In triangle ABC, prove that:
sin^2 (A) = sin^2 (B) + sin^2(C) - 2sinB sinC cosA
triangle sine and cos law---thz
2007-02-04 1:42 am
回答 (1)
2007-02-04 2:01 am
✔ 最佳答案
cos(A-B)-cos(A+B)=2sinAsinBcos(A+B)+cos(A-B)=2cosAcosB
sin^2 B + sin^2C - 2sinB sinC cosA
=sin^2 B + sin^2C - [cos(B-C)-cos(B+C)] cosA
=sin^2 B + sin^2C - [cos(B-C)-cos(180-A)] cosA
=sin^2 B + sin^2C - [cos(B-C)+cosA] cosA
=sin^2 B + sin^2C - cos(B-C)cosA-cos^2A
=sin^2 B + sin^2C + cos(B-C)cos(B+C)-cos^2A
=sin^2 B + sin^2C + 1/2[cos(2B)+cos(2C)]-cos^2A
=sin^2 B + sin^2C + 1/2[2cos^2B-1+2cos^2C-1]-cos^2A
=sin^2 B + sin^2C + cos^2B+cos^2C-1-cos^2A
=1-cos^2A
=sin^2A
收錄日期: 2021-04-25 16:54:54
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