Question

Can anyone prove this please?

#{(b^2-c^2)//a^2} xx sin 2A + {(c^2-a^2)//b^2}xx sin 2B + {(a^2-b^2)//c^2} xx sin 2C# =0

Answer 1

Use the sine law for triangles and some simple trigonometric identities.

Explanation:

From the sine law of triangles

`a/{sin A}= b/{sin B} = c/{sin C}`

we can easily see that

`{b^2 -c^2}/a^2 = {sin^2B-sin^2C}/sin^2A= {(sin B-sinC)(sin B+ sin C)}/{sin^2A} = {2 sin ({B-C}/2)cos({B+C}/2)times 2 sin({B+C}/2)cos({B-C}/2)}/sin^2A = {sin(B-C)sin(B+C)}/sin^2A = {sin(B-C)sin(pi-A)}/sin^2A = sin(B-C)/sinA`

So that

`{b^2 -c^2}/a^2 times sin2A = 2cosAsin(B-C) = 2 cosAsinBcosC-2cosAcosBsinC`

The other two terms can be obtained from this one by simply cyclically permuting `A`, `B` and `C`. Adding the three terms leads to the proof trivially .

Answer 2

Please see below.

Explanation:

The first term of `LHS=(b^2-c^2)/a^2*sin2A`

`=(4R^2[sin^2A-sin^2B])/(4R^2*sin^2A)*sin2A`

`=(sin(B+C)sin(B-C))/sin^2A*sin2A`

`=(sinAsin(B-C))/(sinA*sinA)*2sinA*cosA`

`=2cosAsin(B-C)`

`=sin(A+B-C)-sin(A-B+C)`

`=sin(pi-2C)-sin(pi-2B)=sin2C-sin2B`

Similarly The second term`=sin2A-sin2B` and

The third term`=sin2B-sin2A`

Whole `LHS=sin2C-sin2B+sin2A-sin2C+sin2B-sin2C=0`

Note that `sin^2A-sin^2B=sin(A+B)*sin(A-B)`

Answer 3

Kindly refer to the Explanation.

Explanation:

Prerequisites : In the usual Notation for `DeltaABC,`

Sine-Rule : `a/sinA=2R, or, sinA=a/(2R)`.

Cosine-Rule : `cosA=(b^2+c^2-a^2)/(2bc)`.

We have, `(b^2-c^2)/a^2*sin2A=(b^2-c^2)/a^2*(2sinAcosA)`,

`=(b^2-c^2)/a^2*{2*a/(2R)*(b^2+c^2-a^2)/(2bc)}`,

`={(b^2-c^2)(b^2+c^2-a^2)}/(Rabc)`,

`={(b^2-c^2)(b^2+c^2)-a^2(b^2-c^2)}/(Rabc)`,

`rArr(b^2-c^2)/a^2*sin2A={(b^4-c^4)-a^2(b^2-c^2)}/(Rabc)`.

Obtaining similar expressions for the remaining terms of the left

member and adding them, the result follows.