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Sin(3x),(4x),(5x) or Cos(3x),(4x),(5x)???

futureaussiecto

Can someone please explain to me how to use the double angle formulae to work these out?

Reply 1

saiyamanadingdongbanana

futureaussiecto

Can someone please explain to me how to use the double angle formulae to work these out?

*gets out formulae booklet*

ok, so the single angle formula is;

sin (A + B) = sin A cos B + sin B cos A

which is in the formula book.

now ok, the double angle formula [sin(2x)] is basically;

sin (x + x) = sin x cos x + sin x cos x

which is also

sin (2x) = 2sin x cos x

so, if we were to look at

sin(3x)

this is basically;

sin(1.5x + 1.5x) = ...

so do the subbing in into the single angle formula;

2 sin (1.5x) cos (1.5x)

so therefore;

Sin(3x) = 2 sin 1.5x cos 1.5x

checking values;

let x=2

sin(3*2) = 2(sin (1.5*2) * cos (1.5*2)

which is right

Reply 2

futureaussiecto

see i understand that

but in my book it says

Sin3A

=sin(2A+A)
=Sin2AcosA+cos2ASinA

which threw me

Reply 3

Speleo

No you have to find it in terms of sinx.
Basically, use the double angle formulae to express sin3x in terms of sin2x and sinx, then use the double angle formulae again to get the sin2x down to sinx.

Reply 4

Dekota

futureaussiecto

see i understand that

but in my book it says

Sin3A

=sin(2A+A)
=Sin2AcosA+cos2ASinA

which threw me

Yes that is true.

Reply 5

Dez

That formula from your book is just using the addition formulae instead of the double angle ones, perfectly acceptable, and a bit easier for odd numbers I guess. A and B are just numbers, if you get 5A just let it be C or something and carry on. Nothing sacred about letters.

Reply 6

evariste_3086

futureaussiecto

Can someone please explain to me how to use the double angle formulae to work these out?

sin(2x)=sin(x+x)=sinxcosx+cosxsinx=2sinxcosx
cos(2x)=cos(x+x)=cosxcosx-sinxsinx=cos^2x-sin^2x
sin(3x)=sin(2x+x)=sin2xcosx+cos2xsinx,,,now use expressions for cos(2x) and sin(2x)
cos(3x)=cos(2x+x) etc
sin(4x)=sin(2x+2x) or sin(3x+x) etc

Reply 7

saiyamanadingdongbanana

futureaussiecto

see i understand that

but in my book it says

Sin3A

=sin(2A+A)
=Sin2AcosA+cos2ASinA

which threw me

ahh

ok, i did this;

sin (1.5x + 1.5x) = ...

all they did was say

sin (2x + x) = ...

and worked it out like that instead.

in both cases, the 3x was split up into two parts, but they were split differently.

Reply 8

Libertine

The idea is you break it down so there's only terms involving sinx or cosx, no sin2x or sin3x etc. So just use the addition formulae to get it in terms of sin2x and cos2x, and the double-angle formulae to break it down further. It just involves a lot of substitution.

Reply 9

futureaussiecto

so sin(1x) is just sin x?

if so, isnt sin(3x) just sin(2x) [which is 2sinxcosx} + sinx??

Reply 10

Dekota

futureaussiecto

so sin(1x) is just sin x?

Yes/

if so, isnt sin(3x) just sin(2x) [which is 2sinxcosx} + sinx??

No/

Reply 11

Libertine

futureaussiecto

if so, isnt sin(3x) just sin(2x) [which is 2sinxcosx} + sinx??

No:

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

Reply 12

futureaussiecto

Libertine

No:

sin(3x)
= sin (2x + x)
= sin2x.cosx + cos2x.sinx

ok from that it looks like

sin2x = sin2xcosx
and
sinx = cos2xsinx

:eek:

Reply 13

Dekota

futureaussiecto

ok from that it looks like

sin2x = sin2xcosx
and
sinx = cos2xsinx

:eek:

sin (2x + x) does not expand to sin2x + sinx

Never do that with trig functions.

Reply 14

futureaussiecto

Dekota

sin (2x + x) does not expand to sin2x + sinx

Never do that with trig functions.

yeh i know, i was just saying from the prev quote that when it said

sin(2x + x) = sin2x.cosx + cos2x.sinx

that (2x) is the sin2x.cosx bit

and

(x) is the cos2x.sinx

Reply 15

Libertine

futureaussiecto

yeh i know, i was just saying from the prev quote that when it said

sin(2x + x) = sin2x.cosx + cos2x.sinx

that (2x) is the sin2x.cosx bit

and

(x) is the cos2x.sinx

But it's not. The whole expression on the right is equal to the whole expression on the right. The bits before and after the addition signs don't correspond.