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evaluate: sinx. sin2x. sin3x

evaluate:
sinx. sin2x. sin3x

Grade:11

3 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
Hello Student,
Please find answer to your question below

You can solve in many ways which would depend on the options you want.
You can convert all in the form of sinx & cosx.
I am doing it by one way from which you can get the idea.
f(x) = sinx.sin2x.sin3x
f(x) = sin2x(sinx.sin3x)
sina.sinb = \frac{1}{2}(cos(a-b)-cos(a+b))
f(x) = sin2x.\frac{1}{2}(cos(x-3x)-cos(x+3x))
f(x) = sin2x.\frac{1}{2}(cos(2x)-cos(4x))
f(x) = \frac{1}{2}(sin(2x).cos(2x)-sin(2x).cos(4x))
f(x) = \frac{1}{2}(\frac{1}{2}sin(4x)-sin(2x).cos(4x))
lily
16 Points
8 years ago
thank you very much sir.
Shivam
13 Points
5 years ago
I = ? sin x. sin 2x. sin 3x dxI= (1/2) ? [ 2 sin 2x. sin x ] sin 3x dxI= (1/2) ? [ cos ( 2x-x) - cos ( 2x+x) ] sin 3x dxI= (1/2) ? ( cos x - cos 3x ) sin 3x dxI= (1/4) ? [ ( 2 sin 3x cos x ) - ( 2 sin 3x cos 3x ) ] dxI= (1/4) ? { [ sin ( 3x + x ) + sin ( 3x - x ) ] - ( sin 2·3x ) } dxI= (1/4) ? [ ( sin 4x + sin 2x ) - ( sin 6x ) ] dxI= (/4) { [ ( - cos 4x ) / 4 ] + [ ( - cos 2x ) / 2 ] - [ ( - cos 6x ) / 6 ] } + CI= ( -1/16) cos 4x - (1/8) cos 2x + (1/24) cos 6x + C

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