# How will you prove the formula cos(A-B)=cosAcosB+sinAsinB using formula of vector product of two vectors?

Jun 10, 2016

As below

#### Explanation: Let us consider two unit vectors in X-Y plane as follows :

• $\hat{a} \to$ inclined with positive direction of X-axis at angles A
• $\hat{b} \to$ inclined with positive direction of X-axis at angles 90+B, where $90 + B > A$
• Angle between these two vectors becomes
$\theta = 90 + B - A = 90 - \left(A - B\right)$,

$\hat{a} = \cos A \hat{i} + \sin A \hat{j}$
$\hat{b} = \cos \left(90 + B\right) \hat{i} + \sin \left(90 + B\right)$
$= - \sin B \hat{i} + \cos B \hat{j}$
Now
$\hat{a} \times \hat{b} = \left(\cos A \hat{i} + \sin A \hat{j}\right) \times \left(- \sin B \hat{i} + \cos B \hat{j}\right)$
$\implies | \hat{a} | | \hat{b} | \sin \theta \hat{k} = \cos A \cos B \left(\hat{i} \times \hat{j}\right) - \sin A \sin B \left(\hat{j} \times \hat{i}\right)$
Applying Properties of unit vectos $\hat{i} , \hat{j} , \hat{k}$
$\hat{i} \times \hat{j} = \hat{k}$
$\hat{j} \times \hat{i} = - \hat{k}$
$\hat{i} \times \hat{i} = \text{null vector}$
$\hat{j} \times \hat{j} = \text{null vector}$
and
$| \hat{a} | = 1 \mathmr{and} | \hat{b} | = 1 \text{ ""As both are unit vector}$

Also inserting
$\theta = 90 - \left(A - B\right)$,

Finally we get
$\implies \sin \left(90 - \left(A - B\right)\right) \hat{k} = \cos A \cos B \hat{k} + \sin A \sin B \hat{k}$

$\therefore \cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$