Asked by Anonymous
prove sin(A+B)/sin(A-B)=(tanA+tanB)/(tanA-tanB)
Answers
Answered by
Steve
tan(A+B) = (tanA+tanB)/(1-tanAtanB)
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
so,
tanA+tanB = tan(A+B)(1-tanAtanB)
tanA-tanB = tan(A-B)(1+tanAtanB)
divide the two and you have
tan(A+B)/tan(A-B) * (1-tanAtanB)/(1+tanAtanB)
multiply top and bottom by cosAcosB nad you have
tan(A+B)/tan(A-B) * (cosAcosB-sinAsinB)/(cosAcosB+sinAsinB)
That's just
tan(A+B)/tan(A-B) * cos(A+B)/cos(A-B)
= sin(A+B)/sin(A-B)
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
so,
tanA+tanB = tan(A+B)(1-tanAtanB)
tanA-tanB = tan(A-B)(1+tanAtanB)
divide the two and you have
tan(A+B)/tan(A-B) * (1-tanAtanB)/(1+tanAtanB)
multiply top and bottom by cosAcosB nad you have
tan(A+B)/tan(A-B) * (cosAcosB-sinAsinB)/(cosAcosB+sinAsinB)
That's just
tan(A+B)/tan(A-B) * cos(A+B)/cos(A-B)
= sin(A+B)/sin(A-B)
Answered by
timothy
very good
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