Math, asked by angleaayushi21, 8 months ago

prove sin 3a + sin 2a - sin a = 4sin a cos a/2 cos 3a/2​

Answers

Answered by senboni123456
2

Step-by-step explanation:

Let, the argument be 'α',i.e, let a=α

then we have,

 \sin( 3\alpha )  +  \sin(2 \alpha ) -   \sin( \alpha )

 =  \sin(3 \alpha )   + 2 \cos( \frac{2 \alpha   + \alpha }{2} )  \sin( \frac{2 \alpha  -  \alpha }{2} )

 = 2 \sin( \frac{ 3\alpha }{2} )  \cos( \frac{3 \alpha }{2} ) + 2 \cos( \frac{3 \alpha }{2} )  \sin( \frac{ \alpha }{2} )

 = 2 \cos( \frac{3 \alpha }{2} ) ( \sin( \frac{3 \alpha }{2} ) +   \sin( \frac{ \alpha }{2} ) )

 = 2 \cos( \frac{3 \alpha }{2} )(2 \sin( \frac{ \frac{3 \alpha }{2} +  \frac{ \alpha }{2}  }{2} )  \cos( \frac{ \frac{3 \alpha }{2} -  \frac{ \alpha }{2}  }{2} ) )

 = 4 \sin( \alpha )  \cos( \frac{3 \alpha }{2} )  \cos( \frac{ \alpha }{2} )

Hence proved

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