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Posted by 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 - , 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 - , Sunday, February 22, 2009 at 8:19pm

    thanks! i accidentally posted this twice

  • Precalc and trig - , 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 - , 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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