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

Since $\log_b \sin x = a$ , $b^a=\sin x$ .

Let $c = \log_b \cos x$ . Then $b^c=\cos x$ .

Since $\sin^2x+\cos^2x=1$ , $(b^c)^2+(b^a)^2=1$ $b^{2c}+b^{2a}=1$ $b^{2c}=1-b^{2a}$ $\log_b (1-b^{2a}) = 2c$ $c=\boxed{(\textbf{D}) \ \frac 12 \log_b(1-b^{2a})}$

-aopspandy

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1980 AHSME Problems/Problem 18

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