Prove that: sin3x+sin2x-sinx = $$4sinxcos\dfrac{x}{2}cos\dfrac{3x}{2}$$

Asked by Pragya Singh | 1 year ago |  121

#### 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= $$2sin\dfrac{3x}{2}.cos\dfrac{3x}{2}+2cos\dfrac{3x}{2}sin\dfrac{x}{2}$$

$$2cos(\dfrac{3x}{2})[sin(\dfrac{3x}{2})+sin(\dfrac{x}{2})]$$

$$2cos(\dfrac{3x}{2}).2sinxcos(\dfrac{x}{2})$$

L.H.S

=$$4sinxcos (\dfrac{x}{2})cos(\dfrac{3x}{2})$$ =R.H.S

Answered by Abhisek | 1 year ago

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