Question

In quadrilateral \[ABCD\], if
\[\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) + \sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) = 2\] then find the value of \[\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\sin \left( {\dfrac{D}{2}} \right)\] is
A.\[\dfrac{1}{4}\]
B.\[\dfrac{1}{2}\]
C.\[\dfrac{1}{8}\]
D.1

Answer 1

Hint: Here need to find the value of the given trigonometric expression. For that, we will use the sum to product identities of trigonometry. From there, we will get the values of all angles of the quadrilateral. Then we will put the values of all angles in the given trigonometric expression to get the value.

Formula used:
We will use the formula of the product to sum identities of trigonometry is given by \[\sin a \cdot \cos b = \dfrac{{\sin \left( {a + b} \right) + \sin \left( {a - b} \right)}}{2}\] .
Complete step-by-step answer:
Here it is given that \[ABCD\] is a quadrilateral and it is also given that
\[\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) + \sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) = 2\].
We will first simplify this trigonometric equation to get the value of all angels.
Using the identity \[\sin a \cdot \cos b = \dfrac{{\sin \left( {a + b} \right) + \sin \left( {a - b} \right)}}{2}\] in above equation, we get
\[ \Rightarrow \dfrac{{\sin \left( {\dfrac{{A + B}}{2} + \dfrac{{A - B}}{2}} \right) + \sin \left( {\dfrac{{A + B}}{2} - \dfrac{{A - B}}{2}} \right)}}{2} + \dfrac{{\sin \left( {\dfrac{{C + D}}{2} + \dfrac{{C - D}}{2}} \right) + \sin \left( {\dfrac{{C + D}}{2} - \dfrac{{C - D}}{2}} \right)}}{2} = 2\]
On adding and subtracting the terms inside the bracket, we get
\[ \Rightarrow \dfrac{{\sin A + \sin B}}{2} + \dfrac{{\sin C + \sin D}}{2} = 2\]
On further simplification, we get
\[ \Rightarrow \dfrac{{\sin A + \sin B + \sin C + \sin D}}{2} = 2\]
On multiplying 2 on both sides, we get
\[ \Rightarrow 2 \times \dfrac{{\sin A + \sin B + \sin C + \sin D}}{2} = 2 \times 2\]
On multiplying the terms, we get
\[ \Rightarrow \sin A + \sin B + \sin C + \sin D = 4\] …….. \[\left( 1 \right)\]
We know the maximum value of \[\sin \theta \] is 1 or \[\sin \theta \le 1\]
Thus, the sum of three sine functions is equal to 4 only when each of them is equal to 1. So,
\[ \Rightarrow \sin A = \sin B = \sin C = \sin D = 1\]
We know that \[\sin \dfrac{\pi }{2} = 1\].
Therefore,
\[ \Rightarrow A = B = C = D = \dfrac{\pi }{2}\]
Now, we will find the value of \[\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\sin \left( {\dfrac{D}{2}} \right)\] …….. \[\left( 2 \right)\]
Now, we will substitute the value of all angles obtained in equation 2 here.
$\Rightarrow$ \[\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\sin \left( {\dfrac{D}{2}} \right) = \sin \left( {\dfrac{{\dfrac{\pi }{2}}}{2}} \right) \times \sin \left( {\dfrac{{\dfrac{\pi }{2}}}{2}} \right) \times \sin \left( {\dfrac{{\dfrac{\pi }{2}}}{2}} \right) \times \sin \left( {\dfrac{{\dfrac{\pi }{2}}}{2}} \right)\]
On simplifying the terms, we get
$\Rightarrow$ \[\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\sin \left( {\dfrac{D}{2}} \right) = \sin \left( {\dfrac{\pi }{4}} \right) \times \sin \left( {\dfrac{\pi }{4}} \right) \times \sin \left( {\dfrac{\pi }{4}} \right) \times \sin \left( {\dfrac{\pi }{4}} \right)\]
Substituting \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\] in the above equation, we get
$\Rightarrow$ \[\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\sin \left( {\dfrac{D}{2}} \right) = \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }}\]
On multiplying the terms, we get
$\Rightarrow$ \[\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\sin \left( {\dfrac{D}{2}} \right) = \dfrac{1}{4}\]
Hence, the correct option is option A.

Note: We need to know the meaning of the trigonometric identities as we have used the trigonometric identities in this question. Trigonometric identities are defined as the equalities which involve the trigonometric functions. They are true for every value of the occurring variables for which both sides of the equality are defined. We need to keep in mind that all the trigonometric identities are periodic in nature because they repeat their values after a certain interval.