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## How do you divide # (1+7i) / (5-3i) # in trigonometric form?

Shwetank Mauria
Featured 5 months ago

$\frac{1 + 7 i}{5 - 3 i} = \frac{5}{\sqrt{17}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{19}{8}\right)$

#### Explanation:

Let us write the two complex numbers in polar coordinates and let them be

${z}_{1} = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and ${z}_{2} = {r}_{2} \left(\cos \beta + i \sin \beta\right)$

Here, if two complex numbers are ${a}_{1} + i {b}_{1}$ and ${a}_{2} + i {b}_{2}$ ${r}_{1} = \sqrt{{a}_{1}^{2} + {b}_{1}^{2}}$, ${r}_{2} = \sqrt{{a}_{2}^{2} + {b}_{2}^{2}}$ and $\alpha = {\tan}^{- 1} \left({b}_{1} / {a}_{1}\right)$, $\beta = {\tan}^{- 1} \left({b}_{2} / {a}_{2}\right)$

$\left\{{r}_{1} / {r}_{2}\right\} \left\{\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta}\right\}$ or

$\left\{{r}_{1} / {r}_{2}\right\} \left\{\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} \times \frac{\cos \beta - i \sin \beta}{\cos \beta - i \sin \beta}\right\}$

$\left({r}_{1} / {r}_{2}\right) \frac{\left(\cos \alpha \cos \beta + \sin \alpha \sin \beta\right) + i \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right)}{\left({\cos}^{2} \beta + {\sin}^{2} \beta\right)}$ or

$\left({r}_{1} / {r}_{2}\right) \cdot \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$ or

${z}_{1} / {z}_{2}$ is given by $\left({r}_{1} / {r}_{2} , \left(\alpha - \beta\right)\right)$

So for division complex number ${z}_{1}$ by ${z}_{2}$ , take new angle as $\left(\alpha - \beta\right)$ and modulus the ratio ${r}_{1} / {r}_{2}$ of the modulus of two numbers.

Here $1 + 7 i$ can be written as ${r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ where ${r}_{1} = \sqrt{{1}^{2} + {7}^{2}} = \sqrt{50}$ and $\alpha = {\tan}^{- 1} 7$

and $5 - 3 i$ can be written as ${r}_{2} \left(\cos \beta + i \sin \beta\right)$ where ${r}_{2} = \sqrt{{5}^{2} + {3}^{2}} = \sqrt{34}$ and $\beta = {\tan}^{- 1} \left(- \frac{3}{5}\right)$

and ${z}_{1} / {z}_{2} = \frac{\sqrt{50}}{\sqrt{34}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = \alpha - \beta$

Hence, $\tan \theta = \tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{7 - \left(- \frac{3}{5}\right)}{1 + 7 \times \left(- \frac{3}{5}\right)} = \frac{\frac{38}{5}}{- \frac{16}{5}} = - \frac{19}{8}$.

Hence, $\frac{1 + 7 i}{5 - 3 i} = \frac{\sqrt{50}}{\sqrt{34}} \left(\cos \theta + i \sin \theta\right)$

= $\frac{5}{\sqrt{17}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{19}{8}\right)$

## What is the definition of the Inverse Sine Function?

Ratnaker Mehta
Featured 5 months ago

The Inverse Sine Function is denoted by, $a r c \sin$ or, ${\sin}^{-} 1$ and

is defined by,

$a r c \sin x = \theta , x \in \left[- 1 , 1\right] \iff \sin \theta = x , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

#### Explanation:

The Inverse Sine Function is denoted by, $a r c \sin$ or, ${\sin}^{-} 1$ and

is defined by,

$a r c \sin x = \theta , x \in \left[- 1 , 1\right] \iff \sin \theta = x , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

## How do you divide #( i-3) / (5i +2)# in trigonometric form?

Alan P.
Featured 3 months ago

$\frac{i - 3}{5 i + 2} \approx 0.587 \cdot \left(\cos \left(1.630\right) + i \cdot \sin \left(- 1.630\right)\right)$

#### Explanation:

$\textcolor{red}{\text{~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~}}$
$\textcolor{w h i t e}{\text{XX }} \textcolor{b l u e}{a + b i \leftrightarrow r \cdot \left[\cos \left(\theta\right) + i \cdot \sin \left(\theta\right)\right]}$
$\textcolor{w h i t e}{\text{XXX")color(blue)("where } r = \sqrt{{a}^{2} + {b}^{2}}}$
#color(white)("XXX")color(blue)("and "theta={("arctan"(b/a),"if "(a,bi) in "Q I or Q IV"),("arctan"(b/a)+pi,"if " (a,bi) in "Q II or Q III):})#

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

$\textcolor{red}{\text{~~~ Trigonometric Division ~~~}}$
#color(white)("XX ")color(blue)((cos(theta)+i * sin(theta))/(cos(phi)+i * sin(phi)))#

#color(white)("XXX")color(blue)(= [color(green)((cos(theta) * cos(phi) + sin(theta) * sin(phi))] +i * [color(magenta)(sin(theta) * cos(phi)-cos(theta) * sin(phi))]#

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

If $A = i - 3 = - 3 + 1 i$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{A} = \sqrt{{\left(- 3\right)}^{2} + {1}^{2}} = \sqrt{10}$
and
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{A} = \arctan \left(- \frac{1}{3}\right) + \pi \approx 2.819842099$

If $B = 5 i + 2 = 2 + 5 i$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{B} = \sqrt{{2}^{2} + {5}^{2}} = \sqrt{29}$
and
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{B} = \arctan \left(\frac{5}{2}\right) \approx 1.19028995$

$\frac{A}{B} = \frac{i - 3}{5 + 2 i}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{10}}{\sqrt{29}} \cdot \left(\frac{\cos \left({\theta}_{A}\right) + i \sin \left({\theta}_{A}\right)}{\cos \left({\theta}_{B}\right) + i \sin \left({\theta}_{B}\right)}\right)$

#color(white)("XXX")=sqrt(10)/sqrt(29) *([color(green)(cos(theta_A)*cos(theta_B)+sin(theta_A) * sin(theta_B))+i[color(magenta)(sin(theta_A) * cos(theta_B) - cos(theta_A) * sin(theta_B))])#

Plugging in the values for ${\theta}_{A}$ and ${\theta}_{B}$ and using a calculator/spreadsheet:
$\textcolor{w h i t e}{\text{XXX}} \approx - 0.034482759 + 0.586206897 i$

If we call this value $C$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{C} = \sqrt{{\left(- 0.03448 \ldots\right)}^{2} + {\left(0.5862 \ldots\right)}^{2}} \approx 0.58722022$
and (since $C$ is in Quadrant 2)
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{C} = \arctan \left(\frac{0.5862 \ldots}{- 0.03448 \ldots}\right) + \pi \approx 1.62955215$

$\frac{A}{B} = C = {r}_{C} \left(\cos \left({\theta}_{C}\right) + i \cdot \sin \left({\theta}_{C}\right)\right)$

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

Geoff K.
Featured 2 months ago

Use the formula for a circle $\left({x}^{2} + {y}^{2} = {r}^{2}\right)$, and substitute $x = r \cos \theta$ and $y = r \sin \theta$.

#### Explanation:

The formula for a circle centred at the origin is

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

That is, the distance from the origin to any point $\left(x , y\right)$ on the circle is the radius $r$ of the circle.

Picture a circle of radius $r$ centred at the origin, and pick a point $\left(x , y\right)$ on the circle:

graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}

If we draw a line from that point to the origin, its length is $r$. We can also draw a triangle for that point as follows:

graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}

Let the angle at the origin be theta ($\theta$).

Now for the trigonometry.

For an angle $\theta$ in a right triangle, the trig function $\sin \theta$ is the ratio $\text{opposite side"/"hypotenuse}$. In our case, the length of the side opposite of $\theta$ is the $y$-coordinate of our point $\left(x , y\right)$, and the hypotenuse is our radius $r$. So:

$\sin \theta = \text{opp"/"hyp" = y/r" "<=>" } y = r \sin \theta$

Similarly, $\cos \theta$ is the ratio of the $x$-coordinate in $\left(x , y\right)$ to the radius $r$:

$\cos \theta = \text{adj"/"hyp"=x/r" "<=>" } x = r \cos \theta$

So we have $x = r \cos \theta$ and $y = r \sin \theta$. Substituting these into the circle formula gives

$\text{ "x^2" "+" "y^2" } = {r}^{2}$
${\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2} = {r}^{2}$
${r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta = {r}^{2}$

The ${r}^{2}$'s all cancel, leaving us with

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

This is often rewritten with the ${\sin}^{2}$ term in front, like this:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

And that's it. That's really all there is to it. Just as the distance between the origin and any point $\left(x , y\right)$ on a circle must be the circle's radius, the sum of the squared values for $\sin \theta$ and $\cos \theta$ must be 1 for any angle $\theta$.

## #Cos2x+3cosx=-2# How do I solve this?

Monzur R.
Featured 1 week ago

${x}_{1} = n \pi , n = 2 k \pm 1 , k \in \mathbb{Z}$

${x}_{2} = \frac{n}{3} \pi , \mod \left(n , 2\right) = 0$

#### Explanation:

$\cos 2 x + 3 \cos x = - 2$

Use the double angle formula for cosine to expand $\cos 2 x$ and rewrite the equation in standard form

$2 {\cos}^{2} x - 1 + 3 \cos x + 2 = 0$

$2 {\cos}^{2} x + 3 \cos x + 1 = 0$

Let $\cos x = y$

$y = \frac{- 3 \pm \sqrt{9 - 4 \left(2\right) \left(1\right)}}{2 \left(2\right)} = \frac{- 3 \pm 1}{4}$

$y = - 1$ or $y = - \frac{1}{2}$

$\cos x = - 1$

$x = n \pi , n = 2 k \pm 1 , k \in \mathbb{Z}$

$\cos x = - \frac{1}{2}$

${x}_{2} = \frac{n}{3} \pi , \mod \left(n , 2\right) = 0$

## Find the value of # sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}.#?

dk_ch
Featured yesterday

# sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

Taking $\left(2 r - 1\right) \frac{\pi}{8} = \theta$

# {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

$= \left\{{\left[\sin \theta\right]}^{4} + {\left[\cos \theta\right]}^{4}\right\} .$

$= \left\{{\left[{\sin}^{2} \theta + {\cos}^{2} \theta\right]}^{2} - 2 {\sin}^{2} \theta {\cos}^{2} \theta\right\}$

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

$= \left\{{1}^{2} - \frac{1}{2} {\sin}^{2} 2 \theta\right\}$

$= \left\{1 - \frac{1}{2} {\sin}^{2} \left(2 r - 1\right) \frac{\pi}{4}\right\}$

# ={1-1/2sin^2 ((rpi)/2-pi/4))}#

$= \left\{1 - \frac{1}{2} {\left(\pm \frac{1}{\sqrt{2}}\right)}^{2}\right\} \textcolor{red}{\text{*}}$

$= \left\{1 - \frac{1}{4}\right\} = \frac{3}{4}$

$\left[\textcolor{red}{\text{*"=>sin((rpi)/2-pi/4)=pmsin(pi/4) or pmcos(pi/4)" for " r in ZZ^"+}}\right]$
So

# sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

$= {\sum}_{r = 1}^{r = 4} \left\{\frac{3}{4}\right\} = 4 \times \frac{3}{4} = 3$

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