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Transform sum sin3x+2sin2x+sinx in product

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minlux | Honors

Posted September 23, 2013 at 5:41 PM via web

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Transform sum sin3x+2sin2x+sinx in product

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 23, 2013 at 6:09 PM (Answer #1)

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You need to convert the given summation into a product, hence, you may form the following groups, such that:

`sin 3x + 2sin 2x + sin x = (sin 3x + sin x) + (sin 2x + sin 2x)`

You should notice that the term `2sin 2x` is expressed as the summation `sin 2x + sin 2x` .

The selection of terms in the two groups is decided with respect to the summation of arguments, such that:

`3x + x = 2x + 2x = 4x`

You need to convert the summations of sines into a product, using the following conversion formula, such that:

`sin a + sin b = 2 sin (a/2 + b/2)*cos(a/2 - b/2)`

Reasoning by analogy, yields:

`sin 3x + sin x = 2sin ((3x + x)/2)*cos((3x - x)/2)`

`sin 3x + sin x = 2sin(2x)*cos x`

`sin 2x + sin 2x = 2sin ((2x + 2x)/2)*cos((2x - 2x)/2)`

`sin 2x + sin 2x = 2sin(2x)*cos 0`

Since `cos 0 = 1` , yields:

`sin 2x + sin 2x = 2sin(2x)`

Replacing `2sin(2x)*cos x` for `sin 3x + sin x` and `2sin(2x)` for `sin 2x + sin 2x` yields:

`sin 3x + 2sin 2x + sin x = 2sin(2x)*cos x +2sin(2x)*cos 0`

Factoring out `2sin(2x)` yields:

`sin 3x + 2sin 2x + sin x = 2sin(2x)*(cos x + cos 0)`

You need to convert the summation `cos x + cos 0` into a product, such that:

`cos x + cos 0 = 2cos((x+0)/2)*cos((x-0)/2)`

`cos x + cos 0 = 2cos^2(x/2)`

`sin 3x + 2sin 2x + sin x = 2sin(2x)*2cos^2(x/2)`

`sin 3x + 2sin 2x + sin x = 4sin (2x)*cos^2(x/2)`

You may use the double angle formula, such that:

`sin 3x + 2sin 2x + sin x = 4(2 sin x*cos x)*cos^2(x/2)`

`sin 3x + 2sin 2x + sin x = 8sin x*cos x*cos^2(x/2)`

Hence, converting the given summation into a product, under the given conditions, yields `sin 3x + 2sin 2x + sin x = 8sin x*cos x*cos^2(x/2).`

Sources:

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aruv | High School Teacher | Valedictorian

Posted September 23, 2013 at 6:41 PM (Answer #2)

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`sin(3x)+2sin(2x)+sin(x)`

`=sin(3x)+sin(2x)+sin(2x)+sin(x)`

`since`

`sin(a)+sin(b)=2sin((a+b)/2)cos((a-b)/2)`

Therefore

`=2sin((3x+2x)/2)cos((3x-2x)/2)+2sin((2x+x)/2)cos((2x-x)/2)`

`=2sin((5x)/2)cos(x/2)+2sin((3x)/2)cos(x/2)`

`=2cos(x/2){sin((5x)/2)+sin((3x)/2)}`

`=2cos(x/2){2sin(((5x)/2+(3x)/2)/2)cos(((5x)/2-(3x)/2)/2)`

`=4cos(x/2)sin((8x)/4)cos((2x)/4)`

`=4sin(2x)cos(x/2)cos(x/2)`

`=4sin(2x)cos^2(x/2)`

`Thus`

`sin(3x)+2sin(2x)+sin(x)=4sin(2x)cos^2(x/2)`

Which is required product form.

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