几道高一三角函数数学题，过程越详细，给分越高！！！在线等，高手速度来
\u4e00\u9053\u6570\u5b66\u9898\uff01\u5728\u7ebf\u7b49\uff01\u6c42\u9ad8\u624b\u901f\u5ea6\u8be6\u7ec6\u8fc7\u7a0b\uff01Ai+1=Ai+d
Ai+2=Ai+2d
\u539f\u65b9\u7a0b\u7b49\u4ef7\u4e8eAiX^2+2(Ai+d)X+Ai+2d=0,
\u5373AiX^2+2AiX+Ai+2dX+2d=0, Ai(X+1)^2+2d(X+1)=0,
(AiX+Ai+2d)(x+1)=0.
\u65b9\u7a0b\u7684\u6839x=-1\u6216x=-(Ai+2d)/Ai
\u4f7f\u7528\u8fd9\u65b9\u7a0b\u7684\u516c\u5171\u6839\u4e3ax=-1
(2)
\u8bc1\u660e\uff1a\u2235{an}\u4e3a\u7b49\u5dee\u6570\u5217\u516c\u5deed\u22600\uff0can\u22600
\u22342a(n+1)=an+a(n+2)
\u2234\u65b9\u7a0banx2+(an+a(n+2))x+a(n+2)=0\u7684\u4e8c\u4e2a\u4e0d\u540c\u7684\u6839\u5206\u522b\u4e3a\uff1a
X1=-an-a(n+2)-an+a(n+2))/(2an)=-1\uff0cx2=-2a(n+2)/(2an)= -a(n+2)/an
\u5f53\u65b9\u7a0b\u7cfb\u6570\u4e3aa1,a2,a3\u65f6\uff0cx1=-a3/a1
\u5f53\u65b9\u7a0b\u7cfb\u6570\u4e3aa2,a3,a4\u65f6\uff0cx2=-a4/a2
\u5f53\u65b9\u7a0b\u7cfb\u6570\u4e3aa3,a4,a5\u65f6\uff0cx3=-a5/a3
\u5f53\u65b9\u7a0b\u7cfb\u6570\u4e3aa4,a5,a6\u65f6\uff0cx4=-a6/a4
\u2026\u2026
\u5f53\u65b9\u7a0b\u7cfb\u6570\u4e3aan,a(n+1),a(n+2)\u65f6\uff0cxn=-a(n+2)/an
Bn=1/(xn+1)=1/(1-a(n+2)/an)=an/(an-a(n+2))= -an/(2d)
B(n-1)=a(n-1)/(a(n-1)-a(n+1))= -a(n-1)/(2d)
B(n+1)=a(n+1)/(a(n+1)-a(n+3)) = -a(n+1)/(2d)
B(n+1)-bn= -a(n+1)/(2d)+an/(2d)=-1/2
Bn-b(n-1)= -an/(2d)+ a(n-1)/(2d)=-1/2
\u5373\u6570\u5217{bn}\u524d\u540e\u9879\u4e4b\u5dee\u4e3a\u5e38\u6570
\u2234\u6570\u5217{bn}\u4e3a\u7b49\u5dee\u6570\u5217\u3002
\uff081\uff09cos165\u00b0=-cos(45\u00b0-30\u00b0)
=-\u221a6/4+\u221a2/4
(2)cos(-915\u00b0)=cos[-3*360\u00b0+165\u00b0]
cos165\u00b0=-\u221a6/4+\u221a2/4
2.(1)\u5df2\u77e5cos\u03b8=-3/5\uff0c\u03b8\u2208\uff08\u5140/2\uff0c\u5140\uff09\uff0c\u6c42sin(\u03b8+\u5140/3)\u7684\u503c
sin(\u03b8+\u5140/3)=1/2sin\u03b8+\u221a3/2(-3/5)
=1/2*4/5-3\u221a3/10=(4-3\u221a3)/10
\uff082\uff09\u5df2\u77e5sin\u03b1=15/17\uff0c\u03b1\u2208\uff08\u5140/2\uff0c\u5140\uff09\uff0c\u6c42cos\uff08\u5140/3-\u03b1\uff09\u7684\u503c
cos\uff08\u5140/3-\u03b1)=1/2(-8/17)+\u221a3/2*15/17
=(15\u221a3-8)/34
\uff083\uff09\u5df2\u77e5cos\u03b8=-5/13\uff0c\u03b8\u2208\uff08\u5140\uff0c3\u5140/2\uff09\uff0c\u6c42cos\uff08\u03b8+\u5140/6\uff09\u7684\u503c
cos\uff08\u03b8+\u5140/6\uff09=-5/13*\u221a3/2-(-12/13)*1/2
=(12-5\u221a3)/26
sin^4＋cos^2×sin^2＝sin^2
第二行式子右边等于1-cos^2，代进去就是了。
2.等号右边等于(sin-cos)/(sin＋cos)，分子分母同乘以(sin＋cos)，分子完全平方开括号(sin^2+cos^2＝1)，分母平方差开括号得左式
3.左边tan^2-sin^2＝(sin^2-sin^2cos^2)/cos^2＝sin^2（1-cos^2）/cos^2＝sin^4/cos^2＝tan^2sin^2
4.把(cos＋1)^2这个完全平方的括号拆开就是了
5.把右式的2sin^2cos^2拿到左边，左边变成完全平方式，(sin^2+cos^2)^2＝1^2，这就可以了。
二，答案是3/pi＋k×pi。谢谢采纳
原式=sina^2(sina^2+cosa^2)+cosa^2=sina^2+cosa^2=1
原式=（sina^2+cosa^2-2sinacosa)/(cosa^2-sina^2)=(sina-cosa)^2/(sina-cosa)(-cosa-sina)=(sina-cosa)/(-sina-cosa)=(1-tana)/(1+tana)
原式=tana^2(1-cosa^2)=tana^2sina^2
原式=cosa^2+1-2cosa+sina^2=2-2cosa
原式=(sina^2)^2+(cosa^2)^2+2sina^2cosa^2-2sina^2cosa^2=(sina^2+cosa^2)^2-2sina^2cosa^2=1-2sina^2cosa^2
tana=根号3
cosa=-根号3/2
sina=-1/2
所以，cosa-sina=(-根号3+1）/2
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