急!这四个题怎么解 求大神
2个回答
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(1)
tanx =-3/4
=>
(sinx, cosx ) = (3/5 , -4/5) or ( -3/5 , 4/5)
sinx.(sinx - cosx)
= (3/5)^2 + (3/5)(4/5)
=9/25 +12/25
=21/25
ans : A
(2)
tanx = -1/3
tan2x
= 2tanx/( 1- (tanx)^2 )
=( -2/3) /( 1- 1/9)
=-3/4
=> sin2x = -3/5
(sinx+cosx)^2
= 1+ sin2x
= 1- 3/5
=2/5
sinx + cosx = -(1/5)√10
(3)
sinx + cosx =1/5
tanx +1 =(1/5)secx
(tanx +1)^2 =(1/25)(secx)^2
24(tanx)^2 +50tanx + 24 =0
tanx = (-50+14)/48 or (-50-14)/48
= -3/4(rej) or -4/3
ie tanx = -4/3
(4)
sinx+3cosx =1
tanx + 3 = secx
(tanx+3)^2 = (tanx)^2 +1
6tanx +8=0
tanx = -4/3
tanx =-3/4
=>
(sinx, cosx ) = (3/5 , -4/5) or ( -3/5 , 4/5)
sinx.(sinx - cosx)
= (3/5)^2 + (3/5)(4/5)
=9/25 +12/25
=21/25
ans : A
(2)
tanx = -1/3
tan2x
= 2tanx/( 1- (tanx)^2 )
=( -2/3) /( 1- 1/9)
=-3/4
=> sin2x = -3/5
(sinx+cosx)^2
= 1+ sin2x
= 1- 3/5
=2/5
sinx + cosx = -(1/5)√10
(3)
sinx + cosx =1/5
tanx +1 =(1/5)secx
(tanx +1)^2 =(1/25)(secx)^2
24(tanx)^2 +50tanx + 24 =0
tanx = (-50+14)/48 or (-50-14)/48
= -3/4(rej) or -4/3
ie tanx = -4/3
(4)
sinx+3cosx =1
tanx + 3 = secx
(tanx+3)^2 = (tanx)^2 +1
6tanx +8=0
tanx = -4/3
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