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Algebra Difficulty 2.2 Junior Find the answer

If b>1b>1, sin⁡x>0\sin x>0, cos⁡x>0\cos x>0, and log⁡bsin⁡x=a\log_b \sin x = a, then log⁡bcos⁡x\log_b \cos x equals

Pick one

Solution

log⁡bsin⁡x=a\log_b \sin x = a
ba=sin⁡xb^a=\sin x
log⁡bcos⁡x=c\log_b \cos x=c
bc=cos⁡xb^c=\cos x
Since sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1,
(bc)2+(ba)2=1(b^c)^2+(b^a)^2=1
b2c+b2a=1b^{2c}+b^{2a}=1
b2c=1−b2ab^{2c}=1-b^{2a}
log⁡b(1−b2a)=2c\log_b (1-b^{2a}) = 2c
c=(D) 12log⁡b(1−b2a)c=\boxed{\text{(D)} \ \frac 12 \log_b(1-b^{2a})}
-aopspandy

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.