已知tan(a+π/4)=-1/2,π/2<a<π,求1.tana的值；2.(sin2a-2cos^2a)/√2sin(a-π/4)的值
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(1)
tan(A+π/4)=-1/2
[tanA+tan(π/4)]/(1-tanA.tanπ/4)=-1/2
(tanA+1)/(1-tanA)=-1/2
2tanA+2 = -1+tanA
tanA = -3
A=arctan(-3) =108.43°
(2)
tanA = -3
=> sinA = 3/√10 and cosA=-1/√10
[sin2A - 2(cosA)^2]/√2sin(A-π/4)
=[2sinA.cosA - 2(cosA)^2]/√2[sinA.cos(π/4)-cosA.sin(π/4)]
=[2(-3/10) - 2(1/10)]/[3/√10 +1/√10]
=-(8/10). (√10/4)
=-2√10/10
tan(A+π/4)=-1/2
[tanA+tan(π/4)]/(1-tanA.tanπ/4)=-1/2
(tanA+1)/(1-tanA)=-1/2
2tanA+2 = -1+tanA
tanA = -3
A=arctan(-3) =108.43°
(2)
tanA = -3
=> sinA = 3/√10 and cosA=-1/√10
[sin2A - 2(cosA)^2]/√2sin(A-π/4)
=[2sinA.cosA - 2(cosA)^2]/√2[sinA.cos(π/4)-cosA.sin(π/4)]
=[2(-3/10) - 2(1/10)]/[3/√10 +1/√10]
=-(8/10). (√10/4)
=-2√10/10
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