Question
Answer
$$\dfrac{2-i}{7}(cos\cdot3t+isin\cdot3t)=\dfrac{-3i^3nst-3ciost+6i^2nst+6cost}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-i}{7}(cos\cdot3t+isin\cdot3t)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-i}{7}(cos\cdot3t+i^2ns\cdot3t) \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{2-i}{7}(3cost+3i^2nst) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-3i^3nst-3ciost+6i^2nst+6cost}{7}\end{aligned} $$ | |
| ① | $$ i s i n = i^{1 + 1} n s = i^2 n s $$ |
| ② | Multiply $ \dfrac{2-i}{7} $ by $ 3cost+3i^2nst $ to get $ \dfrac{-3i^3nst-3ciost+6i^2nst+6cost}{7} $. Step 1: Write $ 3cost+3i^2nst $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{2-i}{7} \cdot 3cost+3i^2nst & \xlongequal{\text{Step 1}} \frac{2-i}{7} \cdot \frac{3cost+3i^2nst}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 2-i \right) \cdot \left( 3cost+3i^2nst \right) }{ 7 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 6cost+6i^2nst-3ciost-3i^3nst }{ 7 } = \frac{-3i^3nst-3ciost+6i^2nst+6cost}{7} \end{aligned} $$ |
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