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If an angle $\mathrm{B}$ is complement of an angle $\mathrm{A}$, what are the greatest and least values of $\cos A$ cos $B$ respectively ?
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Verified Answer
The correct answer is:
\frac{1}{2},-\frac{1}{2}
Since, A and B are complementary angles, then A $+\mathrm{B}=90^{\circ}$
Now, $\cos A \cos B=\cos A \cos \left(90^{\circ}-\mathrm{A}\right)$
$=\cos \mathrm{A} \sin \mathrm{A}=\frac{1}{2} \sin 2 \mathrm{~A}$
Since, $-1 \leq \sin 2 \mathrm{~A} \leq 1$
Hence, $-\frac{1}{2} \leq \frac{1}{2} \sin 2 \mathrm{~A} \leq \frac{1}{2}$.
Thus, greatest and least values of $\cos \mathrm{A} \cos \mathrm{B}$ are $\frac{1}{2}$ and $-\frac{1}{2}$.
Now, $\cos A \cos B=\cos A \cos \left(90^{\circ}-\mathrm{A}\right)$
$=\cos \mathrm{A} \sin \mathrm{A}=\frac{1}{2} \sin 2 \mathrm{~A}$
Since, $-1 \leq \sin 2 \mathrm{~A} \leq 1$
Hence, $-\frac{1}{2} \leq \frac{1}{2} \sin 2 \mathrm{~A} \leq \frac{1}{2}$.
Thus, greatest and least values of $\cos \mathrm{A} \cos \mathrm{B}$ are $\frac{1}{2}$ and $-\frac{1}{2}$.
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