# Double Angle Formulae!!!

Just found this forum and wow it's good

Probably really simple how to do this but I'm stuck on what to do:-

By expressing sin3A as sin(2A + A), find an expression for sin3A in terms of sin A

Any help would be greatly appreciated

Thanks
sin(x+y)=sinxcosy+sinycosx

Let x=2A and y=A

=>sin(3A)=sin(2A+A)=sin2AcosA+sinAcos2A

* sin2A=2sinAcosA cos2A=(1-2sin²A)

=>sin(3A)=(2sinAcosA)cosA+sinA(1-2sin²A)

=>sin(3A)=2sinAcos²A+sinA-2sin³A

* cos²A=(1-sin²A)

=>sin(3A)=2sinA(1-sin²A)+sinA-2sin³A

=>sin(3A)=2sinA-2sin³A+sinA-2sin³A

=>sin(3A)=3sinA-4sin³A

=>sin(3A)=sinA(3-4sin²A)

Newton.
needmathshelp
Just found this forum and wow it's good

Probably really simple how to do this but I'm stuck on what to do:-

By expressing sin3A as sin(2A + A), find an expression for sin3A in terms of sin A

Any help would be greatly appreciated

Thanks

sin(2A+A)=sin2AcosA+sinacos2a
sin(2A+A)=2sinAcosAcosA+sinA(1-2 (sinA)^2)
sin(2a+a)=2 sinA( 1- (sinA)^2))+inA - 2 (sinA)~3

etc...now juat do a little tidying up......

mrm.
needmathshelp
Just found this forum and wow it's good

Probably really simple how to do this but I'm stuck on what to do:-

By expressing sin3A as sin(2A + A), find an expression for sin3A in terms of sin A

Any help would be greatly appreciated

Thanks

You have a standard formula:

sin(A+B) = sinAcosB + cosAsinB
Substitute
sin(2A+A)=sin2AcosA+cos2AsinA

Sin(2A) = 2sinAcosA
Cos(2A) = 1-2sin2A

So, we have

2sinAcosAcosA + (1-2sin2A)sinA
= 2sinA(1-sin2A) + sinA-2sin3A
= 2sinA - 2sin3A + sinA - 2sin3A
= 3sinA - 4sin3A.
= sinA (3 - 4sin2A)
Newton

Newton.

I didn't, I was busy typing, got distracted by the phone and completed my post - by which time two people have already answered, and within a minute I've got you jumping on my back. Marvellous.
Wow cheers guys, it's all clear now

And thanks for not making fun of me cause i didn't get it where as you probably found it easy.
sin(3A) = sin(2A+A) = sin2AcosA+cos2AsinA

Substituting
Sin(2A) = 2sinAcosA
Cos(2A) = 1-2sin²A

You get

2sinAcos²A + (1-2sin²A)sinA
= 2sinA(1-sin²A) + sinA-2sin³A
= 2sinA - 2sin³A + sinA - 2sin³A
= 3sinA - 4sin³A

Keep 'em coming
sin(3A)
= sin(2A+A) = sin2AcosA+cos2AsinA

Sub into equations above:
Sin(2A) = 2sinAcosA
Cos(2A) = 1-2sin²A

2sinAcos²A + (1-2sin²A)sinA
= 2sinA(1-sin²A) + sinA-2sin³A
= 2sinA - 2sin³A + sinA - 2sin³A
= sinA(3 - 4sin²A)