c) $$ \cos 240^{\circ}+\sin 210^{\circ}+\frac{\sin 310^{\circ}}{\cos 140^{\circ}} $$ d) $$ \frac{\sin ^{2}\left(-100^{\circ} ight)}{\cos 120^{\circ} \sin 80^{\circ} \cos 370^{\circ}} $$ e) $$ \frac{\sin \left(-640^{\circ} ight)}{\sin ^{2} 217^{\circ}-\cos \left(-397^{\circ} ight) \sin 307^{\circ}} imes \frac{1}{\sin 260^{\circ}} $$ 12. Simplify: a) $$ \left[\sin (- heta)+\cos \left(360^{\circ}+ heta ight) ight]\left[\cos \left( heta-90^{\circ} ight)+\cos heta ight] $$ b) $$ \frac{\sin \left(90^{\circ}-x ight) \sin \left(90^{\circ}+x ight)}{\left(\sin 90^{\circ}-\sin x ight)\left(\sin 90^{\circ}+\sin x ight)} $$ c) $$ \frac{\cos 140^{\circ}-\sin \left(90^{\circ}-x ight)}{\sin 410^{\circ}+\cos (-x)} $$ d) $$ \cos \left(x-90^{\circ} ight) \sin \left(x-180^{\circ} ight)+\frac{\cos \left(360^{\circ}+x ight)}{\sin \left(90^{\circ}-x ight)} $$ e) $$ \frac{\cos \left(90^{\circ}+x ight) \cos (-x) \sin (-x)}{\sin \left(x-90^{\circ} ight) an \left(360^{\circ}-x ight) \cos x} $$ f) $$ 1-\frac{\sin ^{2}\left(180^{\circ}+x ight)}{1-\cos ^{2}\left(180^{\circ}+x ight)} $$

Question

c) $$ \cos 240^{\circ}+\sin 210^{\circ}+\frac{\sin 310^{\circ}}{\cos 140^{\circ}} $$ d) $$ \frac{\sin ^{2}\left(-100^{\circ}ight)}{\cos 120^{\circ} \sin 80^{\circ} \cos 370^{\circ}} $$ e) $$ \frac{\sin \left(-640^{\circ}ight)}{\sin ^{2} 217^{\circ}-\cos \left(-397^{\circ}ight) \sin 307^{\circ}} 	imes \frac{1}{\sin 260^{\circ}} $$ 12. Simplify: a) $$ \left[\sin (-	heta)+\cos \left(360^{\circ}+	hetaight)ight]\left[\cos \left(	heta-90^{\circ}ight)+\cos 	hetaight] $$ b) $$ \frac{\sin \left(90^{\circ}-xight) \sin \left(90^{\circ}+xight)}{\left(\sin 90^{\circ}-\sin xight)\left(\sin 90^{\circ}+\sin xight)} $$ c) $$ \frac{\cos 140^{\circ}-\sin \left(90^{\circ}-xight)}{\sin 410^{\circ}+\cos (-x)} $$ d) $$ \cos \left(x-90^{\circ}ight) \sin \left(x-180^{\circ}ight)+\frac{\cos \left(360^{\circ}+xight)}{\sin \left(90^{\circ}-xight)} $$ e) $$ \frac{\cos \left(90^{\circ}+xight) \cos (-x) \sin (-x)}{\sin \left(x-90^{\circ}ight) 	an \left(360^{\circ}-xight) \cos x} $$ f) $$ 1-\frac{\sin ^{2}\left(180^{\circ}+xight)}{1-\cos ^{2}\left(180^{\circ}+xight)} $$

UpStudy AI · Accepted Answer

c) We are given
$$
\cos 240^{\circ}+\sin 210^{\circ}+\frac{\sin 310^{\circ}}{\cos 140^{\circ}}.
$$
• Notice that
$$
\cos 240^{\circ} = \cos(180^{\circ}+60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2},
$$
$$
\sin 210^{\circ} = \sin(180^{\circ}+30^{\circ}) = -\sin 30^{\circ} = -\frac{1}{2},
$$
$$
\sin 310^{\circ} = \sin(360^{\circ}-50^{\circ}) = -\sin 50^{\circ},
$$
$$
\cos 140^{\circ} = \cos(180^{\circ}-40^{\circ}) = -\cos 40^{\circ}.
$$
• Thus,
$$
\frac{\sin 310^{\circ}}{\cos 140^{\circ}} = \frac{-\sin 50^{\circ}}{-\cos 40^{\circ}} = \frac{\sin 50^{\circ}}{\cos 40^{\circ}}.
$$
But since
$$
\sin 50^{\circ} = \cos(90^{\circ}-50^{\circ}) = \cos 40^{\circ},
$$
we have
\[

c) $$ \cos 240^{\circ} + \sin 210^{\circ} + \frac{\sin 310^{\circ}}{\cos 140^{\circ}} = -\frac{1}{2} - \frac{1}{2} + \frac{\sin 50^{\circ}}{-\cos 40^{\circ}} = -1 - \frac{\cos 40^{\circ}}{\cos 40^{\circ}} = -2 $$

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