若tan(a+π/8)=2,则tan(a-π/8)=

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若tan(a+π/8)=2,则tan(a-π/8)=若tan(a+π/8)=2,则tan(a-π/8)=若tan(a+π/8)=2,则tan(a-π/8)=tan(a+π/8)=tan(a-π/8+π

若tan(a+π/8)=2,则tan(a-π/8)=
若tan(a+π/8)=2,则tan(a-π/8)=

若tan(a+π/8)=2,则tan(a-π/8)=
tan(a+π/8)
=tan(a-π/8+π/4)
=[tan(a-π/8)+tan(π/4)]/ [1-tan(a-π/8)tan(π/4)]
=[tan(a-π/8)+1]/[1-tan(a-π/8)]
因为tan(a+π/8)=2
所以[tan(a-π/8)+1]/[1-tan(a-π/8)]=2
所以tan(a-π/8)+1=2-2tan(a-π/8)
解得tan(a-π/8)=1/3

tan(a-π/8)=tan(a+π/8-π/4)=﹛tan(a+π/8)+tanπ/4﹜/1-tan(a+π/8)tanπ/4=1/3