Question
Question: In any quadrilateral ABCD, prove that (a) sin(A+B)+sin(C+D)=0 (b) cos(A+B)=cos(C+D)...
In any quadrilateral ABCD, prove that
(a) sin(A+B)+sin(C+D)=0
(b) cos(A+B)=cos(C+D)
Solution
Hint: Try to simplify the left-hand side of the equation given in the question by the application of the formula sinA+sinB=2sin(2A+B)cos(2A−B) followed by the use of the fact that the sum of all the angles of a quadrilateral is 360∘ .
Complete step-by-step answer:
Before starting with the solution, let us draw a diagram for better visualisation.
Now starting with the left-hand side of the first equation that is given in the question. According to the formula: 2sin(2X+Y)cos(2X−Y)=sin(X)+sin(Y) , we get
sin(A+B)+sin(C+D)
=2sin(2A+B+C+D)cos(2A+B−C−D)
Now as ABCD is a quadrilateral, we can say:
∠A+∠B+∠C+∠D=360∘
So, substituting the value of A+B+C+D in our expression. On doing so, we get
2sin(2360∘)cos(2A+B−C−D)
=2sin180∘cos(2A+B−C−D)
We know sin180∘=0 . Using this in our expression, we get
=2×0×cos(2A+B−C−D)
=0
Hence, we have proved the first equation.
Now starting with the left-hand side of the second equation that is given in the question. As ABCD is a quadrilateral, we can say:
∠A+∠B+∠C+∠D=360∘
∠A+∠B=360∘−(∠C+∠D)
So, substituting the value of A+B in our expression. On doing so, we get
cos(A+B)
=cos(360∘−(C+D))
We know cos(360∘−x)=cosx . Using this in our expression, we get
=cos(360∘−(C+D))
=cos(C+D)
Hence, we have proved the second equation, as well.
Note: Be careful about the calculation and the signs while opening the brackets. Also, you need to learn the properties related to quadrilateral and other polygons, as they are used very often.