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Question: If \(A + B + C = 180\) then prove that \[\sin 2A + \sin 2B + \sin 2C = 4\sin A \times \sin B \times ...

If A+B+C=180A + B + C = 180 then prove that sin⁡2A+sin⁡2B+sin⁡2C=4sin⁡A×sin⁡B×sin⁡C\sin 2A + \sin 2B + \sin 2C = 4\sin A \times \sin B \times \sin C .

Explanation

Solution

The given question is about trigonometry we are given the relation of angles that are using in the statement. As well as we are also given a trigonometric equation, that we have to prove equal using the relation between the angles given. We will use trigonometric identities the angles given. We will use trigonometric identities for this and further we will prove that equal.

Complete step-by-step answer:
In the given question, we are given the relation between the three angles A,B,CA,B,C and that is
A+B+C=1800 ......................(1)A + B + C = {180^0}{\text{ }}......................{\text{(1)}}
Also we were given a trigonometric equation and we are asked to prove that equation equal which means left hand side is equal to right hand side. Which we will prove equal by using some trigonometric identities and the given relation between 33 angles given in (1)(1)
So, we have to prove
sin⁡2A+sin⁡2B+sin⁡2C=4sin⁡Asin⁡Bsin⁡C\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C
Firstly, we will take left hand side and we will apply identities here and then this term will become equal to Right hand side
So taking Left hand side
sin⁡1A+sin⁡2B+sin⁡2C\sin 1A + \sin 2B + \sin 2C
Here we will apply formula for first two terms which is
sin⁡x+sin⁡y=2sin⁡((x+y)2)cos⁡((x−y)2)\sin x + \sin y = 2\sin \left( {\dfrac{{\left( {x + y} \right)}}{2}} \right)\cos \left( {\dfrac{{\left( {x - y} \right)}}{2}} \right) and sin⁡2x=2sin⁡xcos⁡x,\sin 2x = 2\sin x\cos x, this is applied on third term So we get 2sin⁡(2A+2B2)cos⁡(2A−2B2)+2sin⁡Ccos⁡C2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right) + 2\sin C\cos C
Taking 22 common from angles of first term and cancelling it with 22in denominator, we get
2sin⁡(A+B)cos⁡(A−B)+2sin⁡Ccos⁡C .............(2)2\sin (A + B)\cos (A - B) + 2\sin C\cos C{\text{ }}.............{\text{(2)}}
Also from (1), A+B+C=1800A + B + C = {180^0}
Which become A+B−1800−CA + B - {180^0} - C
Putting it in (2) we get
⇒\Rightarrow 2sin⁡(1800−C)cos⁡(A−B)+2sin⁡Ccos⁡C 2\sin ({180^0} - C)\cos (A - B) + 2\sin C\cos C{\text{ }}
Also sin⁡(1800−x)=sin⁡x\sin ({180^0} - x) = \sin x
So, 2sin⁡Ccos⁡(A−B)+2sin⁡Ccos⁡C 2\sin C\cos (A - B) + 2\sin C\cos C{\text{ }}
Taking 2sin⁡C2\sin C common from both terms, we get
⇒\Rightarrow 2sin⁡C[cos⁡(A−B)+cos⁡C ]2\sin C\left[ {\cos (A - B) + \cos C{\text{ }}} \right]
Writing CC as 1800−(A+B){180^0} - (A + B) we get
⇒\Rightarrow 2sin⁡C[cos⁡(A−B)+cos⁡(1800−(A+B))]2\sin C\left[ {\cos (A - B) + \cos \left( {{{180}^0} - \left( {A + B} \right)} \right)} \right]
⇒2sin⁡C[cos⁡(A+B)+cos⁡(A+B)]\Rightarrow 2\sin C\left[ {\cos (A + B) + \cos (A + B)} \right]
Because cos⁡(1800−x)=cos⁡x\cos ({180^0} - x) = \cos x
So, we get after applying cos⁡(x−y)−cos⁡(x+y)=2sin⁡xsin⁡y\cos (x - y) - \cos (x + y) = 2\sin x\sin y
⇒2sin⁡C(2sin⁡Asin⁡B)\Rightarrow 2\sin C\left( {2\sin A\sin B} \right)
On multiplying, wet get 4sin⁡Asin⁡Bsin⁡C4\sin A\sin B\sin C
Which is the right hand side of the given equal that has to be proved.

Note: The Identities used were for two different angles whereas in the relation given between the angles there were three angles given firstly, we had right that three into two and one and then we had applied the identities on the two separately and on the third one separately. The identities used in the question are proved we can also prove them if we want to do so.