Question
Answer
$$-2(cos\cdot5p\dfrac{i}{6}+isin\cdot5p\dfrac{i}{6})=\dfrac{10i^3nps+10ciops}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-2(cos\cdot5p\frac{i}{6}+isin\cdot5p\frac{i}{6})& \xlongequal{ }-2(cos\cdot5p\frac{i}{6}+5i^2nps\frac{i}{6}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2(cos\cdot5p+5i^2nps)\frac{i}{6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2 \cdot \frac{5i^3nps+5ciops}{6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{10i^3nps+10ciops}{6}\end{aligned} $$ | |
| ① | Use the distributive property. |
| ② | Multiply $5cops+5i^2nps$ by $ \dfrac{i}{6} $ to get $ \dfrac{5i^3nps+5ciops}{6} $. Step 1: Write $ 5cops+5i^2nps $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 5cops+5i^2nps \cdot \frac{i}{6} & \xlongequal{\text{Step 1}} \frac{5cops+5i^2nps}{\color{red}{1}} \cdot \frac{i}{6} \xlongequal{\text{Step 2}} \frac{ \left( 5cops+5i^2nps \right) \cdot i }{ 1 \cdot 6 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5ciops+5i^3nps }{ 6 } = \frac{5i^3nps+5ciops}{6} \end{aligned} $$ |
| ③ | Multiply $2$ by $ \dfrac{5i^3nps+5ciops}{6} $ to get $ \dfrac{ 10i^3nps+10ciops }{ 6 } $. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2 \cdot \frac{5i^3nps+5ciops}{6} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{5i^3nps+5ciops}{6} \xlongequal{\text{Step 2}} \frac{ 2 \cdot \left( 5i^3nps+5ciops \right) }{ 1 \cdot 6 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 10i^3nps+10ciops }{ 6 } \end{aligned} $$ |
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