To compare the trigonometric expressions \( \sin 45^\circ \) and \( \cos \left( \frac{\pi}{6} \right) \), we can first find the values of each expression.
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Calculate \( \sin 45^\circ \): \[ \sin 45^\circ = \frac{\sqrt{2}}{2} \]
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Calculate \( \cos \left( \frac{\pi}{6} \right) \): \[ \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \]
Now we have:
- \( \sin 45^\circ = \frac{\sqrt{2}}{2} \)
- \( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \)
Next, we compare \( \frac{\sqrt{2}}{2} \) and \( \frac{\sqrt{3}}{2} \):
To do this, we can compare \( \sqrt{2} \) and \( \sqrt{3} \):
- \( \sqrt{2} \approx 1.414 \)
- \( \sqrt{3} \approx 1.732 \)
Since \( \sqrt{2} < \sqrt{3} \), we conclude that: \[ \frac{\sqrt{2}}{2} < \frac{\sqrt{3}}{2} \]
Thus, we can write: \[ \sin 45^\circ < \cos \left( \frac{\pi}{6} \right) \]
The answer is: \[ \sin 45^\circ < \cos \left( \frac{\pi}{6} \right) \]