If $b>1$, $\sin x>0$, $\cos x>0$, and $\log_b \sin x = a$, then $\log_b \cos x$ equals
$\text{(A)} \ 2\log_b(1-b^{a/2}) ~~\text{(B)} \ \sqrt{1-a^2} ~~\text{(C)} \ b^{a^2} ~~\text{(D)} \ \frac 12 \log_b(1-b^{2a}) ~~\text{(E)} \ \text{none of these}$
\[\log_b \sin x = a\]
\[b^a=\sin x\]
\[\log_b \cos x=c\]
\[b^c=\cos x\]
Since $\sin^2x+\cos^2x=1$, 
\[(b^c)^2+(b^a)^2=1\]
\[b^{2c}+b^{2a}=1\]
\[b^{2c}=1-b^{2a}\]
\[\log_b (1-b^{2a}) = 2c\]
\[c=\boxed{\text{(D)} \ \frac 12 \log_b(1-b^{2a})}\]
-aopspandy