$6 \cdot 43$ If $b>1, \sin x>0, \cos x>0$ and $\log _{b} \sin x=a$, then $\log _{b} \cos x$ equals
(A) $2 \log _{b}\left(1-b^{\frac{a}{2}}\right)$.
(B) $\sqrt{1-a^{2}}$.
(C) $b^{a^{2}}$.
(D) $\frac{1}{2} \log _{b}\left(1-b^{2 a}\right)$.
(E) None of these.
(31st American High School Mathematics Examination, 1980)
[Solution] Let $\log _{b} \sin x=a$, then $\sin x=b^{a}$, so $\sin ^{2} x=b^{2 a}$. Therefore, $\cos x=\left(1-b^{2 a}\right)^{\frac{1}{2}}$, and thus $\log _{b} \cos x=\frac{1}{2} \log _{b}\left(1-b^{2 a}\right)$. Hence, the answer is $(D)$.