求证:(1)tan(x/2+π/4)+tan(x/2 - π/4)=2tan x(2)(1+sin 2φ)/(cos φ
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求证:
(1)tan(x/2+π/4)+tan(x/2 - π/4)=2tan x
(2)(1+sin 2φ)/(cos φ+sin φ)=cos φ+sin φ
已知sin(α+β)=1/2,sin(α - β)=1/3 :
(1)求证:sin αcos β=5cos αsin β;
(2)求证:tan α=5tan β
(1)tan(x/2+π/4)+tan(x/2 - π/4)=2tan x
(2)(1+sin 2φ)/(cos φ+sin φ)=cos φ+sin φ
已知sin(α+β)=1/2,sin(α - β)=1/3 :
(1)求证:sin αcos β=5cos αsin β;
(2)求证:tan α=5tan β
(1)tan(x/2+π/4)+tan(x/2 - π/4)
=[tanx/2+tanπ/4]/[1-tanx/2*tanπ/4]+[tanx/2-tanπ/4]/[1+tanx/2*tanπ/4]
=[tanx/2+1]/[1-tanx/2]+[tanx/2-1]/[1+tanx/2]
=2[tan^2x/2+1]/[1-tan^2x/2]=2tan x
(2)(1+sin 2φ)/(cos φ+sin φ)=[cos^2 φ+sin^2 φ+2cos φsin φ]/(cos φ+sin φ)
=(cos φ+sin φ)^2/(cos φ+sin φ)=(cos φ+sin φ)
sin(α+β)=1/2,sin(α - β)=1/3 :
sin(α+β)=sinacosβ+cosasinβ=1/2,
sin(α - β)=sinacosβ-cosasinβ=1/3 :
3[sinacosβ+cosasinβ]=2[sinacosβ-cosasinβ]
sin αcos β=5cos αsin β
同除5cos αcos β得
tan α=5tan β
=[tanx/2+tanπ/4]/[1-tanx/2*tanπ/4]+[tanx/2-tanπ/4]/[1+tanx/2*tanπ/4]
=[tanx/2+1]/[1-tanx/2]+[tanx/2-1]/[1+tanx/2]
=2[tan^2x/2+1]/[1-tan^2x/2]=2tan x
(2)(1+sin 2φ)/(cos φ+sin φ)=[cos^2 φ+sin^2 φ+2cos φsin φ]/(cos φ+sin φ)
=(cos φ+sin φ)^2/(cos φ+sin φ)=(cos φ+sin φ)
sin(α+β)=1/2,sin(α - β)=1/3 :
sin(α+β)=sinacosβ+cosasinβ=1/2,
sin(α - β)=sinacosβ-cosasinβ=1/3 :
3[sinacosβ+cosasinβ]=2[sinacosβ-cosasinβ]
sin αcos β=5cos αsin β
同除5cos αcos β得
tan α=5tan β
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