已知0<a<π⼀2,sina=4⼀5,求【(sina)^2+sin2a】⼀【(cosa)^2+cos2a】;tan(a-5π⼀4)

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2025-05-21 14:05:55
推荐回答(1个)
回答1:

0[(sina)^2+sin2a]/[(cosa)^2+cos2a]
=[16/25+2*(4/5)*(3/5)]/[(9/25+2*(3/5)^2-1]
=(8/5)/(2/25)
=20.
tana=sina/cosa=4/3.
tan(a-5π/4)=(tana-tan5π/4)/(1+tana*tan5π/4)
=(4/3-1)/(1+4/3)
=1/7.