## Precalculus (10th Edition)

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We know that for any acute angle $\theta$, $\sin(\theta)=\cos(90^{o}-\theta)$. Hence, $\sin(38^{o})-\cos(52^{o}) \\=\cos(90^{o}-38^{o})-\cos(52^{o}) \\=\cos(52^{o})-\cos(52^{o}) \\=0$