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If $p=\sin \left(989^{\circ}\right) \cos \left(991^{\circ}\right)$, then which one of the following is correct?
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Verified Answer
The correct answer is:
$p$ is finite and negative
Given, $p=\sin \left(989^{\circ}\right) \cos \left(991^{\circ}\right)$
Which can be written as $=\sin \left(1080^{\circ}-91^{\circ}\right) \cos \left(1080^{\circ}-89^{\circ}\right)$
$=-\sin 91^{\circ} \cos 89^{\circ}$
$=-\sin \left(90^{\circ}+1^{\circ}\right) \cos 89^{\circ}$
$=-\cos 1^{\circ} \cos 89^{\circ}$
As $\cos 1^{\circ}$ and $\cos 89^{\circ}$ are positive. therefore their product is also $+\mathrm{ve}$ Hence, $p$ is finite and negative.
Which can be written as $=\sin \left(1080^{\circ}-91^{\circ}\right) \cos \left(1080^{\circ}-89^{\circ}\right)$
$=-\sin 91^{\circ} \cos 89^{\circ}$
$=-\sin \left(90^{\circ}+1^{\circ}\right) \cos 89^{\circ}$
$=-\cos 1^{\circ} \cos 89^{\circ}$
As $\cos 1^{\circ}$ and $\cos 89^{\circ}$ are positive. therefore their product is also $+\mathrm{ve}$ Hence, $p$ is finite and negative.
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