Exercise 4.2.8 solution
From the graph, we see that \(\cos(-A) = \cos(A)\) and that \(\sin(-A)=-\sin(A)\). It follows that since \(\tan x = \dfrac{\sin x}{\cos x}\) that \(\tan(-A)=\dfrac{\sin(-A)}{\cos(-A)}=\dfrac{-\sin A}{\cos A}=-\tan A\).
From the graph, we see that \(\cos(-A) = \cos(A)\) and that \(\sin(-A)=-\sin(A)\). It follows that since \(\tan x = \dfrac{\sin x}{\cos x}\) that \(\tan(-A)=\dfrac{\sin(-A)}{\cos(-A)}=\dfrac{-\sin A}{\cos A}=-\tan A\).