已知tan(π+a)=-1⼀3求[sin(π-2a)+cos^2a]⼀[2cos2a+sin2a+2]

2024-12-31 16:28:31
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回答1:

tan(π+a)=-1/3
tana=-1/3

[sin(π-2a)+cos^2a]/[2cos2a+sin2a+2]
=[sin(2a)+cos^2a]/[2cos2a+sin2a+2]
=[2sinacosa+cos^2a]/[4cos^2a+2sinacosa]
=[2sina+cosa]/[4cosa+2sina]
=[2tana+1]/[4+2tana]
=(2*(-1/3)+1)/(4+2*(-1/3))
=1/10