推 Proakis :謝謝大大 還有一題能教教我嗎 09/15 10:33
→ Proakis :眼花了..原來兩題都解了 09/15 10:33
推 Proakis :為什麼(1-sin^2(y))跟(1-cos^2(x))=1 09/15 10:36
推 sincere617 :在樓下那邊 ^^ 09/15 10:38
推 qna :不等於1阿 展開前後消掉了] 09/15 17:31
※ 引述《Proakis (John G. Proakis)》之銘言：
Prove the identity
cos(x+y)cos(x-y)=cos^2 (x) - sin^2 (y)
cos(x+y)cos(x-y) = (cosxcosy - sinxsiny)(cosxcosy + sinxsiny)
= cos^2(x)cos^2(y) - sin^2(x)sin^2(y)
= cos^2(x)(1-sin^2(y)) - (1-cos^2(x))sin^2(y)
= cos^2(x) - sin^2(y)
※ 引述《Proakis (John G. Proakis)》之銘言：
Prove the identity
tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)
Let x-y = A y-z = B z-x = C
We have A+B+C = 0
tanA + tanB
And tanC = tan(-A-B) = - tan(A+B) = - --------------
1 - tanAtanB
<=> tanC ( 1 - tanAtanB ) = - ( tanA +tanB )
<=> tanA + tanB +tanC = tanAtanBtanC
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◆ From: 140.112.244.138
※ 編輯: FAlin 來自: 140.112.244.138 (09/15 10:32)