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## How do you convert r(2 - cosx) = 2 to rectangular form?

Douglas K.
Featured 7 months ago

The explanation is written with the assumption that you meant $\theta$ for the argument of the cosine function.

#### Explanation:

Use the distributive property :

$2 r - r \cos \left(\theta\right) = 2$

$2 r = r \cos \left(\theta\right) + 2$

Substitute x for $r \cos \left(\theta\right)$ and $\sqrt{{x}^{2} + {y}^{2}}$ for r:

$2 \sqrt{{x}^{2} + {y}^{2}} = x + 2$

Square both sides:

$4 \left({x}^{2} + {y}^{2}\right) = {x}^{2} + 4 x + 4$

$4 {x}^{2} + 4 {y}^{2} = {x}^{2} + 4 x + 4$

$3 {x}^{2} - 4 x + 4 {y}^{2} = 4$

Add $3 {h}^{2}$ to both sides:

$3 {x}^{2} - 4 x + 3 {h}^{2} + 4 {y}^{2} = 3 {h}^{2} + 4$

Remove a factor of the 3 from the first 3 terms:

$3 \left({x}^{2} - \frac{4}{3} x + {h}^{2}\right) + 4 {y}^{2} = 3 {h}^{2} + 4$

Use the middle in the right side of the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ and middle term of the equation to find the value of h:

$- 2 h x = - \frac{4}{3} x$

$h = \frac{2}{3}$

Substitute the left side of the pattern into the equation:

$3 {\left(x - h\right)}^{2} + 4 {y}^{2} = 3 {h}^{2} + 4$

Substitute $\frac{2}{3}$ for h and insert a -0 into the y term:

$3 {\left(x - \frac{2}{3}\right)}^{2} + 4 {\left(y - 0\right)}^{2} = 3 {\left(\frac{2}{3}\right)}^{2} + 4$

$3 {\left(x - \frac{2}{3}\right)}^{2} + 4 {\left(y - 0\right)}^{2} = \frac{16}{3}$

Divide both sides by $\frac{16}{3}$

$3 {\left(x - \frac{2}{3}\right)}^{2} / \left(\frac{16}{3}\right) + 4 {\left(y - 0\right)}^{2} / \left(\frac{16}{3}\right) = 1$

${\left(x - \frac{2}{3}\right)}^{2} / \left(\frac{16}{9}\right) + {\left(y - 0\right)}^{2} / \left(\frac{16}{12}\right) = 1$

Write the denominators as squares:

${\left(x - \frac{2}{3}\right)}^{2} / {\left(\frac{4}{3}\right)}^{2} + {\left(y - 0\right)}^{2} / {\left(\frac{4}{\sqrt{12}}\right)}^{2} = 1$

The is standard Cartesian form of the equation of an ellipse with a center at $\left(\frac{2}{3} , 0\right)$; its semi-major axis is $\frac{4}{3}$ units long and is parallel to the x axis and its semi-minor axis is $\frac{4}{\sqrt{12}}$ units long and is parallel to the y axis.

## How do you graph  y=1+5cos(x-pi)?

Rina Takahashi
Featured 6 months ago

Here is the graph of the function (it's not drawn to scale):

I will explain how I graph it below:

#### Explanation:

Your given equation is $y = 1 + 5 \cos \left(x - \pi\right)$ .

I would write the given equation in the standard form according to $y = A \cos B \left(x - C\right) + D$

You get: $y = 5 \cos \left(x - \pi\right) + 1$

This way, it is easier to define your amplitude, phase shift, vertical shift, and period multiplier.

In order to graph, we need to know the following:

1. Amplitude , which is $| A |$
2. Period Multiplier , which is B
3. Phase Shift , which is C
4. Vertical Shift, which is D
5. Period, which is $P e r i o d = \frac{2 \pi}{B}$

Let's get all our required values from our equation: $y = 5 \cos \left(x - \pi\right) + 1$

1. Amplitude = $| 5 |$ = 5
2. Period Multiplier = B = 1
3. Phase Shift = $- \pi$ = $\pi$ units to the right
4. Vertical Shift = $+ 1$ = 1 units up
5. Period = $P e r i o d = \frac{2 \pi}{1}$ = $2 \pi$

It is important to note that B is not apparent in this equation because it is equal to 1. The actual equation should really be $y = 5 \cos 1 \left(x - \pi\right) + 1$ .

Our period multiplier is important for determining our period, and it also affects our phase shift. However, we do not have to worry about it for this equation since it's just equal to 1.

Since we have both an amplitude and a vertical shift, we get new minimum and maximum y values for the graph:

Maximum Y Value = Vertical Shift + Amplitude = $1 + 5 = 6$

Minimum Y Value = Vertical Shift - Amplitude = $1 - 5 = - 4$

Most importantly, this graph is a positive cosine graph because our equation is 5cos.. and not -5cos.. . A positive cosine graph always starts at the maximum value.

Also, our vertical shift is +1, so our new x-axis start has shifted up by 1. We are starting at y = 1 and not y = 0.

As with our phase shift, it is $\pi$ units to the right. How I like to think of the phase shift is that it's just a new start to the graph. You can treat the phase shift as a place to actually begin your graph.

So.. with all this information, we can now actually graph the equation of $y = 5 \cos \left(x - \pi\right) + 1$ .

Here are some visual cues.

My art is pretty bad, sographing the graph on a website should be much more different.

I hope this helps.

## 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 $y = x$.

## Question on combination of Logarithm and Trigonometry (there are two parts)?

dk_ch
Featured 2 months ago

Q-1

Let $x = 4 \sin 9 \cos 9 = 2 \sin 18 = 2 \times \frac{\sqrt{5} - 1}{4} = \frac{1}{2} \left(\sqrt{5} - 1\right)$

Again

${t}_{1} = 4 \sin 63 \cos 63 = 2 \sin 126 = 2 \sin \left(180 - 54\right) = 2 \sin 54 = \frac{1}{2} \left(\sqrt{5} + 1\right)$

So $x \times {t}_{1} = \frac{1}{2} \left(\sqrt{5} - 1\right) \times \frac{1}{2} \left(\sqrt{5} + 1\right) = \frac{1}{4} \times 4 = 1$

Otherwise without using exact values

$x \cdot {t}_{1} = 4 \sin 9 \cos 9 \cdot 4 \sin 63 \cos 63$

$\implies x \cdot {t}_{1} = 2 \sin 18 \cdot 2 \sin 126$

$\implies x \cdot {t}_{1} = 4 \sin 18 \cdot \sin \left(180 - 54\right)$

$\implies x \cdot {t}_{1} = \frac{2}{\cos} 18 \cdot 2 \sin 18 \cdot \cos 18 \cdot \sin 54$

$\implies x \cdot {t}_{1} = \frac{1}{\cos} 18 \cdot 2 \sin 36 \cdot \sin 54$

$\implies x \cdot {t}_{1} = \frac{\cos \left(54 - 36\right) - \cos \left(54 + 36\right)}{\cos} 18$

$\implies x \cdot {t}_{1} = \frac{\cos 18 - \cos 90}{\cos} 18$

$\implies x \cdot {t}_{1} = \frac{\cos 18 - 0}{\cos} 18 = 1$

Hence $x = \frac{1}{t} _ 1$

Now

${\log}_{{t}_{1}} 4 \sin 9 \cos 9$

$= {\log}_{{t}_{1}} x$

$= {\log}_{{t}_{1}} \left(\frac{1}{t} _ 1\right)$

$= {\log}_{{t}_{1}} {\left({t}_{1}\right)}^{- 1}$

$= - {\log}_{{t}_{1}} \left({t}_{1}\right) = - 1$

Q-2

$l = {\left(\frac{{\cot}^{2} x {\cos}^{2} x}{{\cot}^{2} x - {\cos}^{2} x}\right)}^{2}$

$\implies l = {\left(\frac{{\cot}^{2} x {\cos}^{2} x}{\frac{1}{\tan} ^ 2 x - \frac{1}{\sec} ^ 2 x}\right)}^{2}$

$\implies l = {\left(\frac{{\cot}^{2} x \cdot {\tan}^{2} x \cdot {\cos}^{2} x \cdot {\sec}^{2} x}{{\sec}^{2} x - {\tan}^{2} x}\right)}^{2}$

$\implies l = {\left(\frac{1 \cdot 1}{1}\right)}^{2} = 1$

Now

m=a^(log_(sqrta)abs(2cos(y/2))

=>m=a^(log_(sqrta)abs(2cos((4pi)/2))

=>m=a^(log_(sqrta)abs(2cos(2pi))

=>m=a^(log_(sqrta)2

=>m=((sqrta)^2)^(log_(sqrta)2

=>m=(sqrta)^(2log_(sqrta)2

$\implies m = {\left(\sqrt{a}\right)}^{{\log}_{\sqrt{a}} {2}^{2}} = 4$

So

${l}^{2} + {m}^{2} = {1}^{2} + {4}^{2} = 17$

## sin^(-1)(cos(2x)) is?

Luke Phillips
Featured 1 month ago

$\text{arcsin} \left(\cos \left(2 x\right)\right) = \left\{\begin{matrix}2 x + \frac{\pi}{2} \left(1 - 4 n\right) & \left(n - \frac{1}{2}\right) \pi \le x \le n \pi \\ - 2 x + \frac{\pi}{2} \left(1 + 4 n\right) & n \pi \le x \le \left(n + \frac{1}{2}\right) \pi\end{matrix}\right.$ for $n \in \mathbb{Z}$.

#### Explanation:

Note that,
$\sin \left(\frac{\pi}{2} + x\right) = \sin \left(\frac{\pi}{2}\right) \cos \left(x\right) + \cos \left(\frac{\pi}{2}\right) \cos \left(x\right)$,
$\sin \left(\frac{\pi}{2} + x\right) = \cos \left(x\right)$.

We see that,
$\cos \left(2 x\right) = \sin \left(\frac{\pi}{2} + 2 x\right)$.

Due to the periodicity of $\sin \left(x\right)$ we have that $\sin \left(\pi - x\right) = \sin \left(x\right)$, and $\sin \left(x\right) = \sin \left(x + 2 n \pi\right)$ with $n \in \mathbb{Z}$. (This can be seen from expanding $\sin \left(\pi - x\right)$ using the sine addition formulae).

Define

$f \left(x\right) = \text{arcsin} \left(\cos \left(2 x\right)\right)$.

Then we have

$f \left(x\right) = \text{arcsin} \left(\sin \left(\frac{\pi}{2} + 2 x + 2 n \pi\right)\right)$.

Look at the function $\text{arcsin} \left(x\right)$ pictured below.

graph{arcsin(x) [-10, 10, -5, 5]}

it takes in an argument between 0 and 1 and outputs the principle value of $x$ in $- \frac{\pi}{2} \le x \le \frac{\pi}{2}$ that would make $\sin \left(x\right)$ output that value.

So, we have that $\frac{\pi}{2} + 2 x + 2 n \pi$ needs to be between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$.

I.e.,

$2 x + \pi \left(2 n + \frac{1}{2}\right) < \frac{\pi}{2}$,
$2 x < \pi \left(\frac{1}{2} - \frac{1}{2} - 2 n\right)$,
$2 x < - 2 n \pi$,
$x < - n \pi$.

And,

$2 x + \pi \left(2 n + \frac{1}{2}\right) > - \frac{\pi}{2}$,
$2 x > - \pi \left(1 + 2 n\right)$,
$x > - n \pi - \frac{1}{2} \pi$

As $n$ is an arbitrary integer we can easily let $n = - n$ and see that for $\left(n - \frac{1}{2}\right) \pi < x < n \pi$ that $f \left(x\right) = 2 x + \frac{\pi}{2} \left(1 - 4 n\right)$.

Notice that for $n \in \mathbb{Z}$ as we require this leaves some portions of $f \left(x\right)$ undefined.

We can use $\sin \left(\pi - x\right) = \sin \left(x\right)$ and define $f \left(x\right)$ as,

$f \left(x\right) = \text{arcsin} \left(\sin \left(\pi - \frac{\pi}{2} - 2 x - 2 n \pi\right)\right)$,
$f \left(x\right) = \text{arcsin} \left(\sin \left(\frac{\pi}{2} - 2 x - 2 n \pi\right)\right)$,
$f \left(x\right) = \text{arcsin} \left(\sin \left(\frac{\pi}{2} - 2 x + 2 n \pi\right)\right)$.

We need to solve the same inequalities as before,

$\frac{\pi}{2} - 2 x + 2 n \pi > - \frac{\pi}{2}$ gives $x < \frac{\pi}{2} + n \pi$.
$\frac{\pi}{2} - 2 x + 2 n \pi < \frac{\pi}{2}$ gives $x > n \pi$.

Then we see that for $n \pi < x < \frac{\pi}{2} + n \pi$, $f \left(x\right) = - 2 x + \frac{\pi}{2} \left(4 n + 1\right)$. Now $f \left(x\right)$ is defined for all $x$.

Then,

$f \left(x\right) = \left\{\begin{matrix}2 x + \frac{\pi}{2} \left(1 - 4 n\right) & \left(n - \frac{1}{2}\right) \pi \le x \le n \pi \\ - 2 x + \frac{\pi}{2} \left(1 + 4 n\right) & n \pi \le x \le \left(n + \frac{1}{2}\right) \pi\end{matrix}\right.$ for $n \in \mathbb{Z}$.

## What cosine function represents an amplitude of 3, a period of π, no horizontal shift, and a vertical shift of?

Alan P.
Featured 1 week ago

In order to answer this I have assumed a vertical shift of $+ 7$

$\textcolor{red}{3 \cos \left(2 \theta\right) + 7}$

#### Explanation:

The standard cos function $\textcolor{g r e e n}{\cos \left(\gamma\right)}$ has a period of $2 \pi$

If we want a period of $\pi$ we need to replace $\gamma$ with something that will cover the domain "twice as fast" e.g. $2 \theta$.
That is $\textcolor{m a \ge n t a}{\cos \left(2 \theta\right)}$ will have a period of $\pi$.

To get an amplitude of $3$ we need to multiply all values in the Range generated by $\textcolor{m a \ge n t a}{\cos \left(2 \theta\right)}$ by $\textcolor{b r o w n}{3}$ giving
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b r o w n}{3 \cos \left(2 \theta\right)}$

There is to be no horizontal shift, so the argument for $\cos$ will not be modified by any further addition/subtraction.

In order to achieve the vertical shift (that I assumed would be $\textcolor{red}{+ 7}$ [substitute your own value]) we will need to add $\textcolor{red}{7}$ to all values in our modified range:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{3 \cos \left(2 \theta\right) + 7}$

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