sinx cosx=√2 2,求sinx的四次方
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解题思路:考查三角恒等变换解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/read
y=sinxcosx-cos^2x=1/2sin2x-1/2(1+cos2x)=1/2(sin2x-cos2x-1)=1/2[√2*sin(2x-派/4)-1]=√2/2*sin(2x-派/4)-1/
y=sinxcosx+sinx+cosx=1/2(2sinxcosx+1-1)+sinx+cosx=1/2(sinx+cosx)^2-1/2+(sinx+cosx)=1/2[(sinx+cosx)^2
1.因为tanx>0,所以sinx和cosx同号,不妨令二者同正sinx/cosx=3,(sinx)^2+(cosx)^2=1联立,得sinx=3√10/10,cosx=√10/10,所以2sinxc
f(x)=sin(π/3+x)cos(π/3-x)-sinxcosx+1/4=1/2[sin(π/3+x+π/3-x)+sin(π/3+x-π/3+x)]-sinxcosx+1/4=1/2(sin2π
s(dx)/(sinxcosx)=s(sin²x+cos²x)/(sinxcosx)dx=s(sinx/cosx)+(cosx/sinx)dx=s(sinx/cosx)dx+s(c
sinxcosx=(1/2)*sin(2x)=(1/2)*{2tanx/[1+(tanx)^2]}=2/5
①(sinx+cosx)/(sinx-cosx)=2(sinx+cosx)=2*(sinx-cosx)sinx+cosx=2sinx-2cosxsinx=3cosxtanx=sinx/cosx=3②(
(sinX+cosX)平方=2所以sinX平方+cosX平方+2sinXcosX=2因为sinX平方+cosX平方=1所以sinXcosX=0.5
y=sin^x+2sinxcosx=1/2-cos2x/2+sin2x=根号下(5/4)*[2sin2x/根号5-cos2x/根号5]+1/2设cosa=2/根号5,sina=-1/根号5上式=根号下
sinxcosx=sinxcosx/(sin^2x+cos^2x)=tanx/(1+tan^2x)=3/10再问:还是不明白,能再详细点吗?再答:sin²x+cos²x=1sinx
sinx+cosx=1两边平方(sinx+cosx)²=1sin²x+cos²x+2sinxcosx=12sinxcosx=0∴sinx=0,cosx=1或cosx=0,
分两部分求2sin2x=4sinxcosx注:sin2x=2sinxcosx=4sinxcosx/{(cosx)^2+(sinx)^2}注:{(cosx)^2+(sinx)^2=1=4tanx/{1+
sin2x=2sinxcosx/1=2sinxcosx/(sin^2x+cos^2x)(分子分母同除以cos^2x)=2tanx/(tan^2x+1)=2*(-4)/[(-4)^2+1]=-8/17s
sinx+cosx=t√2sin(x+∏/4)=t-√2≤t≤√21+2sinxcosx=t²sinxcosx=(t²-1)/2y=1+sinx+cosx+sinxcosx=1+t
设t=sinx+cosx=2sin(x+π4),则t∈[-2,2].由(sinx+cosx)2=t2⇒sinxcosx=t2-12.∴y=1+t+t2-12=12(t+1)2.∴ymax=12(2+1
1、sin(x-45)=sinxcos45-cosxsin45=√2/2*(sinx-cosx)=√2/4sinx-cosx=1/2平方sin²x+cos²x-2sinxcosx=
令u=sinx+cosx=√2sin(x+π/4)∈[-√2,√2]u²=sin²x+cos²x+2sinxcosx=1+2sinxcosx∴4sinxcosx=2(u&
f(x)=√3cos²x+sinxcosx=√3(1+cos2x)/2+1/2sin2x=√3/2cos2x+1/2sin2x+√3/2=sinπ/3cos2x+cosπ/3sin2x+√3