sin(a-π/4)+cos(a+π/4)怎么解?

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sin(a-π/4)+cos(a+π/4)怎么解?
sin(a-π/4)+cos(a+π/4)怎么解?

sin(a-π/4)+cos(a+π/4)怎么解?
原式=sin(a-π/4)+sin[π/2-(a+π/4)]
=sin(a-π/4)+sin(π/4-a)
=sin(a-π/4)-sin(a-π/4)
=0.

sin(a-π/4)+cos(a+π/4)
=[sinacosπ/4-cosasinπ/4]+[cosacosπ/4-sinasinπ/4]
=√2/2sina-√2/2cosa+√2/2cosa-√2/2sina
=0
方法二:
sin(a-π/4)+cos(a+π/4)
=sin(a-π/4)+sin[π/2-(a+π/4)]
=sin(a-π/4)+sin(π/4-a)
=sin(a-π/4)-sin(a-π/4)
=0.