Answer
$$siniln\frac{a-ib}{a+ib}=\frac{-bi^3ln^2s+ai^2ln^2s}{bi+a}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}siniln\frac{a-ib}{a+ib}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}sin\frac{-bi^2ln+ailn}{bi+a} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-bi^3ln^2s+ai^2ln^2s}{bi+a}\end{aligned} $$ | |
| ① | Multiply $iln$ by $ \dfrac{a-bi}{a+bi} $ to get $ \dfrac{-bi^2ln+ailn}{bi+a} $. Step 1: Write $ iln $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} iln \cdot \frac{a-bi}{a+bi} & \xlongequal{\text{Step 1}} \frac{iln}{\color{red}{1}} \cdot \frac{a-bi}{a+bi} \xlongequal{\text{Step 2}} \frac{ iln \cdot \left( a-bi \right) }{ 1 \cdot \left( a+bi \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ ailn-bi^2ln }{ a+bi } = \frac{-bi^2ln+ailn}{bi+a} \end{aligned} $$ |
| ② | Multiply $ins$ by $ \dfrac{-bi^2ln+ailn}{bi+a} $ to get $ \dfrac{ -bi^3ln^2s+ai^2ln^2s }{ bi+a } $. Step 1: Write $ ins $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} ins \cdot \frac{-bi^2ln+ailn}{bi+a} & \xlongequal{\text{Step 1}} \frac{ins}{\color{red}{1}} \cdot \frac{-bi^2ln+ailn}{bi+a} \xlongequal{\text{Step 2}} \frac{ ins \cdot \left( -bi^2ln+ailn \right) }{ 1 \cdot \left( bi+a \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -bi^3ln^2s+ai^2ln^2s }{ bi+a } \end{aligned} $$ |
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