已知角a∈(π/4,π/2 ),且(4cosa- 3sina )(2cosa- 3sina)=0 (1)tan( a+π/4)(2)co已知角a∈(π/4,π/2 ),且(4cosa- 3sina )(2cosa- 3sina)=0 (1)tan( a+π/4)(2)cos(π/3-2a)
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已知角a∈(π/4,π/2 ),且(4cosa- 3sina )(2cosa- 3sina)=0 (1)tan( a+π/4)(2)co已知角a∈(π/4,π/2 ),且(4cosa- 3sina )(2cosa- 3sina)=0 (1)tan( a+π/4)(2)cos(π/3-2a)
已知角a∈(π/4,π/2 ),且(4cosa- 3sina )(2cosa- 3sina)=0 (1)tan( a+π/4)(2)co
已知角a∈(π/4,π/2 ),且(4cosa- 3sina )(2cosa- 3sina)=0 (1)tan( a+π/4)(2)cos(π/3-2a)
已知角a∈(π/4,π/2 ),且(4cosa- 3sina )(2cosa- 3sina)=0 (1)tan( a+π/4)(2)co已知角a∈(π/4,π/2 ),且(4cosa- 3sina )(2cosa- 3sina)=0 (1)tan( a+π/4)(2)cos(π/3-2a)
由(4cosa- 3sina )(2cosa- 3sina)=0
得 4cosa- 3sina=0或2cosa- 3sina=0
即tana=4/3或2/3
又角a∈(π/4,π/2 ) 则tana∈(1,+∞)
故tana=4/3 则cosa=3/5,sina=4/5
tan( a+π/4)=(tana+tanπ/4)/(1-tana.tanπ/4 )=-7
cos(π/3-2a)=cosπ/3.cos2a +sinπ/3.sin2a
=(cosa.cosa-sina.sina)/2 + √3 sina.cosa
=(24√3 - 7)/50
=0说明有两组解,一组是tana=4/3,另一组是tana=2/3,根据取值范围tana=4/3
剩下的套公式就可以求得了。