化簡:(sinα-cosα)^2+sin4α/2cos2α 化簡:sin^2x+sin^2(x+2π/3)+sin^2(x-2π/3)

數(shù)學(xué)人氣：930 ℃時間：2020-01-30 04:01:21

優(yōu)質(zhì)解答

(sinα-cosα)²+sin4α/2cos2α
=1-2sinαcosα+2sin2αcos2α/2cos2α
=1-sin2α+sin2α
=1
sin²x+sin²(x+2π/3)+sin²(x-2π/3)
=sin²x+sin²(x+π-π/3)+sin²(x+π/3-π)
=sin²x+sin²(π+x-π/3)+sin²(-π+x+π/3)
=sin²x+sin²(x-π/3)+sin²(x+π/3)
=sin²x+(1/4)sin²x-(√3/2)sinxcosx+(3/4)cos²x+(1/4)sin²x+(√3/2)sinxcosx+(3/4)cos²x
=sin²x+(1/2)sin²x+(3/2)cos²x
=(3/2)sin²x+(3/2)cos²x
=(3/2)(sin²x+cos²x)
=3/2