已知cos[x+(π/4)]=3/5,7π/12
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已知cos[x+(π/4)]=3/5,7π/12已知cos[x+(π/4)]=3/5,7π/12已知cos[x+(π/4)]=3/5,7π/12sin2x=2sinxcosx(sin2x+2sin^2
已知cos[x+(π/4)]=3/5,7π/12
已知cos[x+(π/4)]=3/5,7π/12
已知cos[x+(π/4)]=3/5,7π/12
sin 2x = 2sinxcosx
(sin 2x+2sin^2 x)/(1-tan x)
= 2sin x(cos x+sin x)/(1-tan x)
= 2sin x(cos x+sin x)/(1-(sin x/cos x))
= 2sin x(cos x+sin x)×cos x/(cos x-sin x)
= sin2x(cos x+sin x)/(cos x-sin x)
(中间你可以随意省略!)
cos[x+(π/4)]=3/5
=> cosxsin(π/4)-sinxcos(π/4)=3/5
=> (cos x-sin x)×√2/2=3/5 (√2/2是二分之根号二)
=> cos x-sin x = 3√2/5 ① (3√2/5 是五分之三倍根号二)
=> (cos x-sin x)^2 = 18/25
=> 1 - 2sinxcosx = 18/25
=> 2sinxcosx =7/25
=> sin2x = 7/25 ②
然后再回到 (sin 2x+2sin^2 x)/(1-tan x)
= sin2x(cos x+sin x)/(cos x-sin x)
现在就只有 （cos x+ sinx）不知道 等于多少了 .
(cos x+sin x)^2=1+2sinxcosx=1+sin2x
所以 cos x +sin x = √(1+7/25) = √(32/25) ③
由 ① ② ③
原式等于= 28/75
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