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a maths請寫step..**急

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(a) Prove that if tanAtan(A+B) = k , then (k+1)cos(2A + B) = (1-k)cosB (b) Find the general solution of tanAtan(A + 60° ) = 2

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(a) tanAtan(A+B) = k (sinA/cosA)[sin(A+B)/cos(A+B)]=k sinAsin(A+B)=kcosAcos(A+B) -1/2[cos(2A+B)-cos(-B)]=k/2[cos(2A+B)+cos(-B)] -[cos(2A+B)-cosB]=k[cos(2A+B)+cosB] -cos(2A+B)+cosB=kcos(2A+B)+kcosB (k+1)cos(2A + B) = (1-k)cosB...(*) (b) sub B=60,k=2 in (*) (2+1)cos(2A + 60) = (1-2)cos60 3cos(2A + 60) = -1/2 cos(2A + 60) = -1/6 2A + 60=360n+99.594 or 360n-99.594 2A=360n+39.594 or 360n-159.594 A= 180n+19.797 or 180n-79.797

其他解答:

a maths請寫step..**急 (a) Prove that if tanAtan(A+B) = k , then (k+1)cos(2A + B) = (1-k)cosB (b) Find the general solution of tanAtan(A + 60° ) = 2 *********************************************************************************** (a) tanAtan(A+B) = k (sinA/cosA)[sin(A+B)/cos(A+B)]=k sinAsin(A+B)=kcosAcos(A+B) -1/2[cos(2A+B)-cos(-B)]=k/2[cos(2A+B)+cos(-B)] -[cos(2A+B)-cosB]=k[cos(2A+B)+cosB] -cos(2A+B)+cosB=kcos(2A+B)+kcosB (k+1)cos(2A + B) = (1-k)cosB...(*) (b) sub B=60,k=2 in (*) (2+1)cos(2A + 60) = (1-2)cos60 3cos(2A + 60) = -1/2 cos(2A + 60) = -1/6 2A + 60=360n+99.594 or 360n-99.594 2A=360n+39.594 or 360n-159.594 A= 180n+19.797 or 180n-79.797