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## How do you graph r = 12cos(theta)?

Douglas K.
Featured 6 months ago

Set your compass to a radius of 6, put the center point at $\left(6 , 0\right)$, and draw a circle.

#### Explanation:

Multiply both sides of the equation by r:

${r}^{2} = 12 r \cos \left(\theta\right)$

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

${x}^{2} + {y}^{2} = 12 x$

The standard form of this type of equation (a circle) is:

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

To put the equation in this form, we need to complete the square for both the x and y terms. The y term is easy, we merely subtract $0$ in the square:

${x}^{2} + {\left(y - 0\right)}^{2} = 12 x$

To complete the square for the x terms, we add $- 12 x + {h}^{2}$ both sides of the equation:

${x}^{2} - 12 x + {h}^{2} + {\left(y - 0\right)}^{2} = {h}^{2}$

We can use the pattern, ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ to find the value of h:

${x}^{2} - 2 h x + {h}^{2} = {x}^{2} - 12 x + {h}^{2}$

$- 2 h x = - 12 x$

$h = 6$

Substitute 6 for h into the equation of the circle:

${x}^{2} - 12 x + {6}^{2} + {\left(y - 0\right)}^{2} = {6}^{2}$

We know that the first three terms are a perfect square with h = 6:

${\left(x - 6\right)}^{2} + {\left(y - 0\right)}^{2} = {6}^{2}$

This is a circle with a radius of 6 and a center point $\left(6 , 0\right)$

## How do you evaluate  e^( ( 7 pi)/6 i) - e^( ( 19 pi)/8 i) using trigonometric functions?

Rajat Chatterjee
Featured 6 months ago

$2 \sin \left(\frac{29 \pi}{24}\right) {e}^{\left(\frac{25 \pi}{24}\right) i}$

#### Explanation:

we know euler's theorem states that ${e}^{i \theta} = \cos \theta + i \sin \theta$
so,${e}^{7 \frac{\pi}{6} i} - {e}^{19 \frac{\pi}{8} i}$
=$\cos \left(7 \frac{\pi}{6}\right) + i \sin \left(7 \frac{\pi}{6}\right) - \left(\cos \left(19 \frac{\pi}{8}\right) + i \sin \left(19 \frac{\pi}{8}\right)\right)$
=$\cos \left(7 \frac{\pi}{6}\right) - \cos \left(19 \frac{\pi}{8}\right) + i \left(\sin \left(7 \frac{\pi}{6}\right) - \sin \left(19 \frac{\pi}{8}\right)\right)$
now $\cos c - \cos d = 2 \sin \left(\frac{c + d}{2}\right) \sin \left(\frac{d - c}{2}\right)$
$\sin c - \sin d = 2 \cos \left(\frac{c + d}{2}\right) \sin \left(\frac{c - d}{2}\right)$
so we get $\cos \left(\frac{7 \pi}{6}\right) - \cos \left(\frac{19 \pi}{8}\right)$
=$2 \sin \left(\frac{85 \pi}{24}\right) \sin \left(\frac{29 \pi}{24}\right)$
again similarly we get $i \left(\sin \left(\frac{7 \pi}{6}\right) - \sin \left(\frac{19 \pi}{8}\right)\right)$
=$2 i \cos \left(\frac{85 \pi}{24}\right) \sin \left(\frac{29 \pi}{24}\right)$
so,${e}^{\frac{7 \pi}{6}} - {e}^{\frac{19 \pi}{8}}$=$2 \sin \left(\frac{29 \pi}{24}\right) \left(\sin \left(\frac{85 \pi}{24}\right) - i \cos \left(\frac{85 \pi}{24}\right)\right)$
=$2 \sin \left(\frac{29 \pi}{24}\right) \left[\sin \left(2 \pi + \left(\frac{37 \pi}{24}\right)\right) - i \cos \left(2 \pi + \left(\frac{37 \pi}{24}\right)\right)\right]$
=$2 \sin \left(\frac{29 \pi}{24}\right) \left[\sin \left(\frac{37 \pi}{24}\right) - i \cos \left(\frac{37 \pi}{24}\right)\right]$
=$2 \sin \left(\frac{29 \pi}{24}\right) \left[\cos \left(\frac{25 \pi}{24}\right) + i \sin \left(\frac{25 \pi}{24}\right)\right]$
=$2 \sin \left(\frac{29 \pi}{24}\right) {e}^{\left(\frac{25 \pi}{24}\right) i}$.

## The angle of elevation of a cliff from a fixed point A is theta. After going up a distance of 'k' m towards the top of the cliff, at an angle of phi, it is found that the angle of elevation is alpha. Find height of cliff in terms of k?

dk_ch
Featured 5 months ago

Let CB be the cliff . From point A the angle of elevation of the peak C of the cliff is $\angle C A B = \theta$'. After going up a distance , $A D = k$ $m$ towards the top of the cliff at an angle,$\angle D A E = \phi$,it ls found the angle of elevation $\angle C D F = \alpha$.

DF and DE are perpendiculars drawn from D on CB and AB.

Now $D E = k \sin \phi \mathmr{and} A E = k \cos \phi$

Let
h="CB the height of the cliff" and b=BA

For $\Delta C A B , \text{ } \frac{C B}{B A} = \tan \theta$

$\implies \frac{b}{h} = \cot \theta \implies b = h \cot \theta$

Now $D F = B E = B A - A E = b - k \cos \phi$

$C F = C B - F B = C B - D E = h - k \sin \phi$

For $\Delta C D F , \text{ } \frac{C F}{D F} = \tan \alpha$

$\implies \frac{h - k \sin \phi}{b - k \cos \phi} = \tan \alpha$

$\implies \frac{h - k \sin \phi}{h \cot \theta - k \cos \phi} = \tan \alpha$

$\implies h - h \tan \alpha \cot \theta = k \sin \phi - k \cos \phi \tan \alpha$

$\implies h = \frac{k \left(\sin \phi - \cos \phi \tan \alpha\right)}{1 - \tan \alpha \cot \theta}$

$\implies h = \frac{k \sin \theta \left(\sin \phi \cos \alpha - \cos \phi \sin \alpha\right)}{\sin \theta \cos \alpha - \cos \theta \sin \alpha}$

$\implies h = \frac{k \sin \theta \sin \left(\phi - \alpha\right)}{\sin \left(\theta - \alpha\right)}$

## How do you graph y=5+tan(x+pi)?

Alan P.
Featured 3 months ago

Start from a "basic cycle" for the $\tan$ function to obtain the results below.

#### Explanation:

Starting with the "basic cycle" for $\tan \left(\theta\right)$ i.e. for $\theta \in \left[- \frac{\pi}{2} , + \frac{\pi}{2}\right]$

Then consider what values of $x$ would place $\left(x + \pi\right)$ in this same range, i.e. $\left(x + \pi\right) \in \left[- \frac{\pi}{2} , + \frac{\pi}{2}\right]$
(Yes; I know: because $\tan$ has a cycle length of $\pi$ we could recognize that the cycle will repeat at exactly the same place, but let's do this for the more general case.)
$\left(x + \pi\right) \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] \textcolor{w h i t e}{\text{X"rarrcolor(white)("X}} x \in \left[- \frac{3 \pi}{2} , - \frac{\pi}{2}\right]$
Giving us the "basic cycle" for $\tan \left(x + \pi\right)$:

Adding $5$ to this to get $y = 5 + \tan \left(x + \pi\right)$ simply shifts the points upward $5$ units:

In the image below, I have added the "non-basic cycles" as well as the "basic cycle" used for analysis:

## How do you find the maximum of the graph y=-2cos(x-pi/2)?

reudhreghs
Featured 1 month ago

$\left(\frac{3 \pi}{2} , 2\right)$

#### Explanation:

The maximum and minimum of any $\cos$ graph on its own will be $+ 1$ or $- 1$. It doesn't matter so much what's inside the $\cos$ function, only what's outside.

Multiply these by $- 2$, the outside of the equation, to get maximum/minimum points of $- 2$ and $+ 2$.

The maximum is therefore $+ 2$.

To find at which point this occurs, work backwards.

$+ 2 = - 2 \cos \left(x - \frac{\pi}{2}\right)$

Divide both sides by $- 2$

$- 1 = \cos \left(x - \frac{\pi}{2}\right)$

Do a backwards $\cos$ (${\cos}^{-} 1$) to get rid of the forwards $\cos$

${\cos}^{-} 1 \left(- 1\right) = x - \frac{\pi}{2}$
$\pi = x - \frac{\pi}{2}$
$x = \frac{3 \pi}{2}$

Now we have our $x$ and $y$ values:

$\left(\frac{3 \pi}{2} , 2\right)$

## Trig math problem help?

dk_ch
Featured yesterday

The given function representing the height $h$ in cm of the person above the ground with time $t$ is given by

$h = 15 \cos \left(\frac{2 \pi}{3} t\right) + 65$

(a) h is a cosine function of t, So it will have maximum value when $\cos \left(\frac{2 \pi}{3} t\right) = 1 = \cos 0 \implies t = 0$ s
and the maximum height of the swing becomes

${h}_{\max} = 15 \cos \left(\frac{2 \pi}{3} \times 0\right) + 65 = 80 c m$

(b) it takes $t = 0$sec to reach maximum height after the person starts.

c) The minimum height of the swing will be achieved when $\cos \left(\frac{2 \pi}{3} t\right) = - 1$

Minimum height

${h}_{\min} = 15 \times \left(- 1\right) + 65 = 50 c m$

(d). Again $\cos \left(\frac{2 \pi}{3} t\right) = - 1 = \cos \left(\pi\right) \implies t = \frac{3}{2} = 1.5$s

Here $t = 1.5$s represents the minimum time required to achieve the minimum height after start.

e) For $h = 60$ the equation becomes

$60 = 15 \cos \left(\frac{2 \pi}{3} t\right) + 65$

$\implies 60 - 65 = 15 \cos \left(\frac{2 \pi}{3} t\right)$

$\implies \cos \left(\frac{2 \pi}{3} t\right) = - \frac{1}{3}$

$\implies t = \left[\frac{1}{120} \left(2 n \times 180 \pm {\cos}^{-} 1 \left(- \frac{1}{3}\right)\right)\right] \sec \text{ where } n \in \mathbb{Z}$

$\implies t = \left(3 n \pm 0.912\right)$sec

when $n = 0 , t = 0.912 s$ this is the minimum time to reach at height $h = 60 c m$ after start. just after that height $h = 60 c m$ will be achieved again for $n = 1$

$\implies t = \left(3 \times 1 - 0.912\right) = 2.088$sec

So in $\left(2.088 - 0.912\right) = 1.176 s$ sec (the minimum time difference) within one cycle the swing will be less than 60 cm above the ground.

f) The height of the swing at 10 s can be had by inserting $t = 10$s in the given equation

${h}_{10} = 15 \cos \left(\frac{2 \pi}{3} \times 10\right) + 65 = 57.5$cm

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