高一三角函数证明题
\u9ad8\u4e00\u4e09\u89d2\u51fd\u6570\u8bc1\u660e\u9898\u5df2\u77e5sin\u03b1=asin\u03b2,tan\u03b1=btan\u03b2,\u03b1\u4e3a\u9510\u89d2,\u6c42\u8bc1:(cos\u03b1)^2=(a^2-1)/(b^2-1).
(sin\u03b2)^2=(sin\u03b1)^2/a^2, (cos\u03b2)^2=1-(sin\u03b2)^2=[a^2-(sin\u03b1)^2]/a^2.
(tan\u03b2)^2=(sin\u03b1)^2/[a^2-(sin\u03b1)^2].
(tan\u03b1)^2=b^2*(tan\u03b2)^2=b^2(sin\u03b1)^2/[a^2-(sin\u03b1)^2].
1/(cos\u03b1)^2=b^2/[a^2-(sin\u03b1)^2].
b^2(cos\u03b1)^2=a^2-1+(cos\u03b1)^2.
(cos\u03b1)^2=(a^2-1)/(b^2-1).
cos\u03b1=\u221a[(a^2-1)/(b^2-1)]
\u6b64\u9898\uff08\u03b1=\u03b8\uff0c\u03b2=\u03b3\uff09
sinA/(1+cosA)
=[2sin(A/2)cos(A/2)]/[1+2cos²(A/2)-1]
=[2sin(A/2)cos(A/2)]/[2cos²(A/2)]
=sin(A/2)/cos(A/2)
=tan(A/2)
(1-cosA)/sinA
={1-[1-2sin²(A/2)]}/[2sin(A/2)cos(A/2)]
=2sin²(A/2)/[2sin(A/2)cos(A/2)]
=sin(A/2)/cos(A/2)
=tan(A/2)
证明：因为
4cosA/2• cosB/2•cosC/2=4cosA/2• cosB/2•cos[（π-（A+B））/2]
=4cosA/2• cosB/2•cos[π/2-(A+B)/2]
=4cosA/2• cosB/2•sin(A/2+B/2)
=4cosA/2• cosB/2•(sinA/2•cosB/2+ sinB/2•cosA/2)
=4cosA/2• cosB/2•sinA/2•cosB/2+4cosA/2• cosB/2• sinB/2•cosA/2
=2sinAcos2B/2+2sinBcos2A/2
=2sinA [(1+cosB)/2]+ 2sinB [(1+cosA)/2]
= sinA + sinB + sinA cosB + sinB cosA
= sinA + sinB +sin(A+B)
又因为：A+B+C=π
所以：sin(A+B)=sin(π-C)
=sinC
所以：4cosA/2• cosB/2•cosC/2= sinA+sinB+sinC
原题得证
2. 在ΔABC中，若A+B=120`,求证a/(b+c) +b/(a+c)= 1
证明：因为
a/(b+c) +b/(a+c)= 1
可化简为：[a(a+c)+b(b+c)]/( a+c)( b+c)=1
( a2+ac+b2+bc)/(ab+ac+bc+c2)=1
a2+ac+b2+bc =ab+ac+bc+c2
a2+b2 =ab+c2
c2= a2+b2-ab
又因为：A+B=120` A+B+C=π
所以：C=60`
由余弦定理可知：c2= a2+b2-2abcosC
既有：c2= a2+b2-2abcos60`
所以：c2= a2+b2-ab
原题得证
3. 在ΔABC中，若a•cos2 C/2+c cos2 A/2=3b/2,求证a+c=2b
证明：因为
a•cos2 C/2+c cos2 A/2=3b/2
所以有：a•[(cosC+1)/2]+c •[(cosA+1)/2] =3b/2
a+c+a•cosC + c•cosA =3b
因为:
a=2RsinA
b=2RsinB
c=2RsinC
所以：a+c+a•cosC + c•cosA=a+c+2RsinA•cosC+2RsinC•cosA
=a+c+2Rsin(A+C)
又因为：A+B+C=π
所以：B=π-( A+C)
所以：a+c+a•cosC + c•cosA= a+c+2Rsin(A+C)
= a+c+2RsinB
即：a+c+a•cosC + c•cosA= a+c+b =3b
所以：a+c=2b
原题得证
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