Understanding cos 10° - sin 10°

The question asks us to determine the value or the nature (positive, negative, zero, or one) of the expression $\cos 10^\circ - \sin 10^\circ$. To do this, we need to compare the values of $\cos 10^\circ$ and $\sin 10^\circ$.

Comparing cos θ and sin θ in the First Quadrant

The angles $10^\circ$ is in the first quadrant ($0^\circ < \theta < 90^\circ$). In the first quadrant, both sine and cosine values are positive. However, their relative sizes change.

• For angles $\theta$ between $0^\circ$ and $45^\circ$ (i.e., $0^\circ \le \theta < 45^\circ$), the value of $\cos \theta$ is greater than the value of $\sin \theta$.
• At $\theta = 45^\circ$, $\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$.
• For angles $\theta$ between $45^\circ$ and $90^\circ$ (i.e., $45^\circ < \theta \le 90^\circ$), the value of $\sin \theta$ is greater than the value of $\cos \theta$.

Evaluating cos 10° - sin 10°

The angle given is $10^\circ$. This angle lies in the range $0^\circ \le \theta < 45^\circ$ because $0^\circ \le 10^\circ < 45^\circ$.

According to the comparison rule for the first quadrant, when the angle $\theta$ is between $0^\circ$ and $45^\circ$, we have $\cos \theta > \sin \theta$.

Applying this to $\theta = 10^\circ$:

$\cos 10^\circ > \sin 10^\circ$

If we subtract $\sin 10^\circ$ from both sides of this inequality, we get:

$\cos 10^\circ - \sin 10^\circ > 0$

This inequality tells us that the value of $\cos 10^\circ - \sin 10^\circ$ is positive.

Let's consider some known values to verify the trend:

• $\cos 0^\circ = 1$, $\sin 0^\circ = 0$. $\cos 0^\circ - \sin 0^\circ = 1 - 0 = 1$ (positive).
• $\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$, $\sin 30^\circ = \frac{1}{2} = 0.5$. $\cos 30^\circ - \sin 30^\circ \approx 0.866 - 0.5 = 0.366$ (positive).
• $\cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707$, $\sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707$. $\cos 45^\circ - \sin 45^\circ = 0$ (zero).
• $\cos 60^\circ = \frac{1}{2} = 0.5$, $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$. $\cos 60^\circ - \sin 60^\circ \approx 0.5 - 0.866 = -0.366$ (negative).

Since $10^\circ$ is between $0^\circ$ and $45^\circ$, $\cos 10^\circ$ is greater than $\sin 10^\circ$, making their difference positive.

Therefore, the value of $\cos 10^\circ - \sin 10^\circ$ is positive.

The final answer is $\boxed{positive}$.

Revision Table: Sine and Cosine in First Quadrant

Additional Information: Relationship between Sine and Cosine

The graphs of $y = \sin \theta$ and $y = \cos \theta$ are useful for visualizing their relationship. In the first quadrant, the graph of $y = \cos \theta$ starts at 1 (at $\theta=0^\circ$) and decreases to 0 (at $\theta=90^\circ$). The graph of $y = \sin \theta$ starts at 0 (at $\theta=0^\circ$) and increases to 1 (at $\theta=90^\circ$).

The two graphs intersect at $\theta = 45^\circ$. Before this intersection point (for $\theta < 45^\circ$), the cosine graph is above the sine graph, meaning $\cos \theta > \sin \theta$. After the intersection point (for $\theta > 45^\circ$), the sine graph is above the cosine graph, meaning $\sin \theta > \cos \theta$. This visual confirms the relationship used to solve the problem.