tana+1/tana=5/2,a∈(π/4,π/2),求cos2a和sin(2a+π/4)?如题.
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tana+1/tana=5/2,a∈(π/4,π/2),求cos2a和sin(2a+π/4)?如题.tana+1/tana=5/2,a∈(π/4,π/2),求cos2a和sin(2a+π/4)?如题.
tana+1/tana=5/2,a∈(π/4,π/2),求cos2a和sin(2a+π/4)?如题.
tana+1/tana=5/2,a∈(π/4,π/2),求cos2a和sin(2a+π/4)?
如题.
tana+1/tana=5/2,a∈(π/4,π/2),求cos2a和sin(2a+π/4)?如题.
a∈(π/4,π/2),
所以tana>1
tana+1/tana=5/2
tana=2,tana=1/2
所以tana=2
cos2a
=cos²-sin²a
=(cos²-sin²a)/(cos²+sin²a)
上下除以cos²a
=(1-tan²a)/(1+tan²a)
=-3/5
1/sin2a=(sin²a+cos²a)/2sinacosa
上下除以cos²a
=(tan²a+1)/2tana
=5/4
sin2a=4/5
所以sin(2a+π/4)
=sin2acosπ/4+cos2asinπ/4
=√2/10
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