Question

如图，高一数学。求过程

Answer 1

Answer 2

解：
sin(π/8 +α)=⅓
cos[2(π/8 +α)]=1-2sin²(π/8 +α)=1-2×⅓²=7/9
cos(2α+ π/4)=7/9
cos(2α)cos(π/4)-sin(2α)sin(π/4)=7/9
(√2/2)cos(2α)-(√2/2)sin(2α)=7/9
cos(2α)-sin(2α)=7√2/9
7√2/9>1，因此cos(2α)>0，sin(2α)<0
cos(2α)=sin(2α)+7√2/9
sin²(2α)+cos²(2α)=1
sin²(2α)+[sin(2α)+7√2/9]²=1
162sin²(2α)+126√2sin(2α)+17=0
△=(126√2)²-4×162×17=20736
√20736=144
sin(2α)=(144-126√2)/(2×162)或sin(2α)=-(144+126√2)/(2×162)
sin(2α)=(72-63√2)/162或sin(2α)=-(72+63√2)/162

Answer 3

你先告诉我答案是不是12分之派和二分之根号三，整整一年没做过数学题了，刚居然还去翻了以前的笔记，不知道对不

追答

错了，后面一个没有，只有12分之派

卧槽你那究竟是π／6还是π/8

追问

不是

Answer 4

我试试看看