Flag Trigonometry> evaluate: sinx. sin2x. sin3x...
question mark

evaluate:
sinx. sin2x. sin3x

lily , 10 Years ago
Grade 11
anser 3 Answers
Jitender Singh

Last Activity: 10 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

Last Activity: 10 Years ago

thank you very much sir.

Shivam

Last Activity: 7 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

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments