Trigonometry> evaluate: sinx. sin2x. sin3x...

evaluate:
sinx. sin2x. sin3x

lily , 11 Years ago

Grade 11

3 Answers

Jitender Singh

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.
$f(x) = sinx.sin2x.sin3x$
$f(x) = sin2x(sinx.sin3x)$
$sina.sinb = \frac{1}{2}(cos(a-b)-cos(a+b))$
$f(x) = sin2x.\frac{1}{2}(cos(x-3x)-cos(x+3x))$
$f(x) = sin2x.\frac{1}{2}(cos(2x)-cos(4x))$
$f(x) = \frac{1}{2}(sin(2x).cos(2x)-sin(2x).cos(4x))$
$f(x) = \frac{1}{2}(\frac{1}{2}sin(4x)-sin(2x).cos(4x))$

Last Activity: 11 Years ago

lily

thank you very much sir.

Last Activity: 11 Years ago

Shivam

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

Last Activity: 8 Years ago