高中数学三角函数三道题
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1.tanx=tan\uff08\uff08x+pi\u2215 4\uff09-pi\u2215 4\uff09\u5c55\u5f00
2.cos10\u00b0=cos\uff0830\u00b0-20\u00b0\uff09\u4ee3\u5165
3.cos(a-pi\u2215 4)=0.6 \u77e5\u9053a\u8303\u56f4 sin(a-pi\u2215 4)=0.8
sin\uff08b+3pi\u2215 4)=5\u2215 13 cos\uff08b+3pi\u2215 4)=-12\u2215 13
sin\uff08a+b+pi\u2215 2\uff09=sin\uff08\uff08a-pi\u2215 4\uff09+\uff08b+3pi\u2215 4\uff09\uff09\u53ef\u6c42
sin\uff08a+b\uff09=sin\uff08a+b+pi\u2215 2\uff09-pi\u2215 2\uff09\uff09\u53ef\u6c42
令t= sinx+cosx=√2sin(x+∏/4)，∴t∈[-√2, √2]
则，t^2=(sinx+cosx)^2=1+2sinxcosx
则,sinxcosx=(t^2-1)/2
∴y=t+(t^2-1)/2=(1/2)t^2+t-(1/2)
令y=g(t)= (1/2)t^2+t-(1/2), t∈[-√2, √2]
对称轴是t=-1，开口向上
∴最大值y=g(√2)= (1/2)( √2)^2+√2-(1/2)
=√2+1/2
最小值y=g(-1)= (1/2)( -1)^2+(-1)-(1/2)
=-1
所以,值域是[-1, √2+1/2]
(2) y=cos(2x/5)+sin(2x/5)= √2sin[(2x/5)+∏/4]
∴最小正周期是T=2∏/(2/5)=5∏
∴相邻两条对称轴之间距离是d= T/2=5∏/2
(3) ∵4tan(a/2)=1-〔tan(a/2)〕^2
∴2tan(a/2)/{1- [tan (a/2)]^2}=1/2
∴tana=2tan(a/2)/{1- [tan (a/2)]^2}=1/2
又∵0<a<л/2，且 sina/cosa=tana,(sina)^2+(cosa)^2=1
∴sina=√5/5,cosa=2√5/5
∴sin2a=2sinacosa=4/5，cos2a=3/5
又∵3sinb=sin(2a+b)=sin2acosb+sinbcos2a
3sinb=(4/5)cosb+(3/5)sinb
化简，得
12sinb=4cosb
∴tanb=sinb/cosb=1/3
∴tan(a+b)=(tana+tanb)/(1-tanatanb)
=[(1/2)+(1/3)]/[1-(1/6)]
=1
∴a+b=∏/4
1.设 t=sinx+cosx 则-sqr(2)<t<sqr(2)
原式＝（t^2-1)/2+t 再用二次函数的求解方法求解
2. y=3^0.5cos(2x/5)+sin(2x/5)
=2(sin(pi/3)cos(2x/5)+cos(pi/3)sin(2x/5))
=2sin(2x/5+pi/3)
2x/5+pi/3=kpi+pi/2为函数的对称轴
即x=(5/2)kpi+(5/12)pi
其中k为整数
则相邻两对称轴的距离为(5/2)pi
3.4tan(a/2)=1-tan(a/2)得 2tan(a/2)/(1-tan(a/2))=1/2
tan(a)=1/2
4sinb=sin(2a+b)+sinb=2sin(a+b)cos(a)
2sinb=sin(2a+b)-sinb=2cos(a+b)sin(a)
相除得2tan(a)=tan(a+b)=1 a+b=π/4
很乐意为你解答。
第一题，设sinx+cosx为T，就能做了
1. 设t=sinx+cosx,则-√2<=t<=√2,于是y =(t^2 -1)/2 +t,
-1 <= y <= 1/2+√2.
2. 周期为5л，相邻两条对称轴之间的距离为半个周期5π/2.
3.由0<a,b<л/2，得0<a/2<л/4, 0<a+b<л,
因为 4tan(a/2)=1-〔tan(a/2)）^2,解得tan（a/2）=√5 -2 ,根据倍角公式，得tana=1/2.
又因为 3sinb=sin(2a+b),得3sin[(a+b)-a]=sin[(a+b)+a],展开，得
tan(a+b)=2tana=2*1/2=1,所以a+b=π/4.
见下图，我的计算有很大可能出错，本人向来以幼稚计算错误闻名于世，但思路应该没错
绛旓細鍙堚埖3sinb=sin(2a+b)=sin2acosb+sinbcos2a 3sinb=(4/5)cosb+(3/5)sinb 鍖栫畝锛屽緱 12sinb=4cosb 鈭磘anb=sinb/cosb=1/3 鈭磘an(a+b)=(tana+tanb)/(1-tanatanb)=[(1/2)+(1/3)]/[1-(1/6)]=1 鈭碼+b=鈭/4
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