若tanA=1/5,则2/3+4cos2A的值为多少?

TAN-67
【化学名】(-)-(4aS*,12aR*)-4a-(3-Hydroxyphenyl)-2-methyl-1,2,3,4,4a,5,12,12a-octahydropyrido[3,4-b]acridine
【CAS登记号】【结构式】
【分子式】C23-H24-N2-O
【分子量】344.4556
【原研厂家】Toray (Originator)
【作用类别】ANALGESIC AND ANESTHETIC DRUGS,Analgesic Drugs,Opioid Analgesics,delta-Opioid Agonists
【研发状态】Preclinical
【合成情况】
〖来源〗Chem Pharm Bull
〖合成路线〗
〖标题〗Rational drug design and synthesis of a highly selective nonpeptide delt-opioid agonist,(4aS*,12aR*)-4a-(3-hydroxyphenyl)-2-methyl-1,2,3,4,4a,5,12,12a-octrahydropyrido[3,4-b]acridine (TAN-67)
〖合成方法〗The cyclization of perhydroisoquinoline (I) with o-amino-benzaldehyde (II) in the presence of methanesulfonic acid in refluxing EtOH gave the pyridoacridine (III).Subsequent cleavage of the methyl ether function using n-PrSH and tert-BuOK in boiling DMF provided the target phenol.
〖作者〗Nagase,H.; Kawai,K.; Hayakawa,J.; Wakita,H.; Mizusuna,A.; Matsuura,H.; Tajima,C.; Takezawa,Y.; Endoh,T.
〖参考〗Nagase,H.; Kawai,K.; Hayakawa,J.; Wakita,H.; Mizusuna,A.; Matsuura,H.; Tajima,C.; Takezawa,Y.; Endoh,T.; Rational drug design and synthesis of a highly selective nonpeptide delt-opioid agonist,(4aS*,12aR*)-4a-(3-hydroxyphenyl)-2-methyl-1,2,3,4,4es/10_5_hkavynmfxxnmvbvo.html">eIǚx08Ꚑݐul> li>a href="/databases/10_1_vqrupomukvaqu
〖出处〗Chem Pharm Bull1998,46,(11):1695
〖备注〗The cyclization of perhydroisoquinoline (I) with o-amino-benzaldehyde (II) in the presence of methanesulfonic acid in refluxing EtOH gave the pyridoacridine (III).Subsequent cleavage of the methyl ether function using n-PrSH and tert-BuOK in boiling DMF provided the target phenol (1).
〖来源〗Symp Med Chem
〖合成路线〗
〖标题〗Rational drug design and synthesis of opioid delta-receptor selective nonpeptidic agonists - Optical resolution of TAN-67 and the pharmacological effect of each enantiomer
〖合成方法〗The cyclization of perhydroisoquinoline (I) with o-amino-benzaldehyde (II) in the presence of methanesulfonic acid in refluxing EtOH gave the pyridoacridine (III).Subsequent cleavage of the methyl ether function using n-PrSH and tert-BuOK in boiling DMF provided the target phenol.
〖作者〗Fujii,H.; et al.
〖参考〗Fujii,H.; et al.; Rational drug design and synthesis of opioid delta-receptor selective nonpeptidic agonists - Optical resolution of TAN-67 and the pharmacological effect of each enantiomer.Symp Med Chem 1999,Abst 1P-21
〖出处〗Symp Med Chem1999,():Abst 1P-21

cos4a=2(cos2a)^2-1
代入得到
原式=
[2(cos2a)^2-4cos2a+2]/[2(cos2a)^2+4cos2a+2]
=(1-cos2a)^2(横线代表分号,不好写就这样了)
..-----------
----(cos2a+1)^2
而由万能公式
cos2a=(1-t^2)/(1+t^2).(t=tana)
代入即得
原式=(tana)^4

这个好象不能用证明或几何计算的方法来求
好象是属于用计算器求锐角三角函数值的知识,
你可以先求出
cos2A=3/4
再用计算器求出
2A的度数
从而得到角A的度数
记着cos2A是个整体不能写为COSA=3/8
在计算器上有个SHIFT的键,按之后就能根据3/4来求角的度数了.

等腰直角三角形

用齐次式（最后那个分母是不是应该加个小括号,不加的话没办法用齐次式）
tana=3,平方=9,求tan2a
把原式化成sin2a和cos2a的形式
然后同除以cos2a
带入tan2a就可以了
（在电脑上不会打平方,所以就不给你写具体式子了,步骤都给你写好啦,一般木有问题了^-^)

左边=2cos²2a-1+4cos2a+3
=2(cos²2a+2cos2a+1)
=2(cos2x+1)²
=2(2cos²a-1+1)²
=2[4(cosa)^4]
=8(cosa)^4=右边
命题得证

(3sin2a-4cos2a)/(2tana-1)-sin2a
=(6sina*cosa-4cosa^2+4sina^2)/(2tana-1)-sin2a
=2(2sina-cosa)(sina+2cosa)/((2sina-cosa)/cosa)-sin2a
=2sinacosa+4cosa^2-sin2a
=4cosa^2

tana=-1／2
故tan2a=2tana/[1-(tana)^2]=2×(-1/2)/[1-(-1/2)^2]=-4/3
则
(sin2a+2cos2a)／(4cos2a-4sin2a) 分子分母同除以cos2a
=(tan2a+2)／(4-4tan2a)
=(-4/3+2)／[4-4×(-4/3)]
=1/14

由题设得：（1-tana)/(2+tana)=1.====>1-tana=2+tana.===>tana=-1/2.====>tan2a=2tana/(1-tana*tana)=-1/[1-(1/4)]=-4/3.即tan2a=-4/3.又sin2a/cos2a=tan2a=-4/3.===>3sin2a=-4cos2a.

cos4a+4cos2a+3
=2cos²2a-1+4cos2a+3
=2cos²2a+4cos2a+2
=2[cos(2a)+1]²
=2[(2cos²a-1)+1]²
=8cos⁴a

cos4a=2(cos2a)^2-1
代入得到
原式=
[2(cos2a)^2-4cos2a+2]/[2(cos2a)^2+4cos2a+2]
=(1-cos2a)^2(横线代表分号,不好写就这样了)
..-----------
----(cos2a+1)^2
而由万能公式
cos2a=(1-t^2)/(1+t^2).(t=tana)
代入即得
原式=(tana)^4

(1-tana)/(2+tana)=1
1-tana=2+tana
tana=-1/2
sin2a=2tana/[1+(tana)^2]=4/5
cos2a=[1-(tana)^2]/[1+(tana)^2]=-3/5
3sin2a =-4cos2a
3tan2A=-4cos2A 抄错了吧你