Question

Prove that: $\sin (\mathrm{C}+\mathrm{D})\mathrm{x}\sin (\mathrm{C}-\mathrm{D})={\sin }^2\mathrm{C}-{\sin }^2\mathrm{D}$

Answer 1

As we have $\sin (\mathrm{C}+\mathrm{D})\mathrm{x}\sin (\mathrm{C}-\mathrm{D})={\sin }^2\mathrm{C}-{\sin }^2\mathrm{D}$

We know that formulae for$\sin (C+D)=\sin (C+D)=\sin \left(C\right)\cos \left(D\right)+\cos \left(C\right)\sin \left(D\right)$

Also $\sin (-D)=-\sin \left(D\right)$

And $\cos (-D)=\cos \left(D\right)$, so

$\sin (C-D)=\sin \left(C\right)\cos \left(D\right)-\cos \left(C\right)\sin \left(D\right)$

Therefore $\sin (C+D)\cdot \sin (C-D)$

$$\begin{gathered} =(\sin C\cos D+\cos C\sin D)(\sin C\cos D-\cos C\sin D) \\ ={\left(\sin C\cos D\right)}^2-{\left(\cos C\sin D\right)}^2\left[(a+b)(a-b)=a^2-b^2\right] \end{gathered}$$

$$\begin{gathered} ={\sin }^{2}C{\cos }^{2}D-{\sin }^{2}D{\cos }^{2}C \\ ={\sin }^{2}C(1-{\sin }^{2}D)-{\sin }^{2}D(1-{\sin }^{2}C) \end{gathered}$$

Now we know that $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{}$( by Pythagoras theorem)

$$\begin{gathered} ={\sin }^{2}C-{\sin }^{2}D-{\sin }^{2}C{\sin }^{2}D+{\sin }^{2}D{\sin }^{2}C \\ ={\sin }^{2}C-{\sin }^{2}D-0 \\ ={\sin }^{2}C-{\sin }^{2}D \end{gathered}$$

Hence LHS= RHS proved.