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What is $\sin 105^{\circ}+\cos 105^{\circ}$ equal to?
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The correct answer is:
$\frac{1}{\sqrt{2}}$
\sin 105^{\circ}+\cos 105^{\circ} \\
=\frac{\sin \left(\frac{2 x}{2}\right)}{\cos \left(\frac{2 x}{2}\right)}=\frac{\sin x}{\cos x}=\tan x . \\
=\left(\sin 60^{\circ} \cdot \sin 45^{\circ}+\cos 60^{\circ} \cdot \cos 45^{\circ}\right)+\left(\cos 60^{\circ} \cos 45^{\circ}\right. \\
\left.-\sin 60^{\circ} \sin 45^{\circ}\right) \\
=\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} \\
=\frac{\sin \left(\frac{2 x}{2}\right)}{\cos \left(\frac{2 x}{2}\right)}=\frac{\sin x}{\cos x}=\tan x . \\
=\left(\sin 60^{\circ} \cdot \sin 45^{\circ}+\cos 60^{\circ} \cdot \cos 45^{\circ}\right)+\left(\cos 60^{\circ} \cos 45^{\circ}\right. \\
\left.-\sin 60^{\circ} \sin 45^{\circ}\right) \\
=\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} \\
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