若tana=1/2,求值:(1)sina+cosa/3sina-2cosa; (2)2sin²a-sina·cosa+cos²a.,是同除以cos吗?
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![若tana=1/2,求值:(1)sina+cosa/3sina-2cosa; (2)2sin²a-sina·cosa+cos²a.,是同除以cos吗?](/uploads/image/z/3712205-29-5.jpg?t=%E8%8B%A5tana%EF%BC%9D1%EF%BC%8F2%2C%E6%B1%82%E5%80%BC%EF%BC%9A%EF%BC%881%EF%BC%89sina%EF%BC%8Bcosa%EF%BC%8F3sina%EF%BC%8D2cosa%EF%BC%9B%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%E3%80%80%EF%BC%882%EF%BC%892sin%26%23178%3Ba%EF%BC%8Dsina%C2%B7cosa%EF%BC%8Bcos%26%23178%3Ba%EF%BC%8E%2C%E6%98%AF%E5%90%8C%E9%99%A4%E4%BB%A5cos%E5%90%97%3F)
若tana=1/2,求值:(1)sina+cosa/3sina-2cosa; (2)2sin²a-sina·cosa+cos²a.,是同除以cos吗?
若tana=1/2,求值:(1)sina+cosa/3sina-2cosa;
(2)2sin²a-sina·cosa+cos²a.
,是同除以cos吗?
若tana=1/2,求值:(1)sina+cosa/3sina-2cosa; (2)2sin²a-sina·cosa+cos²a.,是同除以cos吗?
第一题可以同除以cosa的.
1.
(sina+cosa)/(3sina-2cosa)
=(tana+1)/(3tana -2)
=(1/2 +1)/(3/2 -2)
=-3
2.
tana=1/2>0,sina,cosa同号.
tana=sina/cosa=1/2
cosa=2sina
cos²a+sin²a=1
(2sina)²+sin²a=1
5sin²a=1
sin²a=1/5
2sin²a-sinacosa+cos²a
=2sin²a-sina(2sina)+(2sina)²
=4sin²a
=4(1/5)
=4/5
(1)原式=(sina+cosa/3sina-2cosa)/cosa
=tana+1/3tana-2
= 1/2+2/3-2
=5/6
(2)原式=(2sin²a-sina·cosa+cos²a)/cos²a
=2*1/2*1/2-1/2+1
=1
1.tana=sina/cosa=1/2
所以2sina=cosa
原式=(sina+2sina)/(3sina-4sina)=-3
2.原式=(2sin^2a-sinacosa+cos^a)/1=(2sin^2a-sinacosa+cos^a)/(sin^2a+cos^a)
同除以cos^2a
=(2tan^2a-tana+1)/(tan^2a+1)=4/5