Question

What is formula of
\[2\sin A\cos B\]
\[2\cos A\sin B\]
\[2\sin A\sin B\]
\[2\cos A\cos B\]

Answer 1

Hint: In the above given question, we are given four trigonometric expressions of the two combined trigonometric functions of the sine and cosine functions respectively. We have to determine the suitable formula which can be used for each of the four given trigonometric expressions. In order to approach the solution, we have to recall the formulae of the sum and difference of two angles for the sine and cosine functions respectively.

Complete answer:
Since we know that the value of sine and cosine functions for the sum and difference of two angles \[A\] and \[B\] is given by the four formulae written below.
\[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]
\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]
\[\cos \left( {A + B} \right) = \sin A\sin B - \cos A\cos B\]
\[\cos \left( {A - B} \right) = \sin A\sin B + \cos A\cos B\]
Now we can obtain the values of the four above given trigonometric expressions by using these four above written formulae.
Now, adding the formula 1 and 2 gives us,
\[ \Rightarrow \sin \left( {A + B} \right) + \sin \left( {A - B} \right) = \sin A\cos B + \cos A\sin B + \sin A\cos B - \cos A\sin B\]
That is,
\[ \Rightarrow \sin \left( {A + B} \right) + \sin \left( {A - B} \right) = \sin A\cos B + \sin A\cos B\]
Hence,
\[ \Rightarrow \sin \left( {A + B} \right) + \sin \left( {A - B} \right) = 2\sin A\cos B\]
Now, subtracting the formula 2 from 1 gives us,
\[ \Rightarrow \sin \left( {A + B} \right) - \sin \left( {A - B} \right) = \sin A\cos B + \cos A\sin B - \sin A\cos B + \cos A\sin B\]
That is,
\[ \Rightarrow \sin \left( {A + B} \right) - \sin \left( {A - B} \right) = \cos A\sin B + \cos A\sin B\]
Hence,
\[ \Rightarrow \sin \left( {A + B} \right) - \sin \left( {A - B} \right) = 2\cos A\sin B\]
Now, adding the formula 3 and 4 gives us,
\[ \Rightarrow \cos \left( {A + B} \right) + \cos \left( {A - B} \right) = \sin A\sin B - \cos A\cos B + \sin A\sin B + \cos A\cos B\]
That is,
\[ \Rightarrow \cos \left( {A + B} \right) + \cos \left( {A - B} \right) = \sin A\sin B + \sin A\sin B\]
Hence,
\[ \Rightarrow \cos \left( {A + B} \right) + \cos \left( {A - B} \right) = 2\sin A\sin B\]
Now, subtracting the formula 3 from 4 gives us,
\[ \Rightarrow \cos \left( {A - B} \right) - \cos \left( {A + B} \right) = \sin A\sin B + \cos A\cos B - \sin A\sin B + \cos A\cos B\]
That is,
\[ \Rightarrow \cos \left( {A - B} \right) - \cos \left( {A + B} \right) = \cos A\cos B + \cos A\cos B\]
Hence,
\[ \Rightarrow \cos \left( {A - B} \right) - \cos \left( {A + B} \right) = 2\cos A\cos B\]
These are the required values of the four above given trigonometric expressions.
Therefore the formula of the four above given trigonometric expressions are,
\[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\]
\[2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)\]
\[2\sin A\sin B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\]
\[2\cos A\cos B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)\]

Note:
In order to remember these formulae, you must remember the four fundamental formulae for the sine and cosine functions of the sum and difference of two different angles respectively.
For the convenience to remember these four formulae, always keep in mind that for a sine function the terms are products of both sine and cosine terms and the plus or minus sign is unchanged. Whereas for a cosine function the terms are products of only like terms i.e. either only sine or cosine terms and the plus or minus sign is reversed.