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

RS

= 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

= LS

Q.E.D.

- 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
how does

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

so from

cosx[cos^2acosx – sin^2acosx – 2sinacosasinx]

= cos^2acos^2x – sin^2acos^2x - 2sinacosasinxcosx

= cos^2x(cos^2a – sin^2a) - 2sinacosasinxcosx

= LS

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