A.maths Urgent Hw!!!!!!! 20pts
Prove the identities.
1 )sin(x+y)sin(x-y) = (sinx)^2-(siny)^2
2 )cot(A+B) = (cotAcotB-1)/(cotA+cotB)
回答 (3)
1) LHS=sin(x+y)sin(x-y)
= -1/2(cos2x-cos2y)
= -1/2(1-2sin^2x-1+2sin^2y)
= -1/2(-2sin^2x+2sin^2y)
= -1/2{-2(sin^2x-sin^2y)
= sin^2x-sin^2y
**cos2A=1-2sin^2A
2) RHS=(cotAcotB-1)/(cotA+cotB)
=(1/tanAtanB-1)/{(tanA+tanB)/tanAtanB)
={(1-tanAtanB)/(tanAtanB)}{(tanAtanB)/(tanA+tanB)}
=(1-tanAtanB)/(tanA+tanB)
=cot(A+B)
**tan(A+B)=(tanA+tanB)/(1-tanAtanB)
**cot(A+B)=1/tan(A+B)
1) L.H.S.=sin(x+y)sin(x-y)
=(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)
=sin^2xcos^2y-sinxcosycosxsiny+sinxcoscossiny-cos^2x sin^2y
=sin^2xcos^2y-cos^2xsin^2y
=sin^2x(1-sin^2y)-sin^2y(1-sin^2x)
=sin^2x-sin^2xsin^2y-sin^2y+sin^2xsin^2y
=sin^2-sin^2y
=R.H.S.
2)R.H.S=(cotAcotB-1)/(cotA+cotB)
=[1/(tanAtanB)-(tanAtanB)/(tanAtanB)]/[(tanBtanA)/(tanAtanB)
=(1-tanAtanB)/(tanAtanB)*(tanAtanB)/(tanB+tanA)
=(1-tanAtanB)/(tanA+tanB)
=L.H.S.
收錄日期: 2021-04-13 20:18:52
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