Question

How will you prove the formula $\sin (A + B) = \sin A\cos B + \cos A\sin B$ using the formula of scalar product of two vectors?

Answer 1

Hint : By multiplying the magnitudes of two vectors with the cosine of the angle between them, the scalar product is obtained. Orthogonal vectors' scalar product vanishes, while antiparallel vectors' scalar product is negative. A vector perpendicular to both vectors is the vector product of two vectors.

Complete Step By Step Answer:
Consider the following two unit vectors in the $X - Y$ plane:
$\widehat a$ represents inclined with positive direction of $X$-axis at angles $A$
$\widehat b$ represents inclined with positive direction of $X$-axis at angles $90 - B$ , where $90 - B > A$
Angle between these two vectors becomes
$\theta = 90 - B - A = 90 - (A + B)$
\widehat a = \cos A\widehat i + \sin A\widehat j \\\ \widehat b = \cos (90 - B)\widehat i + \sin (90 - B)\widehat j \\\ = \sin B\widehat i + \cos B\widehat j \\\
Now we have to take the scalar product of $\widehat a$ and $\widehat b$
\widehat a.\widehat b = \left( {\cos A\widehat i + \sin A\widehat j} \right).\left( {\sin B\widehat i + \cos B\widehat j} \right) \\\ \Rightarrow \left| {\widehat a} \right|\left| {\widehat b} \right|\cos \theta = \sin A\cos B\left( {\widehat j.\widehat j} \right) + \cos A\sin B\left( {\widehat i.\widehat i} \right) \\\
The Properties of unit vectors $\widehat i,\widehat j,\widehat k$
\widehat i.\widehat j = 0 \\\ \widehat j.\widehat i = 0 \\\ \widehat i.\widehat i = 1 \\\ \widehat j.\widehat j = 1 \\\
And
$\left| {\widehat a} \right| = 1$ and $\left| {\widehat b} \right| = 1$
We know that,
$\theta = 90 - (A + B)$,
On substituting theta, we get
\Rightarrow \cos (90 - (A + B)) = \sin A\cos B + \cos A\sin B \\\ \therefore \sin (A + B) = \sin A\cos B + \cos A\sin B \\\

Note :
Three vectors' scalar triple product is known as
$a.(b \times c) = b.(c \times a) = c.(a \times b)$
Its determinant is the matrix whose columns are the three vectors' Cartesian coordinates.
The vector triple product (VTP) is defined as follows:
$a \times (b \times c) = b(c.a) - c(a.b)$
Keep in mind the vectors are dotted together when recalling this identity, also known as Lagrange's formula: "$BAC$ minus $CAB$." In physics, this formula can be used to make vector calculations simpler.