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How to prove trig identity (sin3x/sinx)-(cos3x/cosx)=2  

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sunithasrivas... | High School Teacher | (Level 1) Honors

Posted April 25, 2013 at 1:55 AM via web

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How to prove trig identity

(sin3x/sinx)-(cos3x/cosx)=2

 

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oldnick | (Level 1) Valedictorian

Posted April 25, 2013 at 2:11 AM (Answer #1)

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`(sin3x)/sinx - (cos3x)/cosx=2`

 

`sin3xcosx-cos3xsinx=2sinxcosx`

`sin(3x-x)= sin2x`

`sin2x= sin 2x`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted April 25, 2013 at 3:34 AM (Answer #2)

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`sin(3x)/sinx-cos(3x)/cosx`

`= [sin(3x)cosx]/(sinxcosx)-[cos(3x)sinx]/(sinxcosx)`

`= (sin(3x)cosx-cos(3x)sinx)/(sinxcosx)`

 

We know that;

`sin2A = 2sinAcosA`

`sinAcosA= 1/2sin2A`

 

`sin(A-B) = sinAcosB-cosAsinB`

Let;

`A = 3x`

`B = x`

 

Then;

`sin(3x-x) = sin3xcosx-cos3xsinx`

`sin2x = sin3xcosx-cos3xsinx`

 

Therefore we can say;

`sin(3x)/sinx-cos(3x)/cosx`

`= (sin(3x)cosx-cos(3x)sinx)/(sinxcosx)`

`= (sin(2x))/(1/2sin2x)`

`= 2`

 

So the identity is proved.

`sin(3x)/sinx-cos(3x)/cosx = 2`

 

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