sin[(4n-1/4)π-α]+cos[(4n-1/4)π-α]
=sin(4nπ-π/4-α)+cos(4nπ-π/4-α)
=sin(-π/4-α)+cos(-π/4-α)
=-sin(π/4+α)+cos(π/4+α)
=-sinπ/4cosα-cosπ/4sinα+cosπ/4cosα-sinπ/4sinα
=√2/2(-cosα-sinα+cosα-sinα)
=-√2sinα
化簡：sin[(4n-1/4)π-α]+cos[(4n-1/4)π-α],其中n為非負(fù)整數(shù).
化簡：sin[(4n-1/4)π-α]+cos[(4n-1/4)π-α],其中n為非負(fù)整數(shù).
那個4n-1是分子啊，不是分開的兩個
那個4n-1是分子啊，不是分開的兩個
數(shù)學(xué)人氣：630 ℃時間：2020-06-23 20:21:56
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