trig identities (1 Viewer)

[Sy123]

Best not to care about my method, I did the grunt work, Spiral was able to recognise a good way of doing this.

 

[kazemagic]

Need help again =.=
How do I do question a?

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=cos3xcosx @plus; sin3xsinx \\\\ = cos(2x@plus;x)cos@plus;sin(2x@plus;x)sinx \\\\ = cosx(cos2xcosx-sin2xsinx)@plus;sinx(sin2xcosx@plus;cos2xsinx)\\\\=cos2xcos^2x-sin2xsinxcosx@plus;sin2xsinxcosx@plus;cos2xsin^2x\\\\ =cos2xcos^2x@plus;cos2xsin^2x \\\\ =cos2x(cos^2x@plus;sin^2x) = cos2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?cos3xcosx + sin3xsinx \\\\ = cos(2x+x)cos+sin(2x+x)sinx \\\\ = cosx(cos2xcosx-sin2xsinx)+sinx(sin2xcosx+cos2xsinx)\\\\=cos2xcos^2x-sin2xsinxcosx+sin2xsinxcosx+cos2xsin^2x\\\\ =cos2xcos^2x+cos2xsin^2x \\\\ =cos2x(cos^2x+sin^2x) = cos2x" title="cos3xcosx + sin3xsinx \\\\ = cos(2x+x)cos+sin(2x+x)sinx \\\\ = cosx(cos2xcosx-sin2xsinx)+sinx(sin2xcosx+cos2xsinx)\\\\=cos2xcos^2x-sin2xsinxcosx+sin2xsinxcosx+cos2xsin^2x\\\\ =cos2xcos^2x+cos2xsin^2x \\\\ =cos2x(cos^2x+sin^2x) = cos2x" /></a>

 

[kazemagic]

<a href="http://www.codecogs.com/eqnedit.php?latex=cos3xcosx @plus; sin3xsinx \\\\ = cos(2x@plus;x)cos@plus;sin(2x@plus;x)sinx \\\\ = cosx(cos2xcosx-sin2xsinx)@plus;sinx(sin2xcosx@plus;cos2xsinx)\\\\=cos2xcos^2x-sin2xsinxcosx@plus;sin2xsinxcosx@plus;cos2xsin^2x\\\\ =cos2xcos^2x@plus;cos2xsin^2x \\\\ =cos2x(cos^2x@plus;sin^2x) = cos2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?cos3xcosx + sin3xsinx \\\\ = cos(2x+x)cos+sin(2x+x)sinx \\\\ = cosx(cos2xcosx-sin2xsinx)+sinx(sin2xcosx+cos2xsinx)\\\\=cos2xcos^2x-sin2xsinxcosx+sin2xsinxcosx+cos2xsin^2x\\\\ =cos2xcos^2x+cos2xsin^2x \\\\ =cos2x(cos^2x+sin^2x) = cos2x" title="cos3xcosx + sin3xsinx \\\\ = cos(2x+x)cos+sin(2x+x)sinx \\\\ = cosx(cos2xcosx-sin2xsinx)+sinx(sin2xcosx+cos2xsinx)\\\\=cos2xcos^2x-sin2xsinxcosx+sin2xsinxcosx+cos2xsin^2x\\\\ =cos2xcos^2x+cos2xsin^2x \\\\ =cos2x(cos^2x+sin^2x) = cos2x" /></a>

thanks pikachu

 

[RealiseNothing]

Need help again =.=
How do I do question a?

An easier way would be to notice it's in the expanded compound angle form, and so you only have to contract the expression:

Remember:



In question (a), and

Therefore it is just:



Try to recognise these things.

 

[Timske]

lol

 

[Carrotsticks]

An easier way would be to notice it's in the expanded compound angle form, and so you only have to contract the expression:

Remember:



In question (a), and

Therefore it is just:



Try to recognise these things.

This is useful for questions like:

Simplify:

sin(3x)/cos(x) + cos(3x)/sin(x).