Question

Simplify the following trigonometric expression: $(\sin 3A + \sin A)\sin A + (\cos 3A - \cos A)\cos A$
A) 1
B) 0
C) -1
D) 2

Answer 1

Hint: According to given in the question we have to find the value the given expression $(\sin 3A + \sin A)\sin A + (\cos 3A - \cos A)\cos A$so, first of all we have to solve the term $(\sin 3A + \sin A)$of the expression with the help of the formula as given below:

Formula used: $\sin A + \operatorname{Sin} B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)...................(1)$
So, with the help of the formula (1) we can simplify the value of $(\sin 3A + \sin A)$
Now, same as we have to solve the term $(\cos 3A - \cos A)$ of the expression with the help of the formula as given below:
$ - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right).....................(2)$
So, with the help of the formula (2) we can simplify the value of $(\cos 3A - \cos A)$
Now, we have to substitute both of the simplified values of $(\sin 3A + \sin A)$ and $(\cos 3A - \cos A)$ in the trigonometric expression given in the question to obtain the value of the expression.

Complete step-by-step answer:
Given expression: $(\sin 3A + \sin A)\sin A + (\cos 3A - \cos A)\cos A$……………………(3)
Step 1: First of all we find the value of the term $(\sin 3A + \sin A)$with the help of formula (1) as mentioned in the solution hint.
Hence, on applying the formula (1) to simplify the term,
$ \Rightarrow (\sin 3A + \sin A) = 2\sin \left( {\dfrac{{3A + A}}{2}} \right)\cos \left( {\dfrac{{3A - A}}{2}} \right)$
On solving the obtained expression just above,
$
\Rightarrow (\sin 3A + \sin A) = 2\sin \left( {\dfrac{{4A}}{2}} \right)\cos \left( {\dfrac{{2A}}{2}} \right) \\
\Rightarrow (\sin 3A + \sin A) = 2\sin 2A\cos A
 $
Step 2: Now, same as step 1 we have to find the value of the term $(\cos 3A - \cos A)$with the help of formula (2) as mentioned in the solution hint.
Hence, on applying the formula (2) to simplify the term,
\[ \Rightarrow (\cos 3A - \cos A) = - 2\sin \left( {\dfrac{{3A + A}}{2}} \right)\sin \left( {\dfrac{{3A - A}}{2}} \right)\]
On solving the obtained expression just above,
\[
\Rightarrow (\cos 3A - \cos A) = - 2\sin \left( {\dfrac{{4A}}{2}} \right)\sin \left( {\dfrac{{2A}}{2}} \right) \\
\Rightarrow (\cos 3A - \cos A) = - 2\sin 2A\sin A
 \]
Step 3: Now, we have to substitute the values of the obtained expression $(\sin 3A + \sin A)$and $(\cos 3A - \cos A)$in the given expression (3)
$ = (2\sin 2A\cos A)\sin A + ( - 2\sin 2A\sin A)\cos A$
On solving the obtained expression just above,
$
   = 2\sin 2A\cos A\sin A - 2\sin 2A\cos A\sin A \\
   = 0
 $
Final solution: Hence, with the help of the formula (1) and (2) given in the solution hint we have obtained the value of the given trigonometric expression: $(\sin 3A + \sin A)\sin A + (\cos 3A - \cos A)\cos A$= 0

Hence, option B is the correct answer.

Note: To solve the given trigonometric expression easily we have to simplify the terms of given expression separately and after simplification we can substitute it again in the given expression.
Remember the positive and negative signs while substituting the values of the terms after simplification separately.