# Difference between revisions of "1980 AHSME Problems/Problem 18"

## Problem

If $b>1$, $\sin x>0$, $\cos x>0$, and $\log_b \sin x = a$, then $\log_b \cos x$ equals

$\text{(A)} \ 2\log_b(1-b^{a/2}) ~~\text{(B)} \ \sqrt{1-a^2} ~~\text{(C)} \ b^{a^2} ~~\text{(D)} \ \frac 12 \log_b(1-b^{2a}) ~~\text{(E)} \ \text{none of these}$

## Solution

$$\log_b \sin x = a$$ $$b^a=\sin x$$ $$\log_b \cos x=c$$ $$b^c=\cos x$$ 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