Posted by hayden on Sunday, February 22, 2009 at 6:49pm.
cos^2 (x+a)-sin^2(a)= (cos x)(cos (2a+x))
prove using only side i worked on the left side...
Nasty nasty proof - Precalc and trig - Reiny, Sunday, February 22, 2009 at 8:10pm
I worked on both sides,
LS = (cosxcosa – sinxsina)^2 – sin^2a
= cos^2xcos^2a – 2cosxcosasinxsina + sin^2xsin^2a – sin^2a
= cos^2xcos^2a – 2cosxcosasinxsina + sin^2a(sin^2x – 1)
= cos^2xcos^2a – 2cosxcosasinxsina – sin^2acos^2x
= cos^2x(cos^2a – sin^2a) – 2cosxcosasinxsina
= cosx[cos2acosx – sin2asinx]
= cosx[(cos^2a – sin^2a)cosx – 2sinacosasinx]]
= cosx[cos^2acosx – sin^2acosx – 2sinacosasinx]
= cos^2acos^2x – sin^2acos^2x - 2sinacosasinx
= cos^2x(cos^2a – sin^2a) - 2sinacosasinx
Precalc and trig - hayden, Sunday, February 22, 2009 at 8:19pm
thanks! i accidentally posted this twice
Precalc and trig - hayden, Sunday, February 22, 2009 at 8:43pm
cos^2x(cos^2a – sin^2a) - 2sinacosasinx
cos^2x(cos^2a – sin^2a) – 2cosxcosasinxsina
Precalc and trig - Reiny, Sunday, February 22, 2009 at 11:16pm
thanks for catching that typo, it is so hard to type those complicated trig expressions without making some errors.
on RS it looks like I forgot to multiply cosx by that last term
cosx[cos^2acosx – sin^2acosx – 2sinacosasinx]
= cos^2acos^2x – sin^2acos^2x - 2sinacosasinxcosx
= cos^2x(cos^2a – sin^2a) - 2sinacosasinxcosx
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