已知sinx =-2 根号2 ,求-360°~360°范围的角x

原式=(sin²x+sinxcosx)/(2sin²x+cos²x)1=sin²x+cos²x=(tan²x+tanx)/(2tan

∵sinx/2-2cosx/2=0∴sinx/2=2cosx/2∴tanx/2=2∴tanx=-4/3∵cos2x/根号2cos(π/4+x)sinx=[(cosx)^2-(sinx)^2]/(cos

解析sinx+cosx=√2/2(sinx+cosx)2=1/2(sinx+cosx)2=1+sin2x=1/2sin2x=-1/2(sinx-cosx)2=1-sin2x=3/2sinx-cosx=

用sinx2+cosx2=1再问：可以帮我把顺序写出吗。再问：谢谢。再答：就是3cos2x+cos2x=1再答：然后cosx=正负0.5就行了再问：就只有后面这两个步骤吧再问：还是前面那个也要再问：s

sinx+cosx=√2两边同时平方得sin²x+2sinxcosx+cos²x=21+2sinxcosx=22sinxcosx=1sinxcosx=1/2答案：1/2

sinx=2tan(x/2)/(1+tanx/2tanx/2)=2*sqrt(5)/(1+5)=sqrt(5)/3cosx=(1-tanx/2tanx/2)/(1+tanx/2tanx/2)=-2/3

(sinx-cosx)^2=1/2=sin^2x+cos^2x+2sinx*cosx=1+2sinx*cosx∴sinx*cosx=-1/4sin^3x-cos^3x=(sinx-cosx)(sin^

sinx/（1-1/tanx）+cosx/(1-tanx)=sin^2x/(sinx-cosx)+cos^2x/(cosx-sinx)=(sin^2x-cos^2x)/(sinx-cosx)=sinx

设a=sinx,b=cosx,题中的隐含条件是a^2+b^2=1由a+b=√2/2得(a+b)^2=1/2,a^2+2ab+b^2=1/2,2ab=-1/2,ab=-1/4(a^2+b^2)^2=1,

sinx-cosx=√2/2（sinx-cos）²=sin²x+cos²x-2sinxcosx=1-2sinxcosx=1/2∴2sinxcosx=1/2∴（sinx+c

分析,f(x)=a*b-√3=2sinx*cosx+2√3sin²x-√3=sin(2x)-√3cos(2x)=2sin(2x-π/3)当,2kπ+π/2≦2x-π/3≦2kπ+3π/2∴k

[sinx+cos(π+x)]/[sinx+sin(π/2-x)]=(sinx-cosx)/(sinx+cosx)=(tanx-1)/(tanx+1)=1-[2/(tanx+1)]sinx=√3/3∴

已知tanx=根号2则[2cos2(x/2)-sinx-1]/(sinx+cosx)=(cosx-sinx)/(sinx+cosx)=(1-tanx)/(tanx+1)=(1-√2)/(1+√2)=(

(1)a*b=0sin2x-cos2x=0sqr(2)sin(2x-π/4)=0x=π/8+kπ/2,k∈Z(2)f(x)=sqr(2)sin(2x-π/4)x∈(3π/8+kπ,7π/8+kπ),k

cos(x-90)=cos(90-x)=sinx=√2/10∴sinx=√2/10再问：cos(X-45)=根号2/10，X属于90到135度，求sinx和sin（2x+60）再答：=cos(x-45

可是我用sinxcosx=-根号3/4sinx=-根号3/4*cosx你确定移项过来不应该是除么.

sin2x/sinx-cosx=2sinxcosx/sinx-cosx=cosxsinx+cosx=根号2/3根号下(1-cos^2(x))+cosx=根号2/3然后自己求就行啦

没错,f(x)=2sin(2x+π/6)周期T=2π/2=π因为-1≤sin(2x+π/6)≤1f(x)max=2f(x)min=-2

已知sinx+cosx=-根号10/5,求（1）1/sinx+1/cosx(2)tanx的值因为sinx+cosx=-(根号10)/5所以(sinx+cosx)^2=[-(根号10)/5]^2