已知n∈Z化簡sin[(4n-1/4)π-a]+cos[(4n+1/4)π-a]
已知n∈Z化簡sin[(4n-1/4)π-a]+cos[(4n+1/4)π-a]
數(shù)學(xué)人氣：697 ℃時(shí)間：2020-06-23 20:21:55
優(yōu)質(zhì)解答
sin[(4n-1/4)π-a]+cos[(4n+1/4)π-a] =sin[4nπ- （π/4 +a)] +cos[4nπ +（π/4 -a)] =-sin（π/4 +a) +cos（π/4 -a) = -sin[ π/2- （π/4 -a)] +cos（π/4 -a) =-cos（π/4 -a)+cos（π/4 -a) =0
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