$cos (A + B) = cos A cos B - sin A sin B$ prove by vector method.

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Solution

We have to prove $c o s (A + B) = c o s A c o s B - s i n A s i n B$ by vector method.

Let $P (x_{1}, y_{1})$ be the point in the first quadrant which is at a distance of $1$ unit from origin $O$ and let angle made by $O P$ with $X$ axis be $A$ .

Similarly, let $Q (x_{2}, y_{2})$ be the point in the fourth quadrant which is at a distance of $1$ unit from origin $O$ and let angle made by $O Q$ with $X$ axis be $B$ .

$\Rightarrow | \to O P | = | \to O Q | = 1$ and $∠ P O Q = A + B$ .

$\Rightarrow \to O P = c o s A^i + s i n A^j$ and $\to O Q = c o s B^i - s i n B^j$

We know that $\to a \cdot \to b = | \to a | | \to b | c o s θ$ where $θ$ is the angle between the vectors.

$\Rightarrow \to O P \cdot \to O Q = | \to O P | | \to O Q | c o s (A + B)$

$\Rightarrow \to O P \cdot \to O Q = c o s (A + B) . . . (1)$

Since we have $\to O P = c o s A^i + s i n A^j$ and $\to O Q = c o s B^i - s i n B^j$

$\Rightarrow \to O P \cdot \to O Q = (c o s A^i + s i n A^j) \cdot (c o s B^i - s i n B^j)$

$= c o s A c o s B + s i n A (- s i n B)$

$= c o s A c o s B - s i n A s i n B$

$∴ \to O P \cdot \to O Q = c o s A c o s B - s i n A s i n B . . . (2)$

From $(1)$ and $(2)$ we get $c o s (A + B) = c o s A c o s B - s i n A s i n B$

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