1 Answer
Solution :-
We know that,
sinA+sinB=2sin\( ( \dfrac{A+B}{2}).cos( \dfrac{A-B}{2})\)
sinA-sinB=2sin\( ( \dfrac{A-B}{2}).cos( \dfrac{A+B}{2})\)
L.H.S.=sin3x+sin2x-sinx
\( sin3x+[2cos(\dfrac{2x+x}{2})(sin\dfrac{2x-x}{2})]\)
\( sin3x+[2cos(\dfrac{3x}{2})(sin\dfrac{x}{2})]\)
Since we know that, sin2x=2sinxcosx
L.H.S= \( 2sin\dfrac{3x}{2}.cos\dfrac{3x}{2}+2cos\dfrac{3x}{2}sin\dfrac{x}{2}\)
\( 2cos(\dfrac{3x}{2})[sin(\dfrac{3x}{2})+sin(\dfrac{x}{2})]\)
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= \( 2cos(\dfrac{3x}{2}).2sinxcos(\dfrac{x}{2})\)
L.H.S
=\(4sinxcos (\dfrac{x}{2})cos(\dfrac{3x}{2})\) =R.H.S
Answered by Abhisek | 1 year agoRelated Questions
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