prove that sinA.sin(B−C)+sinB.sin(C−A)+sinC.sin(A−B)=0

0 July 30, 2021

sinA.sin(B-C)+sinB.sin(C-A)+sinC.sin(A-B)=0

Solution:-

$= \sin A.\sin (B - C) + \sin B.\sin (C - A) + \sin C.\sin (A - B)$

$= \sin A(\sin B.\cos C - \cos B.\sin C) + \sin B(\sin C.\cos A - \cos C.\sin A)$

$+ \sin C(\sin A.\cos B - \cos A.\sin B)$

$= \sin A.\sin B.\cos C - \sin A.\cos B.\sin C + \sin B.\sin C.\cos A -$

$\sin B.\cos C.\sin A + \sin C.\sin A.\cos B - \sin C.\cos A.\sin B$

$\therefore L.H.S = R.H.S$

$\{ proved\}$

How To Solve : -

In first step the formula of sin(a-b)=sina.cosb-cosa.sinb is putted.
This step is simply easy of multiplication.
Arrange the following in regular manner and + and - terms are canceled out.
Hence the Answer is 0
Finally we proved that sinA.sin(B-C)+sinB.sin(C-A)+sinC.sin(A-B)=0

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prove that sinA.sin(B-C)+sinB.sin(C-A)+sinC.sin(A-B)=0

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