Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\frac{1}{2}+sqrt\cdot\frac{5}{2}(cos\cdot1.893+isin\cdot1.893)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-\frac{1}{2}+sqrt\cdot\frac{5}{2}(cos\cdot1.893+i^2ns\cdot1.893) \xlongequal{ } \\[1 em] & \xlongequal{ }-\frac{1}{2}+sqrt\cdot\frac{5}{2}(1cos+i^2ns) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-\frac{1}{2}+\frac{5qrst}{2}(1cos+i^2ns) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-\frac{1}{2}+\frac{5i^2nqrs^2t+5coqrs^2t}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{5i^2nqrs^2t+5coqrs^2t-1}{2}\end{aligned} $$ | |
| ① | $$ i s i n = i^{1 + 1} n s = i^2 n s $$ |
| ② | Multiply $qrst$ by $ \dfrac{5}{2} $ to get $ \dfrac{ 5qrst }{ 2 } $. Step 1: Write $ qrst $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} qrst \cdot \frac{5}{2} & \xlongequal{\text{Step 1}} \frac{qrst}{\color{red}{1}} \cdot \frac{5}{2} \xlongequal{\text{Step 2}} \frac{ qrst \cdot 5 }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5qrst }{ 2 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{5qrst}{2} $ by $ cos+i^2ns $ to get $ \dfrac{5i^2nqrs^2t+5coqrs^2t}{2} $. 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. $$ \begin{aligned} \frac{5qrst}{2} \cdot cos+i^2ns & \xlongequal{\text{Step 1}} \frac{5qrst}{2} \cdot \frac{cos+i^2ns}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 5qrst \cdot \left( cos+i^2ns \right) }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5coqrs^2t+5i^2nqrs^2t }{ 2 } = \frac{5i^2nqrs^2t+5coqrs^2t}{2} \end{aligned} $$ |
| ④ | Add $ \dfrac{-1}{2} $ and $ \dfrac{5i^2nqrs^2t+5coqrs^2t}{2} $ to get $ \dfrac{5i^2nqrs^2t+5coqrs^2t-1}{2} $. To add expressions with the same denominators, we add the numerators and write the result over the common denominator. $$ \begin{aligned} \frac{-1}{2} + \frac{5i^2nqrs^2t+5coqrs^2t}{2} & = \frac{-1}{\color{blue}{2}} + \frac{5i^2nqrs^2t+5coqrs^2t}{\color{blue}{2}} = \\[1ex] &=\frac{ -1 + \left( 5i^2nqrs^2t+5coqrs^2t \right) }{ \color{blue}{ 2 }}= \frac{5i^2nqrs^2t+5coqrs^2t-1}{2} \end{aligned} $$ |