Identity

**topsquark** · April 10th 2008, 03:36 AM

This is a question that Thelema asked me:

cos(s+t)cos(s-t)=cos^2s +cos^2t-1

so far i am at
[cos(s)cos(t)-sin(s)sin(t)][cos(s)cos(t)+sin(s)sin(t)] = cos^2(s)+cos^2(t)-1

I can multiply the left side but then i dont know
how to get it to equal the right. Can u plz help?

I don't have time to help at the moment so I'm asking if someone else can. Thanks!

-Dan

**angel.white** · April 10th 2008, 03:52 AM

From:
$[cos~s * cos~t-sin~s*sin~t][cos~s*cos~t+sin~s*sin~t] = cos^2s+cos^2t-1$

Distribute (FOIL) and simplify:
$cos^2s*cos^2t - sin^2s*sin^2t~~~~=~~~~ cos^2s+cos^2t-1$

Use $sin^2x = 1-cos^2x$ . We choose this one because our answer has cosines in it, and so we know we need to get rid of our sines.
$cos^2s*cos^2t - (1-cos^2s)(1-cos^2t)~~~~=~~~~ cos^2s+cos^2t-1$

Distribute:
$cos^2s*cos^2t - (1-cos^2t-cos^2s+cos^2s*cos^2t)~~~~=~~~~ cos^2s+cos^2t-1$

Distribute the negative sign:
$cos^2s*cos^2t -1+cos^2t+cos^2s-cos^2s*cos^2t~~~~=~~~~ cos^2s+cos^2t-1$

Simplify:
$cos^2s+cos^2t-1~~~~=~~~~ cos^2s+cos^2t-1$

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