Q1. Prove by vector method:

sin(A+B)=sinAcosB+cosAsinB

Q2. Prove by vector method:

cos(A+B)=cosAcosB-sinAsinB

Let $\overrightarrow{\alpha }$ and $\overrightarrow{\beta }$ be two unit vectors, and A and B be the angles made by them respectively with the X-axis.

So, $$\begin{gathered} \overrightarrow{\alpha }=\cos A\hat{i}+\sin A\hat{j} \\ \overrightarrow{\beta }=\cos B\hat{i}+\sin B\hat{j} \end{gathered}$$

Now, $\overrightarrow{\alpha }.\overrightarrow{\beta }=\left(\cos A\hat{i}+\sin A\hat{j}\right)\left(\cos B\hat{i}+\sin B\hat{j}\right)$

$$\begin{gathered} \Rightarrow \alpha \beta \cos \left(A-B\right)=\cos A\cos B+\sin A\sin B \\ \Rightarrow \cos \left(A-B\right)=\cos A\cos B+\sin A\sin B \left[\because \alpha =1, \beta =1\right] \end{gathered}$$ ------------(1)

Putting -B in place of B in (1):-

$$\begin{gathered} \cos \left(A-\left(-B\right)\right)=\cos A\cos \left(-B\right)+\sin A\sin \left(-B\right) \\ \Rightarrow \mathit{c}\mathit{o}\mathit{s}\left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathit{B}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathit{B} \end{gathered}$$


Similarly, 

$$\begin{gathered} \overrightarrow{\alpha }\times \overrightarrow{\beta }=\left(\cos A\hat{i}+\sin A\hat{j}\right)\times \left(\cos B\hat{i}+\sin B\hat{j}\right) \\ \Rightarrow \overrightarrow{\alpha }\times \overrightarrow{\beta }=\cos A\sin B\hat{k}-\sin A\cos B\hat{k} \end{gathered}$$

$$\begin{gathered} \Rightarrow \alpha \beta \sin \left(A-B\right)\left(-\hat{k}\right)=\left(\sin A\cos B-\cos A\sin B\right)\left(-\hat{k}\right) \\ \Rightarrow \sin \left(A-B\right)=\sin A\cos B-\cos A\sin B -------\left(2\right) \end{gathered}$$

Putting B=-B in (2):-

$$\begin{gathered} \sin \left(A-\left(-B\right)\right)=\sin A\cos \left(-B\right)-\cos A\sin \left(-B\right) \\ \mathbf{\Rightarrow }\mathbf{sin}\left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{=}\mathbf{sin}\mathit{A}\mathbf{cos}\mathit{B}\mathbf{+}\mathbf{cos}\mathit{A}\mathbf{sin}\mathit{B} \end{gathered}$$

Hence Proved.