Question

What is the formula for ${\sin }^{2}A+{\sin }^{2}B$?

Answer 1

Determine the formula for ${\sin }^{2}A+{\sin }^{2}B$.

Put ${\sin }^{2}A=\frac{1-\cos 2A}{2}$and ${\sin }^{2}B=\frac{1-\cos 2B}{2}$ and simplify the given expression:

$$\begin{gathered} \begin{array}{rcl}{\sin }^{2}A+{\sin }^{2}B& =& \frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}\\ & \Rightarrow & {\sin }^{2}A+{\sin }^{2}B=\frac{2-\left(\cos 2A+\cos 2B\right)}{2}\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{C}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{D}\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{\left(}\frac{\mathbf{C}\mathbf{+}\mathbf{D}}{\mathbf{2}}\mathbf{\right)}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{\left(}\frac{\mathbf{C}\mathbf{-}\mathbf{D}}{\mathbf{2}}\mathbf{\right)}\right]\end{array} \\ \begin{array}{rcl}& \Rightarrow & {\sin }^{2}A+{\sin }^{2}B=\frac{2-2\left(\cos \left(A+B\right).\cos \left(A-B\right)\right)}{2}\end{array} \\ \begin{array}{rcl}& \Rightarrow & {\sin }^{2}A+{\sin }^{2}B=1-\left(\cos \left(A+B\right).\cos \left(A-B\right)\right)\end{array} \end{gathered}$$

Hence, the required formula is $\begin{array}{l}{\sin }^{2}A+{\sin }^{2}B=1-\cos \left(A+B\right)\cos \left(A-B\right)\end{array}$.