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Question: In quadrilateral \[ABCD\], if \[\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B...

In quadrilateral ABCDABCD, if
sin(A+B2)cos(AB2)+sin(C+D2)cos(CD2)=2\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(A2)sin(B2)sin(C2)sin(D2)\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.14\dfrac{1}{4}
B.12\dfrac{1}{2}
C.18\dfrac{1}{8}
D.1

Explanation

Solution

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 sinacosb=sin(a+b)+sin(ab)2\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 ABCDABCD is a quadrilateral and it is also given that
sin(A+B2)cos(AB2)+sin(C+D2)cos(CD2)=2\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 sinacosb=sin(a+b)+sin(ab)2\sin a \cdot \cos b = \dfrac{{\sin \left( {a + b} \right) + \sin \left( {a - b} \right)}}{2} in above equation, we get
sin(A+B2+AB2)+sin(A+B2AB2)2+sin(C+D2+CD2)+sin(C+D2CD2)2=2\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
sinA+sinB2+sinC+sinD2=2\Rightarrow \dfrac{{\sin A + \sin B}}{2} + \dfrac{{\sin C + \sin D}}{2} = 2
On further simplification, we get
sinA+sinB+sinC+sinD2=2\Rightarrow \dfrac{{\sin A + \sin B + \sin C + \sin D}}{2} = 2
On multiplying 2 on both sides, we get
2×sinA+sinB+sinC+sinD2=2×2\Rightarrow 2 \times \dfrac{{\sin A + \sin B + \sin C + \sin D}}{2} = 2 \times 2
On multiplying the terms, we get
sinA+sinB+sinC+sinD=4\Rightarrow \sin A + \sin B + \sin C + \sin D = 4 …….. (1)\left( 1 \right)
We know the maximum value of sinθ\sin \theta is 1 or sinθ1\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,
sinA=sinB=sinC=sinD=1\Rightarrow \sin A = \sin B = \sin C = \sin D = 1
We know that sinπ2=1\sin \dfrac{\pi }{2} = 1.
Therefore,
A=B=C=D=π2\Rightarrow A = B = C = D = \dfrac{\pi }{2}
Now, we will find the value of sin(A2)sin(B2)sin(C2)sin(D2)\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)\sin \left( {\dfrac{D}{2}} \right) …….. (2)\left( 2 \right)
Now, we will substitute the value of all angles obtained in equation 2 here.
\Rightarrow sin(A2)sin(B2)sin(C2)sin(D2)=sin(π22)×sin(π22)×sin(π22)×sin(π22)\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(A2)sin(B2)sin(C2)sin(D2)=sin(π4)×sin(π4)×sin(π4)×sin(π4)\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 sinπ4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} in the above equation, we get
\Rightarrow sin(A2)sin(B2)sin(C2)sin(D2)=12×12×12×12\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(A2)sin(B2)sin(C2)sin(D2)=14\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.