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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)

Explanation

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(A+B2)cos(AB2)\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) followed by the use of the fact that the sum of all the angles of a quadrilateral is 360360{}^\circ .

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(X+Y2)cos(XY2)=sin(X)+sin(Y)2\sin \left( \dfrac{X+Y}{2} \right)\cos \left( \dfrac{X-Y}{2} \right)=\sin \left( X \right)+sin\left( Y \right) , we get
sin(A+B)+sin(C+D)\sin \left( A+B \right)+\sin \left( C+D \right)
=2sin(A+B+C+D2)cos(A+BCD2)=2\sin \left( \dfrac{A+B+C+D}{2} \right)\cos \left( \dfrac{A+B-C-D}{2} \right)
Now as ABCD is a quadrilateral, we can say:
A+B+C+D=360\angle A+\angle B+\angle C+\angle D=360{}^\circ
So, substituting the value of A+B+C+D in our expression. On doing so, we get
2sin(3602)cos(A+BCD2)2\sin \left( \dfrac{360{}^\circ }{2} \right)\cos \left( \dfrac{A+B-C-D}{2} \right)
=2sin180cos(A+BCD2)=2\sin 180{}^\circ \cos \left( \dfrac{A+B-C-D}{2} \right)
We know sin180=0\sin 180{}^\circ =0 . Using this in our expression, we get
=2×0×cos(A+BCD2)=2\times 0\times \cos \left( \dfrac{A+B-C-D}{2} \right)
=0=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\angle A+\angle B+\angle C+\angle D=360{}^\circ
A+B=360(C+D)\angle A+\angle B=360{}^\circ -\left( \angle C+\angle D \right)
So, substituting the value of A+B in our expression. On doing so, we get
cos(A+B)
=cos(360(C+D))=\cos \left( 360{}^\circ -\left( C+D \right) \right)
We know cos(360x)=cosx\cos \left( 360{}^\circ -x \right)=\cos x . Using this in our expression, we get
=cos(360(C+D))=\cos \left( 360{}^\circ -\left( C+D \right) \right)
=cos(C+D)=\cos \left( C+D \right)
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.