radians as well as $x = {330}^{\circ}$ or $x = 5.7596$ radians i.e. in $Q 3$ and $Q 4$.

Hence, if $A$ is positive, first identify $\angle \alpha$ radians whose sine is $A$ and solutions are $\alpha$ and $\pi - \alpha$ in radians or $\alpha$ and ${180}^{\circ} - \alpha$ if $\alpha$ is in degrees. The solution would be in $Q 1$ and $Q 2$.

If $A$ is negative, first identify $\angle \alpha$ radians whose sine is $- A$ (note that as $A$ is negative, $- A$ is positive) and solutions are $\pi + \alpha$ and $2 \pi - \alpha$ in radians or ${180}^{\circ} + \alpha$ and ${360}^{\circ} - \alpha$ if $\alpha$ is in degrees. The solution would be in $Q 3$ and $Q 4$.

General solutions are $x = n \pi \pm {\left(- 1\right)}^{n} \alpha$ or $x = n \times {360}^{\circ} \pm {\left(- 1\right)}^{n} \alpha$. Note that here ${\left(- 1\right)}^{n}$ takes care of positive or negative sign before $\alpha$.

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

George C.
Featured 2 weeks 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

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