分类：教育/科学-学习帮助
问题：高一数学已知cos（π/4+x）=3/5，17π/12&lt;x 
已知c（π/4+x）=3/5，17π/12<x<7π/4,求(sin2x+2* sinx*sinx)/(1-tgx)的值.答案是  -28/75请写出详细的过程,谢谢!!!!
回答：已知c(π/4 + x) = 3/5，17π/12<x<7π/4,求(sin2x + 2sinx*sinx)/(1 - tgx)的值。cos(π/4 + x) = 3/5cos(π/4)cosx - sin(π/4)sinx = 3/5√2/2(cosx - sinx) = 3/5cosx - sinx = 3√2/5  因为 17π/12＜x＜7π/417π/12 + π/4＜x + π/4＜7π/4 + π/420π/12＜x + π/4＜2ππ + 2π/3＜x + π/4＜2π......(x + π/4)在第四象限。所以sin(π/4 + x) = -√{1 - [cos(π/4 + x)]^2}= -√{1 - (3/5)^2}= -4/5sin(π/4 + x) = cos(π/4)sinx + sin(π/4)cosx= √2/2(sinx + cosx)√2/2(sinx + cosx) = -4/5sinx + cosx = -4√2/5  上式两边平方得(cosx + sinx)^2 = (-4√2/5)^21 + 2sinxcosx = 14/252sinxcosx = 1/5(sin2x + 2sinx*sinx)/(1 - tgx)= 2sinx(cosx + sinx)/(1 - sinx/cosx)= 2sinxcosx(sinx + cosx)/(cosx - sinx)= 1/5×(-4√2/5)/(3√2/5)= -4/75