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Question: Find the value of \(\sin A \cdot \sin \left( {A + B} \right)\) A. \({\cos ^2}A \cdot \cos B + \si...

Find the value of sin⁡A⋅sin⁡(A+B)\sin A \cdot \sin \left( {A + B} \right)
A. cos⁡2A⋅cos⁡B+sin⁡A⋅sin⁡2B{\cos ^2}A \cdot \cos B + \sin A \cdot {\sin ^2}B
B. sin⁡2A⋅cos⁡B+12cos⁡2A⋅sin⁡B{\sin ^2}A \cdot \cos B + \dfrac{1}{2}\cos 2A \cdot \sin B
C. sin⁡2A⋅cos⁡B+12sin⁡2A⋅sin⁡B{\sin ^2}A \cdot \cos B + \dfrac{1}{2}\sin 2A \cdot \sin B
D. cos⁡2A⋅sin⁡B+cos⁡A⋅cos⁡2B{\cos ^2}A \cdot \sin B + \cos A \cdot {\cos ^2}B

Explanation

Solution

Here, to find the value of the expression sin⁡A⋅sin⁡(A+B)\sin A \cdot \sin \left( {A + B} \right), we need to use the formula of sin⁡(A+B)\sin \left( {A + B} \right), Now, we know that the formula sin⁡(A+B)\sin \left( {A + B} \right) is sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B.
After using this formula we will be using the formula of sin2A. The formula of sin2A is sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A.

Complete step-by-step answer:
In this question, we are given a trigonometric expression and we need to find its value.
The given expression is: sin⁡A⋅sin⁡(A+B)\sin A \cdot \sin \left( {A + B} \right) - - - - - - - - - - - - - (1)
Now, to solve this expression, we need to use some trigonometric formulas and relations.
So, first of all, we have the formula for
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B
Therefore, putting this value in equation (1), we get
⇒sin⁡A⋅sin⁡(A+B)=sin⁡A(sin⁡Acos⁡B+cos⁡Asin⁡B)\Rightarrow \sin A \cdot \sin \left( {A + B} \right) = \sin A\left( {\sin A\cos B + \cos A\sin B} \right)
Now, opening the bracket, we get
⇒sin⁡A⋅sin⁡(A+B)=sin⁡2Acos⁡B+sin⁡Acos⁡Asin⁡B\Rightarrow \sin A \cdot \sin \left( {A + B} \right) = {\sin ^2}A\cos B + \sin A\cos A\sin B - - - - - - - - - - - - - - (2)
Now, we know the formula of sin2A that is
sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A
So, in equation (2), we have sinAcosA. So, if we multiply and divide the equation with 2, we can use the above formula.
Therefore, equation (2) will become
⇒sin⁡A⋅sin⁡(A+B)=sin⁡2Acos⁡B+12(2sin⁡Acos⁡Asin⁡B)\Rightarrow \sin A \cdot \sin \left( {A + B} \right) = {\sin ^2}A\cos B + \dfrac{1}{2}\left( {2\sin A\cos A\sin B} \right)
⇒sin⁡A⋅sin⁡(A+B)=sin⁡2Acos⁡B+12sin⁡2Asin⁡B\Rightarrow \sin A \cdot \sin \left( {A + B} \right) = {\sin ^2}A\cos B + \dfrac{1}{2}\sin 2A\sin B
Hence, we have got sin⁡A⋅sin⁡(A+B)=sin⁡2Acos⁡B+12sin⁡2Asin⁡B\sin A \cdot \sin \left( {A + B} \right) = {\sin ^2}A\cos B + \dfrac{1}{2}\sin 2A\sin B.

So, the correct answer is “Option C”.

Note: The most important part in this question is that we need to identify which formulas can be used for further simplification. Here, as we had sinAcosA, we identified that sin2A formula can be used here so we made some changes and used the sin2A formula.