# How do you write the trigonometric form into a complex number in standard form 1/4(cos225+isin225)?

Aug 17, 2016

$- \frac{\sqrt{2}}{8} - \frac{\sqrt{2}}{8} i$

#### Explanation:

Firstly, consider the trig part inside the bracket.

Now ${225}^{\circ}$ is an angle in the 3rd quadrant where both the sin and cos ratios are $\textcolor{b l u e}{\text{negative}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\cos {225}^{\circ} = - \cos {\left(225 - 180\right)}^{\circ} = - \cos {45}^{\circ}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin {225}^{\circ} = - \sin {\left(225 - 180\right)}^{\circ} = - \sin {45}^{\circ}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

also $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin {45}^{\circ} = \cos {45}^{\circ} = \frac{1}{\sqrt{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow \cos {225}^{\circ} + i \sin {225}^{\circ} = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i$

so now back to the original expression.

$\frac{1}{4} \left(\cos {225}^{\circ} + i \sin {225}^{\circ}\right) = \frac{1}{4} \left(- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i\right)$

distributing gives.

$- \frac{1}{4 \sqrt{2}} - \frac{1}{4 \sqrt{2}} i = - \frac{\sqrt{2}}{8} - \frac{\sqrt{2}}{8} i \text{ in standard form}$