-2cos(x)sin(x) = -sin(2x)

Thanks,

 

Don't have a  teacher, just teaching myself haha.

 

I am 3/5 through the calculus book, nearing integral calculus (it is the next topic). It has started to give me minima/maxima questions (to solve using 2nd derivatives), and trig functions have started popping up (probably reintroducing for integrals).

 

I'll practice deriving these identities to become more familiar with them, and when solving these problems I'll derive each one instead of memorizing to hopefully become more familiar with them.

sin(A+B)=sinACosB-SinBcosA        (This is an identity that you need to memorize.)

 

therefore

sin(2x)=2cos(x)sin(x) 

therefore

-sin(2x)=-2cos(x)sin(x) 

 

Do you want a proof for the original identity - I mean for the first line?

Ah. So are the 'double' and 'half' angle identities both derived from the angle addition/difference identities?

 

So the most important identities to remember would be the sum/difference identities.

 

sin(a + b) = sin(a)cos(b) + cos(a)sin(b) 
sin(a – b) = sin(a)cos(b) – cos(a)sin(b) 
cos(a + b) = cos(a)cos(b) – sin(a)sin(b) 
cos(a – b) = cos(a)cos(b) + sin(a)sin(b)

 

As well as:

sin^2(x) + cos^2(x) = 1 (the other identities are easily derived from this).

 

So most functions with some trig function can be solved using these 2 sets of identities?

 

This function popped up towards the end of my derivatives chapter, and the book on trig barely covered those identities at all! :( (it mentioned the sin(a+b) identity but never used it).

 

Thanks!

Yes, they are the main ones.

A lot can be derived just from these.

Your teacher may push you to remember a whole heap more - I mean ones that you can derive.

It depends how quick you can get as to whether you memorize more or not.

 

You probably haven't done any calculus yet but I am just youwing you the bit where you will need to rearrage trig.

 

In calculus you may be asked to integrate cos^2x

You can't do this unless you change it.

 

\(cos(2x)=sin^2x-cos^2x\\ cos(2x)=1-cos^2x-cos^2x\\ cos(2x)=1-2cos^2x\\ 1-2cos^2x=cos(2x)\\ 1-cos(2x)=2cos^2x\\ cos^2x=\frac{1-cos(2x)}{2}\\ cos^2x=\frac{1}{2}-\frac{cos(2x)}{2}\\\)

The RHS you can integrate.  

Now you will probably be told to memorize this resultant rearrangement.  

I work it out EVERY time.  But my method is slow.

Wake up Melody !

 

Three basic identities in #1 and #3 are incorrect.