已知:sinA/sinB=p,cosA/cosB=q(p不为1且不为-1,q不为0),求tanA*tanB的值
參考答案:解答:
因为
sinA/sinB=p,cosA/cosB=q(p不为1且不为-1,q不为0)
则
(sinA)^2=(psinB)^2=p^2(sinB)^2 (1)式
(cosA)^2=(qcosB)^2=q^2(cosB)^2 =>
1-(sinA)^2 = q^2-q^2(sinB)^2 (2)式
由(1)式,(2)式解得
(sinB)^2 = (1-q^2)/(p^2-q^2)
所以 (cosB)^2=1-(sinB)^2=1-(1-q^2)/(p^2-q^2)=(p^2-1)/(p^2-q^2)
而
tanA*tanB =(sinA/cosA)*(sinB/cosB)
= (psinB/qcosB)*(sinB/cosB) (因sinA/sinB=p,cosA/cosB=q)
= (p/q)*[(sinB)^2/(cosB)^2]
=(p/q)*[(1-q^2)/(p^2-1)]
=p(1-q^2)/[q(p^2-1)]
注:其中^2表示平方