Answer
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}ce^{-x/5}(cos\cdot2 \cdot \frac{x}{5}+isin\cdot2\frac{x}{5}+ge^{-x/5}(cos\cdot2\frac{x}{5}-isin\cdot2\frac{x}{5}))& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}ce^{-x/5}(cos\cdot2 \cdot \frac{x}{5}+i^2ns\cdot2\frac{x}{5}+ge^{-x/5}(cos\cdot2\frac{x}{5}-i^2ns\cdot2\frac{x}{5})) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}ce^{-x/5}((cos+i^2ns)\cdot2 \cdot \frac{x}{5}+ge^{-x/5}(cos-i^2ns)\cdot2\frac{x}{5}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}ce^{-x/5}((cos+i^2ns)\frac{2x}{5}+ge^{-x/5}(cos-i^2ns)\frac{2x}{5}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}ce^{-x/5}(\frac{2i^2nsx+2cosx}{5}+ge^{-x/5}\frac{-2i^2nsx+2cosx}{5})\end{aligned} $$ | |
| ① | $$ i s i n = i^{1 + 1} n s = i^2 n s $$$$ i s i n = i^{1 + 1} n s = i^2 n s $$ |
| ② | Use the distributive property. |
| ③ | Use the distributive property. |
| ④ | Multiply $2$ by $ \dfrac{x}{5} $ to get $ \dfrac{ 2x }{ 5 } $. 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. $$ \begin{aligned} 2 \cdot \frac{x}{5} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{x}{5} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x }{ 1 \cdot 5 } \xlongequal{\text{Step 3}} \frac{ 2x }{ 5 } \end{aligned} $$ |
| ⑤ | Multiply $2$ by $ \dfrac{x}{5} $ to get $ \dfrac{ 2x }{ 5 } $. 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. $$ \begin{aligned} 2 \cdot \frac{x}{5} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{x}{5} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x }{ 1 \cdot 5 } \xlongequal{\text{Step 3}} \frac{ 2x }{ 5 } \end{aligned} $$ |
| ⑥ | Multiply $cos+i^2ns$ by $ \dfrac{2x}{5} $ to get $ \dfrac{2i^2nsx+2cosx}{5} $. Step 1: Write $ cos+i^2ns $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} cos+i^2ns \cdot \frac{2x}{5} & \xlongequal{\text{Step 1}} \frac{cos+i^2ns}{\color{red}{1}} \cdot \frac{2x}{5} \xlongequal{\text{Step 2}} \frac{ \left( cos+i^2ns \right) \cdot 2x }{ 1 \cdot 5 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2cosx+2i^2nsx }{ 5 } = \frac{2i^2nsx+2cosx}{5} \end{aligned} $$ |
| ⑦ | Multiply $cos-i^2ns$ by $ \dfrac{2x}{5} $ to get $ \dfrac{-2i^2nsx+2cosx}{5} $. Step 1: Write $ cos-i^2ns $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} cos-i^2ns \cdot \frac{2x}{5} & \xlongequal{\text{Step 1}} \frac{cos-i^2ns}{\color{red}{1}} \cdot \frac{2x}{5} \xlongequal{\text{Step 2}} \frac{ \left( cos-i^2ns \right) \cdot 2x }{ 1 \cdot 5 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2cosx-2i^2nsx }{ 5 } = \frac{-2i^2nsx+2cosx}{5} \end{aligned} $$ |