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

May 30, 2016

I could prove it using the dot product of vectors.

#### Explanation:

Let
$\hat{A} \mathmr{and} \hat{B}$ be two unit vectors in the $x$-$y$ plane such that $\hat{A}$ makes an angle $- A$ and $\hat{B}$ makes an angle $B$ with $x$-axis so that the angle between the two is $\left(A + B\right)$
The unit vectors can be written in Cartesian form as
$\hat{A} = \cos A \hat{i} - \sin A \hat{j}$ and $\hat{B} = \cos B \hat{i} + \sin B \hat{j}$ ....(1)
To prove
cos(A+B)=cosAcosB−sinAsinB

We know that dot product of two vectors is
$\vec{A} \cdot \vec{B} = | \vec{A} | | \vec{B} | \cos \theta$
Inserting our unit vectors in the above; $| \vec{A} | = | \vec{B} | = 1$ and value of $\theta = \left(A + B\right)$, we obtain

$\hat{A} \cdot \hat{B} = \cos \left(A + B\right)$
Using equation (1)
LHS $= \left(\cos A \hat{i} - \sin A \hat{j}\right) \cdot \left(\cos B \hat{i} + \sin B \hat{j}\right)$
From property of dot product we know that only terms containing $\hat{i} \cdot \hat{i} \mathmr{and} \hat{j} \cdot \hat{j} \text{ are} = 1$ and rest vanish.
$\therefore$ LHS$= \cos A \cos B - \sin A \sin B$

Equating LHS with RHS we obtain

cos(A+B)=cosAcosB−sinAsinB

Jun 2, 2016

As follows

#### 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$$

Sin(A+B) =SinA CosB + CosASinB formula can also be obtained
by taking scalar product of $\hat{a} \mathmr{and} \hat{b}$

Now

$\hat{a} \cdot \hat{b} = \left(\cos A \hat{i} + \sin A \hat{j}\right) \cdot \left(\sin B \hat{i} + \cos B \hat{j}\right)$
$\implies | \hat{a} | | \hat{b} | \cos \theta = \sin A \cos B \left(\hat{j} \cdot \hat{j}\right) + \cos A \sin B \left(\hat{i} \cdot \hat{i}\right)$

Applying Properties of unit vectos $\hat{i} , \hat{j} , \hat{k}$
$\hat{i} \cdot \hat{j} = 0$
$\hat{j} \cdot \hat{i} = 0$
$\hat{i} \cdot \hat{i} = 1$
$\hat{j} \cdot \hat{j} = 1$

and

$| \hat{a} | = 1 \mathmr{and} | \hat{b} | = 1$
Also inserting
$\theta = 90 - \left(A + B\right)$,

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

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