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Pre-Calculus Help

Fireball77

Senior member
I have been working on this problem for over 3 days now. I cannot seem to get it. Any ideas of where I might be over looking something.

I have the Triple Angle Formula: 3sinx-4sin^3x
I have the Double Angle Formula: 2sinxcosx
and I have the addition formula: sin(a+b) =sin(a)cos(b) + cos(a)sin(b)

I keep getting stuck, how do i get rid of the cos? I think where my problem is, is the FOIL of the problem after I plug in my angle formulas.

Prove that Sin5x=5sinx-20sin^3x+16sin^5x; x=theta

sin(3x+2x) = sin(3sinx-4sin^3x)cos(2sinxcosx) + cos(3sinx-4sin^3x)sin(2sinxcosx)

and hints or clues, would be highly appreciated.
 
Originally posted by: Fireball77
Prove that Sin5x=5sinx-20sin^3x+16sin^5x; x=theta

sin(3x+2x) = sin(3sinx-4sin^3x)cos(2sinxcosx) + cos(3sinx-4sin^3x)sin(2sinxcosx)

Why?
What's the point?
At least you're approaching it like I would, get someone else to do it. 😉
 
Originally posted by: Fireball77
I have been working on this problem for over 3 days now. I cannot seem to get it. Any ideas of where I might be over looking something.

I have the Triple Angle Formula: 3sinx-4sin^3x
I have the Double Angle Formula: 2sinxcosx
and I have the addition formula: sin(a+b) =sin(a)cos(b) + cos(a)sin(b)

I keep getting stuck, how do i get rid of the cos? I think where my problem is, is the FOIL of the problem after I plug in my angle formulas.

Prove that Sin5x=5sinx-20sin^3x+16sin^5x; x=theta

sin(3x+2x) = sin(3sinx-4sin^3x)cos(2sinxcosx) + cos(3sinx-4sin^3x)sin(2sinxcosx)

and hints or clues, would be highly appreciated.

How did you get that result? I'm pretty sure that's wrong.
 
sin(2x+3x)=sin(2x)cos(3x)+sin(3x)cos(2x)

Use double and triple angle identities from there....be careful, cause you seem to have used the addition formula wrong.
 
Originally posted by: chuckywang
sin(2x+3x)=sin(2x)cos(3x)+sin(3x)cos(2x)

Use double and triple angle identities from there....be careful, cause you seem to have used the addition formula wrong.


damn. i was too slow. what ^^ said
 
sin(3x+2x) = sin(3sinx-4sin^3x)cos(2sinxcosx) + cos(3sinx-4sin^3x)sin(2sinxcosx)

Isnt this wrong..

sin (3x+2x) = sin3xcos2x + cos3xsin 2x
= (3sinx-4sin^3x)(2cos^2x-1) + .......

which is diff from sin(3sinx - 4sin^x)...........

Am i right?
 
Why not try:

sin(5x) = sin(4x + x) = sin(4x)cos(x) + cos(4x)sin(x)
= sin (2 ( 2x))cos(x) + cos(2 * 2(x))sin(x)
= 2sin(2x)(cos(2x)cos(x)+ [cos^2(2x)-sin^2(2x)]sin(x)

See what you get from there.
 
Originally posted by: TuxDave
Originally posted by: Fireball77
I have been working on this problem for over 3 days now. I cannot seem to get it. Any ideas of where I might be over looking something.

I have the Triple Angle Formula: 3sinx-4sin^3x
I have the Double Angle Formula: 2sinxcosx
and I have the addition formula: sin(a+b) =sin(a)cos(b) + cos(a)sin(b)

I keep getting stuck, how do i get rid of the cos? I think where my problem is, is the FOIL of the problem after I plug in my angle formulas.

Prove that Sin5x=5sinx-20sin^3x+16sin^5x; x=theta

sin(3x+2x) = sin(3sinx-4sin^3x)cos(2sinxcosx) + cos(3sinx-4sin^3x)sin(2sinxcosx)

and hints or clues, would be highly appreciated.

How did you get that result? I'm pretty sure that's wrong.



Yeah I have NO idea how you even came up with that...

--Mark
 
Damn, I've got Pre-Cal next year. I hope I don't have to resort to asking ATOT for help. That's like asking a guy looking at Jessica Alba pr0n not to get a hardon.
 
i found my mistake, i should of started with this

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

thanks for all your help guys.
 
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