5.\tan \theta<0이고 \cos \left(\frac{\pi}{2}+\theta\right)=\frac{\sqrt{5}}{5}일 때, \cos \theta의 값은?


1.-\frac{2 \sqrt{5}}{5}

2.-\frac{\sqrt{5}}{5}

3.0

4.\frac{\sqrt{5}}{5}

5.\frac{2 \sqrt{5}}{5}


The given conditions are:


1. \(\tan \theta < 0\)

2. \(\cos \left(\frac{\pi}{2}+\theta\right) = \frac{\sqrt{5}}{5}\)


We know that \(\cos \left(\frac{\pi}{2}+\theta\right) = -\sin \theta\). So, we can rewrite the second condition as \(-\sin \theta = \frac{\sqrt{5}}{5}\), which implies \(\sin \theta = -\frac{\sqrt{5}}{5}\).


We also know that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Given that \(\tan \theta < 0\) and \(\sin \theta = -\frac{\sqrt{5}}{5}\), we can infer that \(\cos \theta > 0\) (since the sign of \(\tan \theta\) is determined by the product of the signs of \(\sin \theta\) and \(\cos \theta\)).


Finally, we can use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to solve for \(\cos \theta\). Substituting \(\sin \theta = -\frac{\sqrt{5}}{5}\) into the identity gives us \(\cos^2 \theta = 1 - \left(-\frac{\sqrt{5}}{5}\right)^2\), and taking the positive square root (since we've determined that \(\cos \theta > 0\)) gives us the value of \(\cos \theta\).


Let's calculate this.


The value of \(\cos \theta\) is \(\frac{2 \sqrt{5}}{5}\), which corresponds to option 5.


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