在高一数学三角函数中给出sinA=五分之三，杂判断他是钝角还是锐角？！ 高一数学三角函数
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\u6b63\u5f26\uff08sin\uff09\u7b49\u4e8e\u5bf9\u8fb9\u6bd4\u659c\u8fb9\uff1bsinA=a/c
\u4f59\u5f26\uff08cos\uff09\u7b49\u4e8e\u90bb\u8fb9\u6bd4\u659c\u8fb9\uff1bcosA=b/c
\u6b63\u5207\uff08tan\uff09\u7b49\u4e8e\u5bf9\u8fb9\u6bd4\u90bb\u8fb9\uff1btanA=a/b
\u4f59\u5207\uff08cot\uff09\u7b49\u4e8e\u90bb\u8fb9\u6bd4\u5bf9\u8fb9\uff1bcotA=b/a
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sin30\u00b0=1/2
sin45\u00b0=\u221a2/2
sin60\u00b0=\u221a3/2
cos30\u00b0=\u221a3/2
cos45\u00b0=\u221a2/2
cos60\u00b0=1/2
tan30\u00b0=\u221a3/3
tan45\u00b0=1
tan60\u00b0=\u221a3[1]
cot30\u00b0=\u221a3
cot45\u00b0=1
cot60\u00b0=\u221a3/3
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sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-sinBcosA
cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
tan(A+B)=(tanA+tanB)/(1-tanAtanB)
tan(A-B)=(tanA-tanB)/(1+tanAtanB)
ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA)
ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA)
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\u534a\u89d2\u516c\u5f0f
sin(A/2)=\u221a((1-cosA)/2) sin(A/2)=-\u221a((1-cosA)/2)
cos(A/2)=\u221a((1+cosA)/2) cos(A/2)=-\u221a((1+cosA)/2)
tan(A/2)=\u221a((1-cosA)/((1+cosA))
tan(A/2)=-\u221a((1-cosA)/((1+cosA))
ctg(A/2)=\u221a((1+cosA)/((1-cosA))
ctg(A/2)=-\u221a((1+cosA)/((1-cosA))
\u548c\u5dee\u5316\u79ef
2sinAcosB=sin(A+B)+sin(A-B) 2cosAsinB=sin(A+B)-sin(A-B) 2cosAcosB=cos(A+B)-sin(A-B) -2sinAsinB=cos(A+B)-cos(A-B) sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2) tanA+tanB=sin(A+B)/cosAcosB tanA-tanB=sin(A-B)/cosAcosB ctgA+ctgBsin(A+B)/sinAsinB - ctgA+ctgBsin(A+B)/sinAsinB \u4e09\u3001\u9510\u89d2\u4e09\u89d2\u51fd\u6570\u56fe\u50cf\u548c\u6027\u8d28
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\u2234a\uff08-2a+10\uff09=2/3
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sinA+cosA=3\u5206\u4e4b2\uff0c\u540c\u65f6\u5e73\u65b9\u5f97
sinA^2+cosA^2+2sinAcosA=4/9
\u6240\u4ee52sinAcosA=-5/9
\u56e0\u4e3a00
\u6240\u4ee5cosA<0\u6240\u4ee5A\u4e3a\u949d\u89d2\uff0c\u5373\u4e09\u89d2\u5f62\u4e3a\u949d\u89d2\u4e09\u89d2\u5f62\u3002
sin在一二象限为+
cos在一二象限分别为+,-
所以是根据cos判断,锐角则大于0,等于0则直角,小于0钝角
设A是锐角，则180°-A是钝角，且sin(180°-A)=sinA=3/5.故无法判断。
不能判断。
还应该有其他条件吧？
你可以把原题拿来
解 因为sinA=3/5 所以A为第一象限角或第三象限角 而三角形的内角只能大于0度小于180度（即为第一.二象限角） 综上所述，A为锐角
呵呵，文字太多，不会打
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