solve sin3x=(sinx)(3-4sin^2x) the whole way through please

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bballbink

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Feb 21, 2009
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i am having alot of trouble, if u can solve this it will make my day.

sin3x=(sinx)(3-4sin^2x)
sin3x= sin(2x+x)
sin2x cosx+ cos2x sinx
 
Re: solve the whole way through please

sin3x=(sinx)(3-4sin^2x)
sin3x= sin(2x+x)
sin2x cosx+ cos2x sinx
2sinx cosx cosx + 1-sin^2x sinx
2sinx cos^2x +1-sin^2x sinx
3sinx cos^x +1-2sin^2x
3sinx+ 1-4sin^2x
sinx(3-4sin^x)

is this right
 
Re: solve the whole way through please

bballbink said:
sin3x=(sinx)(3-4sin^2x)
sin3x= sin(2x+x)
sin2x cosx+ cos2x sinx
= 2 sin(x)cos(x)cosx + [1 - 2sin^2(x)] sin(x)

= 2 sin(x)cos^2(x) + [1 - 2sin^2(x)] sin(x)

= 2 sin(x)[1 - sin^2(x)] + [1 - 2sin^2(x)] sin(x)

Now continue....

Do not perform random operations and expect to be correct


2sinx cosx cosx + 1-sin^2x sinx
2sinx cos^2x +1-sin^2x sinx
3sinx cos^x +1-2sin^2x
3sinx+ 1-4sin^2x
sinx(3-4sin^x)

is this right
Click to expand...

Not quite.... You are doing random operations without paying attention to the result.
 
Re: solve the whole way through please

is there any way you could solve it the whole way through maybe i will understand once i just visualize it. thanks
 
Re: solve the whole way through please

bballbink said:
is there any way you could solve it the whole way through maybe i will understand once i just visualize it. thanks
Click to expand...

Just distribute and multiply - same as any algebra problem.
 
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