[SOLVED] Double Angle Identities

**Evales** · June 26th 2008, 01:32 PM

I have to make use of:
$sin2x = 2sinxcosx$
$cos2x = cos^2x - sin^2x$
$sin3x = 3sinx - 4sin^3x$
$cos3x = 4cos^3x - 3cosx$

In order to prove:
$sin3x = 4sinxcos^2x - sinx$
$sin4x = 8sinxcos^3x - 4sinxcosx$
ect.

It looks like they are connected but I just can't see how they are connected.
Below is some of my working.

$sin3x = 4sinxcos^2x - sinx$
$RHS = 2(2sinxcosx)cosx - sinx$
$RHS = (sin2x)2cosx - sinx$

I am then promptly stumped

Should I be working with LHS, what sort of substitutions should I be doing?

**Moo** · June 26th 2008, 01:39 PM

Hello,

Originally Posted by Evales

I have to make use of:
$sin2x = 2sinxcosx$
$cos2x = cos^2x - sin^2x$
$sin3x = 3sinx - 4sin^3x$
$cos3x = 4cos^3x - 3cosx$

In order to prove:
$sin3x = 4sinxcos^2x - sinx$
$sin4x = 8sinxcos^3x - 4sinxcosx$
ect.

It looks like they are connected but I just can't see how they are connected.
Below is some of my working.

$sin3x = 4sinxcos^2x - sinx$
$RHS = 2(2sinxcosx)cosx - sinx$
$RHS = (sin2x)2cosx - sinx$

I am then promptly stumped

Should I be working with LHS, what sort of substitutions should I be doing?

Hmm for $\sin(3x)$ , you can study both the RHS or the LHS.

$RHS=4sinxcos^2x-sinx=sinx(4cos^2x-1)$

But we know that $cos^2x=1-sin^2x$

--> $RHS=sinx(4(1-sin^2x)-1)=sinx(4-4sin^2x-1)=$ $sinx(3-4sin^2x)=3sinx-4sin^3x=sin(3x)$

I guess it will be the same method for sin(4x) : juggle with the given elements and the basic properties

**bleesdan** · June 26th 2008, 01:41 PM

You may also need trigonometric sum formulas.
I'm fishy on how to add Latex in a forum post, so you get a link!
Trigonometric Addition Formulas -- from Wolfram MathWorld

Then, note that sin(3x)=sin(x+2x)
Mess with the LHS, and remember all of your identities!!!

**Evales** · June 26th 2008, 01:44 PM

Yeh my teacher was saying that I can do it using only the identities given but I ended up using $1 = sin^2x + cos^2x$ Thanks a bunch guys!

**mr fantastic** · June 26th 2008, 03:54 PM

Originally Posted by bleesdan

You may also need trigonometric sum formulas.
I'm fishy on how to add Latex in a forum post, so you get a link!
Trigonometric Addition Formulas -- from Wolfram MathWorld

Then, note that sin(3x)=sin(x+2x)
Mess with the LHS, and remember all of your identities!!!

Any 'messing' to be done happens on the RHS (Right Hand Side).

**nikhil** · June 26th 2008, 10:33 PM

Hi!Evales,hope ur having a great time.
RHS side of the identity is in single angle form and the LHS is in a multiple angle form. Main objective is to convert multiple angle to single angle.
You may use addition formula for example sin3x=sin(x+2x),
sin4x=sin(3x+x) or sin(2x+2x).when you will expand them,the expansion will involve multiple angle(2x or 3x) now again use addition formula to convert the multiple angle to single angle.keep on doing this until all multiple angles are converted into single angle.
NOTE:conversly single angle can also be converted into multiple angle eg sinx=sin(2x-x) or sin(3x-2x) etc.
Hope this will help.

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