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## How do you convert (5pi)/7 from radians to degree?

It is ${900}^{\setminus \circ} / 7$

#### Explanation:

From ${\alpha}^{\setminus \circ} / \alpha = {360}^{\setminus \circ} / \left(2 \cdot \pi\right)$ we get
${\alpha}^{\setminus \circ} = {180}^{\setminus \circ} / \pi \cdot \frac{5}{7} \cdot \pi$
canceling the $\pi$ we get
${\alpha}^{\setminus \circ} = {900}^{\setminus \circ} / 7$

## How do you simplify tan(x+y) to trigonometric functions of x and y?

$\tan \left(x + y\right) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$

#### Explanation:

This can be expanded through the tangent angle addition formula:

$\tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

Thus,

$\tan \left(x + y\right) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$

The tangent addition formula can be found using the sine and cosine angle addition formulas.

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
$\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Since $\tan x = \sin \frac{x}{\cos} x$,

$\tan \left(\alpha + \beta\right) = \sin \frac{\alpha + \beta}{\cos} \left(\alpha + \beta\right) = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}$

This can be written in terms of tangent by dividing both the numerator and denominator by $\cos \alpha \cos \beta$.

$\tan \left(\alpha + \beta\right) = \frac{\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\sin \frac{\alpha}{\cos} \alpha \left(\cos \frac{\beta}{\cos} \beta\right) + \sin \frac{\beta}{\cos} \beta \left(\cos \frac{\alpha}{\cos} \alpha\right)}{\cos \frac{\alpha}{\cos} \alpha \left(\cos \frac{\beta}{\cos} \beta\right) - \sin \frac{\alpha}{\cos} \alpha \left(\sin \frac{\beta}{\cos} \beta\right)}$

Final round of simplification yields:

$\tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

## How do you evaluate  e^( ( 5 pi)/4 i) - e^( ( 3 pi)/8 i) using trigonometric functions?

$2 \cos \left(\frac{13 \pi}{16}\right) \left\{\cos \left(\frac{7 \pi}{16}\right) + i \sin \left(\frac{7 \pi}{16}\right)\right\}$

#### Explanation:

${e}^{\frac{5 \pi}{4} i} - {e}^{\frac{3 \pi}{8} i}$

Euler formula : ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$

${e}^{\frac{5 \pi}{4} i} = \cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)$
${e}^{\frac{3 \pi}{8} i} = \cos \left(\frac{3 \pi}{8}\right) + i \sin \left(\frac{3 \pi}{8}\right)$

${e}^{\frac{5 \pi}{4} i} - {e}^{\frac{3 \pi}{8} i}$
$= \cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right) - \left(\cos \left(\frac{3 \pi}{8}\right) + i \sin \left(\frac{3 \pi}{8}\right)\right)$
$= \cos \left(\frac{5 \pi}{4}\right) - \cos \left(\frac{3 \pi}{8}\right) + i \left(\sin \left(\frac{5 \pi}{4}\right) - \sin \left(\frac{3 \pi}{8}\right)\right)$

color(blue)( cos(C) -cos(D) = 2cos((C+D)/2)cos((C-D)/2)

color(blue)(sin(C) - sin(D) = 2cos((C+D)/2)sin((C-D)/2)

$\cos \left(\frac{5 \pi}{4}\right) - \cos \left(\frac{3 \pi}{8}\right) = 2 \cos \left(\frac{\frac{5 \pi}{4} + \frac{3 \pi}{8}}{2}\right) \cos \left(\frac{\frac{5 \pi}{4} - \frac{3 \pi}{8}}{2}\right)$
cos((5pi)/4)-cos((3pi)/8) = 2cos(1/2((10pi)/8+(3pi)/8))cos((1/2((10pi)/8-(3pi)/8))
cos((5pi)/4)-cos((3pi)/8) = 2cos(1/2((13pi)/8)cos((1/2((7pi)/8))
color(green)(cos((5pi)/4)-cos((3pi)/8) = 2cos((13pi)/16)cos((7pi)/16)

sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((5pi)/4+(3pi)/8)sin(1/2((5pi)/4-(3pi)/8))
sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((10pi)/8+(3pi)/8)sin(1/2((10pi)/8-(3pi)/8))
sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((13pi)/8)sin(1/2((7pi)/8))
color(green)(sin((5pi)/4)-sin((3pi)/8) = 2cos((13pi)/16)sin((7pi)/16)

${e}^{\frac{5 \pi}{4} i} - {e}^{\frac{3 \pi}{8} i}$
$= \cos \left(\frac{5 \pi}{4}\right) - \cos \left(\frac{3 \pi}{8}\right) + i \left(\sin \left(\frac{5 \pi}{4}\right) - \sin \left(\frac{3 \pi}{8}\right)\right)$
$= 2 \cos \left(\frac{13 \pi}{16}\right) \cos \left(\frac{7 \pi}{16}\right) + i \left(2 \cos \left(\frac{13 \pi}{16}\right) \sin \left(\frac{7 \pi}{16}\right)\right)$
$= 2 \cos \left(\frac{13 \pi}{16}\right) \left\{\cos \left(\frac{7 \pi}{16}\right) + i \sin \left(\frac{7 \pi}{16}\right)\right\}$

## How do you express tan theta - cot theta +sintheta  in terms of cos theta ?

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

#### Explanation:

$\tan \left(\theta\right) - \cot \left(\theta\right) + \sin \left(\theta\right)$
We have to write in terms of $\cos \left(\theta\right)$

$\textcolor{b l u e}{\text{Let us start by using the identity}}$
$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$ and $\cot \left(\theta\right) = \cos \frac{\theta}{\sin} \left(\theta\right)$

We get

$\tan \left(\theta\right) - \cot \left(\theta\right) + \sin \left(\theta\right)$
$= \sin \frac{\theta}{\cos} \left(\theta\right) - \cos \frac{\theta}{\sin} \left(\theta\right) + \sin \left(\theta\right)$

$\textcolor{b l u e}{\text{In order to simplify we need to use Least Common Denominator for all the fractions}}$

$= \frac{\sin \left(\theta\right) \sin \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)} - \frac{\cos \left(\theta\right) \cos \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)} + \frac{\sin \left(\theta\right) \cos \left(\theta\right) \sin \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)}$

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

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

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

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

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

## How do you divide  (1-2i) / (6-8i)  in trigonometric form?

First of all we have to convert these two numbers into trigonometric forms.
If $\left(a + i b\right)$ is a complex number, $u$ is its magnitude and $\alpha$ is its angle then $\left(a + i b\right)$ in trigonometric form is written as $u \left(\cos \alpha + i \sin \alpha\right)$.
Magnitude of a complex number $\left(a + i b\right)$ is given by$\sqrt{{a}^{2} + {b}^{2}}$ and its angle is given by ${\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let $r$ be the magnitude of $\left(1 - 2 i\right)$ and $\theta$ be its angle.
Magnitude of $\left(1 - 2 i\right) = \sqrt{{1}^{2} + {\left(- 2\right)}^{2}} = \sqrt{1 + 4} = \sqrt{5} = r$
Angle of $\left(1 - 2 i\right) = T a {n}^{-} 1 \left(\frac{- 2}{1}\right) = {\tan}^{-} 1 \left(- 2\right) = \theta$

$\implies \left(1 - 2 i\right) = r \left(C o s \theta + i \sin \theta\right)$

Let $s$ be the magnitude of $\left(6 - 8 i\right)$ and $\phi$ be its angle.
Magnitude of $\left(6 - 8 i\right) = \sqrt{{6}^{2} + {\left(- 8\right)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 = s$
Angle of $\left(6 - 8 i\right) = T a {n}^{-} 1 \left(\frac{- 8}{6}\right) = T a {n}^{-} 1 \left(- \frac{4}{3}\right) = \phi$

$\implies \left(6 - 8 i\right) = s \left(C o s \phi + i \sin \phi\right)$

Now,
$\frac{1 - 2 i}{6 - 8 i}$

$= \frac{r \left(C o s \theta + i \sin \theta\right)}{s \left(C o s \phi + i \sin \phi\right)}$

=r/s*(Costheta+isintheta)/(Cosphi+isinphi)*(Cosphi-isinphi)/(Cosphi-isinphi

$= \frac{r}{s} \cdot \frac{\cos \theta \cos \phi + i \sin \theta \cos \phi - i \cos \theta \sin \phi - {i}^{2} \sin \theta \sin \phi}{{\cos}^{2} \phi - {i}^{2} {\sin}^{2} \phi}$

$= \frac{r}{s} \cdot \frac{\left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + i \left(\sin \theta \cos \phi - \cos \theta \sin \phi\right)}{{\cos}^{2} \phi + {\sin}^{2} \phi}$

$= \frac{r}{s} \cdot \frac{\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)}{1}$

$= \frac{r}{s} \left(\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)\right)$

Here we have every thing present but if here directly substitute the values the word would be messy for find $\theta - \phi$ so let's first find out $\theta - \phi$.

$\theta - \phi = {\tan}^{-} 1 \left(- 2\right) - {\tan}^{-} 1 \left(- \frac{4}{3}\right)$
We know that:
${\tan}^{-} 1 \left(a\right) - {\tan}^{-} 1 \left(b\right) = {\tan}^{-} 1 \left(\frac{a - b}{1 + a b}\right)$

$\implies {\tan}^{-} 1 \left(- 2\right) - {\tan}^{-} 1 \left(- \frac{4}{3}\right) = {\tan}^{-} 1 \left(\frac{\left(- 2\right) - \left(- \frac{4}{3}\right)}{1 + \left(- 2\right) \left(- \frac{4}{3}\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{- 6 + 4}{3 + 8}\right) = {\tan}^{-} 1 \left(- \frac{2}{11}\right)$

$\implies \theta - \phi = {\tan}^{-} 1 \left(- \frac{2}{11}\right)$

$\frac{r}{s} \left(\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)\right)$

$= \frac{\sqrt{5}}{10} \left(\cos \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right)\right)$

$= \sqrt{\frac{5}{100}} \left(\cos \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right)\right)$

$= \sqrt{\frac{1}{20}} \left(\cos \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right)\right)$

$= \frac{1}{2 \sqrt{5}} \left(\cos \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right)\right)$

You can also do it by another method.
By firstly dividing the complex numbers and then changing it to trigonometric form, which is much easier than this.

First of all let's simplify the given number
$\frac{1 - 2 i}{6 - 8 i}$.

Multiply and divide by the conjugate of the complex number present in the denominator i.e $6 + 8 i$.

$\frac{1 - 2 i}{6 - 8 i} = \frac{\left(1 - 2 i\right) \left(6 + 8 i\right)}{\left(6 - 8 i\right) \left(6 + 8 i\right)} = \frac{6 + 8 i - 12 i - 16 {i}^{2}}{{6}^{2} - {\left(8 i\right)}^{2}}$
$= \frac{6 - 4 i + 16}{36 - \left(- 64\right)} = \frac{22 - 4 i}{36 + 64} = \frac{22 - 4 i}{100} = \frac{22}{100} - \frac{4 i}{100} = \frac{11}{50} - \frac{i}{25}$

$\frac{1 - 2 i}{6 - 8 i} = \frac{11}{50} - \frac{i}{25}$

Let $t$ be the magnitude of $\left(\frac{11}{50} - \frac{i}{25}\right)$ and $\beta$ be its angle.

Magnitude of $\left(\frac{11}{50} - \frac{i}{25}\right) = \sqrt{{\left(\frac{11}{50}\right)}^{2} + {\left(- \frac{1}{25}\right)}^{2}} = \sqrt{\frac{121}{2500} + \frac{1}{625}} = \sqrt{\frac{121 + 4}{2500}} = \sqrt{\frac{125}{2500}} = \sqrt{\frac{1}{20}} = \frac{1}{2 \sqrt{5}} = t$
Angle of $\left(\frac{11}{50} - \frac{i}{25}\right) = T a {n}^{-} 1 \left(\frac{- \frac{1}{25}}{\frac{11}{50}}\right) = {\tan}^{-} 1 \left(- \frac{2}{11}\right) = \beta$

$\implies \left(\frac{11}{50} - \frac{i}{25}\right) = t \left(C o s \beta + i \sin \beta\right)$

$\implies \left(\frac{11}{50} - \frac{i}{25}\right) = \frac{1}{2 \sqrt{5}} \left(C o s \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right) + i \sin \left({\tan}^{-} 1 \left(- \frac{2}{11}\right)\right)\right)$.

## How do you divide ( i+8) / (3i -1 ) in trigonometric form?

$\frac{i + 8}{3 i - 1}$

$= \frac{8 + i}{- 1 + 3 i}$

First of all we have to convert these two numbers into trigonometric forms.
If $\left(a + i b\right)$ is a complex number, $u$ is its magnitude and $\alpha$ is its angle then $\left(a + i b\right)$ in trigonometric form is written as $u \left(\cos \alpha + i \sin \alpha\right)$.
Magnitude of a complex number $\left(a + i b\right)$ is given by$\sqrt{{a}^{2} + {b}^{2}}$ and its angle is given by ${\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let $r$ be the magnitude of $\left(8 + i\right)$ and $\theta$ be its angle.
Magnitude of $\left(8 + i\right) = \sqrt{{8}^{2} + {1}^{2}} = \sqrt{64 + 1} = \sqrt{65} = r$
Angle of $\left(8 + i\right) = T a {n}^{-} 1 \left(\frac{1}{8}\right) = \theta$

$\implies \left(8 + i\right) = r \left(C o s \theta + i \sin \theta\right)$

Let $s$ be the magnitude of $\left(- 1 + 3 i\right)$ and $\phi$ be its angle.
Magnitude of $\left(- 1 + 3 i\right) = \sqrt{{\left(- 1\right)}^{2} + {3}^{2}} = \sqrt{1 + 9} = \sqrt{10} = s$
Angle of $\left(- 1 + 3 i\right) = T a {n}^{-} 1 \left(\frac{3}{-} 1\right) = T a {n}^{-} 1 \left(- 3\right) = \phi$

$\implies \left(- 1 + 3 i\right) = s \left(C o s \phi + i \sin \phi\right)$

Now,
$\frac{8 + i}{- 1 + 3 i}$

$= \frac{r \left(C o s \theta + i \sin \theta\right)}{s \left(C o s \phi + i \sin \phi\right)}$

=r/s*(Costheta+isintheta)/(Cosphi+isinphi)*(Cosphi-isinphi)/(Cosphi-isinphi

$= \frac{r}{s} \cdot \frac{\cos \theta \cos \phi + i \sin \theta \cos \phi - i \cos \theta \sin \phi - {i}^{2} \sin \theta \sin \phi}{{\cos}^{2} \phi - {i}^{2} {\sin}^{2} \phi}$

$= \frac{r}{s} \cdot \frac{\left(\cos \theta \cos \phi + \sin \theta \sin \phi\right) + i \left(\sin \theta \cos \phi - \cos \theta \sin \phi\right)}{{\cos}^{2} \phi + {\sin}^{2} \phi}$

$= \frac{r}{s} \cdot \frac{\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)}{1}$

$= \frac{r}{s} \left(\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)\right)$

Here we have every thing present but if here directly substitute the values the word would be messy for find $\theta - \phi$ so let's first find out $\theta - \phi$.

$\theta - \phi = {\tan}^{-} 1 \left(\frac{1}{8}\right) - {\tan}^{-} 1 \left(- 3\right)$
We know that:
${\tan}^{-} 1 \left(a\right) - {\tan}^{-} 1 \left(b\right) = {\tan}^{-} 1 \left(\frac{a - b}{1 + a b}\right)$

$\implies {\tan}^{-} 1 \left(\frac{1}{8}\right) - {\tan}^{-} 1 \left(- 3\right) = {\tan}^{-} 1 \left(\frac{\left(\frac{1}{8}\right) - \left(- 3\right)}{1 + \left(\frac{1}{8}\right) \left(- 3\right)}\right)$

$= {\tan}^{-} 1 \left(\frac{1 + 24}{8 - 3}\right) = {\tan}^{-} 1 \left(\frac{25}{5}\right) = {\tan}^{-} 1 \left(5\right)$

$\implies \theta - \phi = {\tan}^{-} 1 \left(5\right)$

$\frac{r}{s} \left(\cos \left(\theta - \phi\right) + i \sin \left(\theta - \phi\right)\right)$

$= \frac{\sqrt{65}}{\sqrt{10}} \left(\cos \left({\tan}^{-} 1 \left(5\right)\right) + i \sin \left({\tan}^{-} 1 \left(5\right)\right)\right)$

$= \sqrt{\frac{65}{10}} \left(\cos \left({\tan}^{-} 1 \left(5\right)\right) + i \sin \left({\tan}^{-} 1 \left(5\right)\right)\right)$

$= \sqrt{\frac{13}{2}} \left(\cos \left({\tan}^{-} 1 \left(5\right)\right) + i \sin \left({\tan}^{-} 1 \left(5\right)\right)\right)$

You can also do it by another method.
By firstly dividing the complex numbers and then changing it to trigonometric form, which is much easier than this.

First of all let's simplify the given number
$\frac{i + 8}{3 i - 1}$

$= \frac{8 + i}{- 1 + 3 i}$

Multiply and divide by the conjugate of the complex number present in the denominator i.e $- 1 - 3 i$.

$\frac{8 + i}{- 1 + 3 i} = \frac{\left(8 + i\right) \left(- 1 - 3 i\right)}{\left(- 1 + 3 i\right) \left(- 1 - 3 i\right)} = \frac{- 8 - 24 i - i - 3 {i}^{2}}{{\left(- 1\right)}^{2} - {\left(3 i\right)}^{2}}$
$= \frac{- 8 - 25 i + 3}{1 - \left(- 9\right)} = \frac{- 5 - 25 i}{1 + 9} = \frac{- 5 - 25 i}{10} = - \frac{5}{10} - \frac{25 i}{10} = - \frac{1}{2} - \frac{5 i}{2}$

$\frac{8 + i}{- 1 + 3 i} = - \frac{1}{2} - \frac{5 i}{2}$

Let $t$ be the magnitude of $\left(\frac{1}{10} - \frac{5 i}{2}\right)$ and $\beta$ be its angle.

Magnitude of $\left(- \frac{1}{2} - \frac{5 i}{2}\right) = \sqrt{{\left(- \frac{1}{2}\right)}^{2} + {\left(- \frac{5}{2}\right)}^{2}} = \sqrt{\frac{1}{4} + \frac{25}{4}} = \sqrt{\frac{26}{4}} = \sqrt{\frac{13}{2}} = t$
Angle of $\left(- \frac{1}{2} - \frac{5 i}{2}\right) = T a {n}^{-} 1 \left(\frac{- \frac{5}{2}}{- \frac{1}{2}}\right) = {\tan}^{-} 1 \left(5\right) = \beta$

$\implies \left(- \frac{1}{2} - \frac{5 i}{2}\right) = t \left(C o s \beta + i \sin \beta\right)$

$\implies \left(- \frac{1}{2} - \frac{5 i}{2}\right) = \sqrt{\frac{13}{2}} \left(C o s \left({\tan}^{-} 1 \left(5\right)\right) + i \sin \left({\tan}^{-} 1 \left(5\right)\right)\right)$.

## A triangle has sides A,B, and C. If the angle between sides A and B is (5pi)/8, the angle between sides B and C is pi/4, and the length of B is 19, what is the area of the triangle?

Area $= \frac{1}{2} \cdot 19 \cdot 32.44 \approx 308.13$

#### Explanation: The area of a triangle $= \frac{1}{b} \cdot h$ where $b =$base and $h =$height

In this case $\tan \left(\frac{5 \pi}{8}\right) = \frac{h}{x}$ and $\tan \left(\frac{\pi}{4}\right) = \frac{h}{y}$ where

$x + y = B = 19$
$y = 19 - x$

So $x \tan \left(\frac{5 \pi}{8}\right) = y \tan \left(\frac{\pi}{4}\right)$
$x \tan \left(\frac{5 \pi}{8}\right) = \left(19 - x\right) \tan \left(\frac{\pi}{4}\right)$
$x \left(\tan \left(\frac{5 \pi}{8}\right) + \tan \left(\frac{\pi}{4}\right)\right) = 19 \tan \left(\frac{\pi}{4}\right)$

$\therefore x = 19 \tan \frac{\frac{\pi}{4}}{\tan \left(\frac{5 \pi}{8}\right) + \tan \left(\frac{\pi}{4}\right)}$

$h = \left(19 \tan \frac{\frac{\pi}{4}}{\tan \left(\frac{5 \pi}{8}\right) + \tan \left(\frac{\pi}{4}\right)}\right) \tan \left(\frac{5 \pi}{8}\right)$
$\approx \frac{19 \cdot 1 \cdot 2.414}{2.414 + 1}$
$\approx 32.44$

Area $= \frac{1}{2} \cdot 19 \cdot 32.44 \approx 308.13$

## How do you express cos theta - cos^2 theta + cot^2 theta  in terms of sin theta ?

$\frac{2 {\sin}^{4} \theta - 4 {\sin}^{2} \theta + \sin \theta \sin 2 \theta + 2}{2 {\sin}^{2} \theta}$

#### Explanation:

Write in terms of $\sin \theta$ and $\cos \theta$.

$= \cos \theta - {\cos}^{2} \theta + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$

Find a common denominator.

$= \frac{\cos \theta {\sin}^{2} \theta}{\sin} ^ 2 \theta - \frac{{\cos}^{2} \theta {\sin}^{2} \theta}{\sin} ^ 2 \theta + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$

Combine.

$= \frac{\cos \theta {\sin}^{2} \theta - {\cos}^{2} \theta {\sin}^{2} \theta + {\cos}^{2} \theta}{\sin} ^ 2 \theta$

The following simplification may seem unecessary, but is actually relevant. Its purpose will become clear in the following step.

$= \frac{\sin \theta \left(\textcolor{b l u e}{\cos \theta \sin \theta}\right) - \textcolor{g r e e n}{{\cos}^{2} \theta} {\sin}^{2} \theta + \textcolor{g r e e n}{{\cos}^{2} \theta}}{\sin} ^ 2 \theta$

Use the following identities:

• color(green)(cos^2theta=1-sin^2theta
• 2costhetasintheta=sin2theta=>color(blue)(costhetasintheta=(sin2theta)/2

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

$= \frac{\frac{\sin \theta \sin 2 \theta}{2} - {\sin}^{2} \theta + {\sin}^{4} \theta + 1 - {\sin}^{2} \theta}{\sin} ^ 2 \theta$

$= \frac{\frac{\sin \theta \sin 2 \theta}{2} - 2 {\sin}^{2} \theta + {\sin}^{4} \theta + 1}{\sin} ^ 2 \theta$

$= \frac{2 {\sin}^{4} \theta - 4 {\sin}^{2} \theta + \sin \theta \sin 2 \theta + 2}{2 {\sin}^{2} \theta}$

## A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/12 and the angle between sides B and C is pi/12. If side B has a length of 3, what is the area of the triangle?

$\textcolor{g r e e n}{{\text{area " = 1.125 " units"^2 -> 1 1/8 " units}}^{2}}$

#### Explanation:

$\textcolor{b l u e}{\text{Assumption: }}$

As $\pi$ is used in the angular measure it is assumed that the unit is radians. (Not stated)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Method Plane}}$

Determine $\angle c b a$

Using Sine Rule and $\angle c b a$ determine length of side A
Determine h using $h = A \sin \left(\frac{5 \pi}{12}\right)$
Determine area $h \times \frac{B}{2}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To determine} \angle c b a}$

Sum internal angles of a triangle is ${180}^{0} = \pi \text{ radians}$

$\implies \angle c b a = \pi - \frac{5 \pi}{12} - \frac{\pi}{12}$
$\textcolor{b r o w n}{\angle c b a = \frac{\pi}{2} \to {90}^{o}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine length of A}}$

Using $\frac{B}{\sin \left(b\right)} = \frac{A}{\sin} \left(a\right)$

$\implies \frac{3}{\sin \left(\frac{\pi}{2}\right)} = \frac{A}{\sin} \left(\frac{\pi}{12}\right)$

$\implies A = \frac{3 \times \sin \left(\frac{\pi}{12}\right)}{\sin \left(\frac{\pi}{2}\right)}$

But $\sin \left(\frac{\pi}{2}\right) = 1$

$\textcolor{b l u e}{\implies A = 3 \times \sin \left(\frac{\pi}{12}\right)}$

$\textcolor{b r o w n}{\text{This is an exact value so keep it in this form for now to reduce error}}$ $\textcolor{b r o w n}{\text{on final calculation.}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To determine h}}$

$h = A \sin \left(\frac{5 \pi}{12}\right)$

$\implies h = 3 \times \sin \left(\frac{\pi}{12}\right) \times \sin \left(\frac{5 \pi}{12}\right)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To determine area}}$

$\text{area } = \frac{B}{2} \times h$

$\text{area } = \frac{3}{2} \times 3 \times \sin \left(\frac{\pi}{12}\right) \times \sin \left(\frac{5 \pi}{12}\right)$

but $\sin \left(\frac{\pi}{12}\right) \times \sin \left(\frac{5 \pi}{12}\right) = \frac{1}{4}$

$\text{area } = \frac{3}{2} \times 3 \times \frac{1}{4}$

$\textcolor{g r e e n}{{\text{area " = 1.125 " units"^2 -> 1 1/8 " units}}^{2}}$

## What are the components of the vector between the origin and the polar coordinate (1, (5pi)/4)?

The $x$ component is: $\cos \left(\frac{5 \pi}{4}\right)$
The $y$ component is: $\sin \left(\frac{5 \pi}{4}\right)$

#### Explanation:

Remembering our trigonometry, the vertical component of a vector is given by
$r \cdot \sin \left(\theta\right)$ where $r$ is the length of the line,
and the horizontal component by
$r \cdot \cos \left(\theta\right)$ in the polar coordinate $\left(1 , \frac{5 \pi}{4}\right)$, $r$ is 1, and the angle $\theta = \frac{5 \pi}{4}$.

Hence:
The $x$ component is: $\cos \left(\frac{5 \pi}{4}\right)$
The $y$ component is: $\sin \left(\frac{5 \pi}{4}\right)$

In this case, $\frac{5 \pi}{4}$ is midway in the lower left quadrant, or ${135}^{\circ}$, so both are equal to $- \frac{1}{\sqrt{2}}$