急求助!!F.4 A.maths－bgdwze3的部落格

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急求助!!F.4 A.maths

發問:

1.Show that for and△ABC tanA + tanB + tanC =tanAtanBtanC , where A,B,C are the interior anglesb) Evaluate, indegrees,the angles of△ABC if tanA :tanB: tanC = 1:2:-62 Given that x + ∮ = 兀/4, exprees tan ∮ in term of tanx and hence prove that ( 1 + tan x)(1+tan∮)=2.Deduce that tan兀/8 =厂2 -13. Let cos(x +a) = p... 顯示更多 1.Show that for and△ABC tanA + tanB + tanC =tanAtanBtanC , where A,B,C are the interior angles b) Evaluate, indegrees,the angles of△ABC if tanA :tanB: tanC = 1:2:-6 2 Given that x + ∮ = 兀/4, exprees tan ∮ in term of tanx and hence prove that ( 1 + tan x)(1+tan∮)=2.Deduce that tan兀/8 =厂2 -1 3. Let cos(x +a) = p and sin(x + B) = q. a. Express cosx and sina interms of a,B, p and q. b. Hence show that p^2 + q^2 + 2pqsin(a-B) = cos^2(a - B). THX!!!

最佳解答:

1 tanA + tanB + tanC =tan(180-(B+C)) + tanB + tanC =-tan (B+C) + tanB + tanC =[(tanB+tanC)/(tanBtanC-1)]+ tanB + tanC =(tanB + tanC)[((tanBtanC-1)+1)/(tanBtanC-1)] =tanBtanC(tanB + tanC)/(tanBtanC-1) =tanBtanC[-tan(B+C)] =tanBtanCtanA (b) tanA :tanB: tanC = 1:2:-6 tanB=2tanA; tanC =-6tanA tanA + tanB + tanC =tanAtanBtanC tanA + 2tanA -6 tanA =tanA(2tanA)(-6tanA) -3tanA=-12tan^3A tanA-4tan^3A=0 tanA(1-4tan^2A)=0 tanA=0 (rejected) or tanA=1/2 or tanA=-1/2 (rejected) tanA=1/2 A=26.565 B=45 C=108.435 2 Given that x + ∮ = 兀/4, exprees tan ∮ in term of tanx and hence prove that ( 1 + tan x)(1+tan∮)=2.Deduce that tan兀/8 =厂2 -1 tan (x+∮) =(tanx+tan∮)/(1-tanxtan∮) (tanx+tan∮)/(1-tanxtan∮)=1 (tanx+tan∮)=(1-tanxtan∮) tanx-1=-tan∮(tanx+1) tan∮=(1-tanx)/(1+tanx) ( 1 + tan x)(1+tan∮) =( 1 + tan x)[1+(1-tanx)/(1+tanx)] =( 1 + tan x)[(1+tanx)+(1-tanx)/(1+tanx)] =(1+tanx)+(1-tanx) =2 ( 1 + tan x)(1+tan∮)=2 1+(tanx+tan∮)+tanxtan∮=2 (tanx+tan∮)+tanxtan∮=1 if I let x=∮=兀/8 then tan兀/8+tan兀/8+tan^2兀/8=1 tan^2兀/8+2tan兀/8-1=0 tan兀/8 =1/2[-2+厂(4+4)] =1/2[-2+2厂2] =厂2 -1 3 (a) cos(x +a) = p and sin(x + B) = q. cosxcosa-sinxsina=p sinxcosB+cosxsinB=q So cosxcosacosB-sinxsinacosB=pcosB sinxcosBsina+cosxsinBsina=qsina add together that is cosxcosacosB+cosxsinBsina=pcosB+qsina cosx(cosacosB+sinBsina)=pcosB+qsina cosxcos(a-B)=pcosB+qsina...(1) On the other hand cosxcosasinB-sinxsinasinB=psinB sinxcosBcosa+cosxsinBcosa=qcosa second minus first sinxcosBcosa+sinxsinasinB=qcosa-psinB sinx(cosBcosa+sinasinB)=qcosa-psinB sinxcos(B-a)=qcosa-psinB...(2) cosxcos(a-B)=pcosB+qsina...(1) Now from (1) cosxcos(B-a)=pcosB+qsina...(3) sinx=[qcosa-psinB]/cos(B-a) cosx=[pcosB+qsina]/cos(B-a) (b) sin^2x=[qcosa-psinB]^2/cos(B-a)^2 cos^2x=[pcosB+qsina]^2/cos(B-a)^2 [qcosa-psinB]^2+[pcosB+qsina]^2=cos^2(B-a) q^2+p^2-2pqcosasinB+2sinacosB=cos^2(B-a) p^2 + q^2 + 2pqsin(a-B) = cos^2(a - B).

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