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trig identities (1 Viewer)
- Thread starter kazemagic
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Timske
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<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
what is that?It is Cowpea
An easier way would be to notice it's in the expanded compound angle form, and so you only have to contract the expression:Need help again =.=
How do I do question a?
Remember:
In question (a),
Therefore it is just:
Try to recognise these things.
Carrotsticks
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This is useful for questions like: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.
Simplify:
sin(3x)/cos(x) + cos(3x)/sin(x).