Question

If $\sin A + \sin B + \sin C + \sin D =4$, then find the value of $\sin A . \sin B . \sin C . \sin D$.

Answer 1

Hint: We know that the value of \[{{\sin }}\] function oscillates between the value $-1$ and $1$. Therefore, for the maximum value of \[{{\sin A}}\], we get \[{{\sin A = 1}}\]. Similarly for $\sin A, \sin B, \sin C$ and $\sin D$ we have $\sin A = \sin B = \sin C = \sin D =1$, that’s why the value in the question is $4$. Using this logic we will get the required answer.

Complete step by step answer:
The range of \[{{\sin }}\] is between $-1$ and $1$ , taking $\sin A = \sin B = \sin C = \sin D =1$, on putting the values of $\sin A, \sin B, \sin C$ and $\sin D$ in $\sin A . \sin B . \sin C . \sin D$, we get
$=1 \times 1 \times 1 \times 1$
$=1$
Therefore , the value of $\sin A . \sin B . \sin C . \sin D = 1$.

Note:
Alternate solution:
We consider the value of \[{{\sin A}}\] to be $1$, therefore the value of \[A = {90^ \circ }\] , as \[\operatorname{Sin} {90^ \circ } = 1\], similarly it is applicable for $ \sin B, \sin C$ and $\sin D$. Therefore , the value of \[A + B + C + D = {360^ \circ }\] .
Now , the expression $\sin A . \sin B . \sin C . \sin D$ can be written as
\[ = \dfrac{4}{4}{{(\sin A}}{{\sin B)}}{{(\sin C}}{{\sin D)}}\] ……………………….. (multiplying and dividing by $4$
\[ = \dfrac{1}{4}{{(2 \sin A}}{{\sin B)}}{{.(2 \sin C}}{{\sin D)}}\] , rearranging the expression we get ,
Now , using the trigonometric identities \[{{2 \sin A}}{{\sin B = \cos(A - B) - \cos(A + B)}}\] we get ,
\[ = \dfrac{1}{4}\left[ {{{ \cos (A - B) - cos(A + B)}}} \right].\left[ {{{ \cos (C - D) - cos(C + D)}}} \right]\], Now putting the value of \[A = B = C = D = {90^ \circ }\] , we get
\[ = \dfrac{1}{4}\left[ {{{ \cos (}}{{90}^ \circ }{\text{ - }}{{90}^ \circ }{\text{) - cos(9}}{{\text{0}}^ \circ }{\text{ + 9}}{{\text{0}}^ \circ }{\text{)}}} \right].\left[ {{{ \cos (}}{{90}^ \circ }{\text{ - }}{{90}^ \circ }{\text{) - cos(9}}{{\text{0}}^ \circ }{\text{ + 9}}{{\text{0}}^ \circ }{\text{)}}} \right]\] , on solving we get
\[ = \dfrac{1}{4}\left[ {{{ \cos (}}{{\text{0}}^ \circ }{\text{) - cos(}}{{180}^ \circ }{\text{)}}} \right].\left[ {{{ \cos (}}{{\text{0}}^ \circ }{\text{) - cos(}}{{180}^ \circ }{\text{)}}} \right]\] , on putting the values of \[\cos {0^ \circ } = 1\] and \[\cos {180^ \circ } = - 1\] we get ,
\[ = \dfrac{1}{4}\left[ {{\text{1 - ( - 1)}}} \right].\left[ {1{\text{ - ( - 1)}}} \right]\] , on solving further we get
 \[ = \dfrac{1}{4}\left[ {\text{2}} \right].\left[ 2 \right]\]
\[ = \dfrac{1}{4} \times 4\] , on solving we get
\[ = 1\]
Hence Proved.

$\bullet $ The Trigonometric identities formula should be remembered as these questions are based on Trigonometric identities . Therefore , the solution of the question can be approached using different methods . Here , we have shown two of the methods in one we have solved using the given question , the other one with the expression we have to find the value of . The formulae we have used are of fundamental importance and one should have gone through acquaintance with these formulae .