1、由题意知：f(x)=sin(wx+Ф)的周期为2*(π/4-(-π/4))=π即2π/w=π所以w=2由f(π/4)=sin(2*(-π/4)+Ф)=0 得：Ф=π/2所以f(x)=sin(2x+π/2)2、sinBsinCcosA=sin2A=2sinAcosAsinBsinCcosA-2sinAcosA=0cosA(sinBsi...已知函数f(x)=根号2sin(2x-π/4)+2cos2x-1求f(x)的最大值及其取得最大值时x的集合在△ABC中,a,b,c分别是角A,B,C的对边,已知a=根号3/4,A=π/3),b=f(5π/12),求△ABC的面积继续第2题：cosA=0或sinBsinC-2sinA=0由cosA=0得A=π/2由sinBsinC-2sinA=0得：sinBsinC=2sinA ，sinA=1/2*sinBsinC又：sinA=sin(B+C)=sinBcosC+cosBsinC=1/2*sinBsinCcotC+cotB=1/2(cotB-cotC)²=(cotB+cotC)²-4cotBcotC=1/4-4cotBcotC.....................1式cosA=-cos(B+C)=sinBsinC-cosBcosC所以cotA=cosA/sinA=(sinBsinC-cosBcosC)/(1/2*sinBsinC)=2-2cotBcotCcotBcotC=1-1/2*cotA上式代入1式得：(cotB-cotC)²=(cotB+cotC)²-4cotBcotC=1/4-4(1-1/2*cotA)=2cotA-15/4因为(cotB-cotC)²≥0所以2cotA-15/4≥0cotA≥15/8所以值域为A=π/2 或A≤acot(15/8)-------------------------------------------新问题：f(x)=√2sin(2x-π/4)+2cos2x-1=√2(sin2x*cosπ/4-cos2xsinπ/4)+2cos2x-1=sin2x+cos2x-1=√2(√2/2*sin2x+√2/2cos2x)-1=√2sin(2x+π/4)-1因为-1≤sin(2x+π/4)≤1所以-√2-1≤f(x)≤√2-1其他自己算吧面积先用正弦定理求出B、C，S=1/2absinC