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## What is the angle between <1 , 5 , 9 >  and  < -2 , 3 , 2 > ?

Douglas K.
Featured 3 months ago

$\theta \approx 0.76 r a \mathrm{di} a n s$

#### Explanation:

Let $\overline{A} = < 1 , 5 , 9 >$
Let $\overline{B} = < - 2 , 3 , 2 >$

The $\overline{A} \cdot \overline{B}$ is:

$\overline{A} \cdot \overline{B} = \left(1\right) \left(- 2\right) + \left(5\right) \left(3\right) + \left(9\right) \left(2\right) = 31$

$| \overline{A} | = \sqrt{{1}^{2} + {5}^{2} + {9}^{2}} = \sqrt{107}$

$| \overline{B} | = \sqrt{{\left(- 2\right)}^{2} + {3}^{2} + {2}^{2}} = \sqrt{17}$

Use $\overline{A} \cdot \overline{B} = | \overline{A} | | \overline{B} | \cos \left(\theta\right)$

Solve for $\theta$

$\theta = {\cos}^{-} 1 \left(\frac{\overline{A} \cdot \overline{B}}{| \overline{A} | | \overline{B} |}\right)$

$\theta = {\cos}^{-} 1 \left(\frac{31}{\sqrt{107} \sqrt{17}}\right)$

$\theta \approx 0.76 r a \mathrm{di} a n s$

## How do you convert 3 sqrt 2 - 3 sqrt2i to polar form?

Shwetank Mauria
Featured 3 months ago

In polar form $3 \sqrt{2} - 3 \sqrt{2} i$ is $6 \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

#### Explanation:

A complex number $a + b i$ can be written in polar form as $r \left(\cos \theta + i \sin \theta\right)$ i.e. $a = r \cos \theta$ and $b = r \sin \theta$.

Squaring and adding the last two, we get ${a}^{2} + {b}^{2} = {r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta$

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

and $\frac{b}{a} = \frac{r \sin \theta}{r \cos \theta} = \tan \theta$

Here in the complex number $3 \sqrt{2} - 3 \sqrt{2} i$, we have $a = 3 \sqrt{2}$ and $b = - 3 \sqrt{2}$ and hence $r = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{\left(3 \sqrt{2}\right)}^{2} + {\left(- 3 \sqrt{2}\right)}^{2}}$

= $\sqrt{18 + 18} = \sqrt{36} = 6$ and

$\cos \theta = \frac{3 \sqrt{2}}{6} = \frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{- 3 \sqrt{2}}{6} = - \frac{1}{\sqrt{2}}$

i.e. $\theta = \left(- \frac{\pi}{4}\right)$

Hence in polar form $3 \sqrt{2} - 3 \sqrt{2} i$ is $6 \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right)$

## HS trigonometry question? arc cot(x)+2arc sin (sqrt(3)/2)= pi The rad is only over three. Can someone help me?

sente
Featured 3 months ago

$x = \frac{\sqrt{3}}{3}$

#### Explanation:

$\text{arccot} \left(x\right) + 2 \arcsin \left(\frac{\sqrt{3}}{2}\right) = \pi$

The inverse sine function $\arcsin \left(x\right)$ is defined as the unique value in the interval $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ such that $\sin \left(\arcsin \left(x\right)\right) = x$. On that interval, we have $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ as a well known angle. Thus $\arcsin \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$

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

$\implies \text{arccot} \left(x\right) = \frac{\pi}{3}$

$\implies \cot \left(\text{arccot} \left(x\right)\right) = \cot \left(\frac{\pi}{3}\right)$

$\implies x = \cot \left(\frac{\pi}{3}\right)$

$= \cos \frac{\frac{\pi}{3}}{\sin} \left(\frac{\pi}{3}\right)$

$= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$= \frac{1}{\sqrt{3}}$

$= \frac{\sqrt{3}}{3}$

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

Douglas K.
Featured 3 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 $\left(6 , 0\right)$

## What is the exact value of cos(tan^-1(2)+tan^-1(3)) ?

mason m
Featured 2 weeks ago

$\cos \left({\tan}^{-} 1 \left(2\right) + {\tan}^{-} 1 \left(3\right)\right) = - \frac{1}{\sqrt{2}}$

#### Explanation:

First use the cosine angle-addition formula:

$\cos \left(a + b\right) = \cos \left(a\right) \cos \left(b\right) - \sin \left(a\right) \sin \left(b\right)$

Then the original expression equals:

$= \cos \left({\tan}^{-} 1 \left(2\right)\right) \cos \left({\tan}^{-} 1 \left(3\right)\right) - \sin \left({\tan}^{-} 1 \left(2\right)\right) \sin \left({\tan}^{-} 1 \left(3\right)\right)$

Note that if $\theta = {\tan}^{-} 1 \left(2\right)$, then $\tan \left(\theta\right)$. That is, a right triangle with angle $\theta$ has $\tan \left(\theta\right) = 2$, which is the triangle with the leg opposite $\theta$ being $2$ and the leg adjacent to $\theta$ being $1$. The Pythagorean theorem tells us that the hypotenuse of this triangle is $\sqrt{5}$.

So, when $\theta = {\tan}^{-} 1 \left(2\right)$, we see that:

$\cos \left({\tan}^{-} 1 \left(2\right)\right) = \cos \left(\theta\right) = \text{adjacent"/"hypotenuse} = \frac{1}{\sqrt{5}}$

$\sin \left({\tan}^{-} 1 \left(2\right)\right) = \sin \left(\theta\right) = \text{opposite"/"hypotenuse} = \frac{2}{\sqrt{5}}$

Now let angle $\phi$ be defined by $\phi = {\tan}^{-} 1 \left(3\right)$, such that $\tan \left(\phi\right) = 3$. This is the right triangle with:

$\left\{\begin{matrix}\text{opposite"=3 \\ "adjacent"=1 \\ "hypotenuse} = \sqrt{10}\end{matrix}\right.$

Then:

$\cos \left({\tan}^{-} 1 \left(3\right)\right) = \cos \left(\phi\right) = \text{adjacent"/"hypotenuse} = \frac{1}{\sqrt{10}}$

$\sin \left({\tan}^{-} 1 \left(3\right)\right) = \sin \left(\phi\right) = \text{opposite"/"hypotenuse} = \frac{3}{\sqrt{10}}$

Plugging these into the expression from earlier we get:

$= \frac{1}{\sqrt{5}} \left(\frac{1}{\sqrt{10}}\right) - \frac{2}{\sqrt{5}} \left(\frac{3}{\sqrt{10}}\right)$

Note that $\sqrt{5} \left(\sqrt{10}\right) = \sqrt{50} = 5 \sqrt{2}$:

$= \frac{1 - 6}{5 \sqrt{2}} = - \frac{1}{\sqrt{2}}$

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