a triangle ABC, how to prove sin^2(A)-sin^2(B)=sin(A-B)sinC?

sin(A-B)sinC=sin(A-B)sin(A+B)=(sinAcosB-cosAsinB)(sinAcosB+cosAsinB)

=sin^2Acos^2B-cos^2Asin^2B=sin^2Acos^2B-(1-sin^2A)(1-cos^2B)

=sin^2Acos^2B-(1-sin^2A-cos^2B+sin^2Acos^2B)

= -1+sin^2A+cos^2B= -1+sin^2A+(1-sin^2B)=sin^2A-sin^2B

硬要從左到右需要不少技巧,需加上一些臨時輔佐數字,例如:加1再減1..等等,不是一時想得到的,若是規定如此只好請你將我寫得從右寫到左,至於其他方法目前還沒想到

LHS--->RHS:-

We are learning -0 to +0 and +0 to -0,properly,

For the -0 to +0 refers to +sinC

For the +0 to -0 refers to +sinC

;

Both improperly !?

Yet properly to sinC.

由要證明的等式 sin^2(A)-sin^2(B)=sin(A-B)sinC

來看, 顯然是從右式變到左式比較自然. 也沒有規

定一定要從左式變到右式才正規. 若一定要, 那麼

把右式變成左式的過程反著寫就可以了.

Gong, can you rewrite LHS instead of RHS?