91Ó°ÊÓ

The Pythagorean Identity is a fundamental relation in trigonometry that connects the sine and cosine functions. It is expressed as: \[ \sin^2 \theta + \cos^2 \theta = 1 \]This identity helps to find the cosine value when the sine is known, and vice versa. In our exercise, we start with \(\sin \theta = \frac{1}{2}\) and use the Pythagorean identity to find \(\cos \theta\). By substituting \(\sin \theta\) into the identity, we get the equation: \[ \left( \frac{1}{2} \right)^2 + \cos^2 \theta = 1 \]Solving this, we find \( \cos^2 \theta = \frac{3}{4} \), and thus \( \cos \theta = \pm \frac{\sqrt{3}}{2} \). This reveals that the cosine can have two possible values depending on the quadrant.

The sine function measures the y-coordinate of a point on the unit circle for a given angle \(\theta\). It is a periodic function, meaning it repeats values over regular intervals. Key points to remember about the sine function:

• \(\sin \theta\) ranges from -1 to 1.
• It is positive in the first and second quadrants.
• It is negative in the third and fourth quadrants.

For our problem, \(\sin \theta = \frac{1}{2}\), which implies \(\theta\) could be in either the first or the second quadrant. Sine being positive indicates these quadrant possibilities.

The cosine function measures the x-coordinate of a point on the unit circle for a given angle \(\theta\). Like the sine function, it is periodic and ranges from -1 to 1. Important details about the cosine function include:

• \(\cos \theta\) is positive in the first and fourth quadrants.
• It is negative in the second and third quadrants.

In our exercise, after finding \(\cos^2 \theta = \frac{3}{4} \), we take the square root to find \(\cos \theta = \pm \frac{\sqrt{3}}{2} \). The sign depends on whether \(\theta\) is in the first or second quadrant, making \(\cos \theta\) positive in the first quadrant and negative in the second.

The tangent function is the ratio of the sine function to the cosine function: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]This function helps to find the slope of the angle \(\theta\). Some key points about the tangent function are:

• It can take any real number value.
• It is positive in the first and third quadrants.
• It is negative in the second and fourth quadrants.

For our exercise, we calculated \(\tan \theta\) for both possible values of \(\cos \theta\):\[ \text{For } \cos \theta = \frac{\sqrt{3}}{2} \Rightarrow \tan \theta = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]\[ \text{For } \cos \theta = -\frac{\sqrt{3}}{2} \Rightarrow \tan \theta = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \]Choosing the correct value of \( \tan \theta \) depends on the quadrant.

The coordinate plane is divided into four quadrants determined by the positive and negative signs of the x and y coordinates:

• First Quadrant: Both x and y are positive.
• Second Quadrant: x is negative, y is positive.
• Third Quadrant: Both x and y are negative.
• Fourth Quadrant: x is positive, y is negative.

Each trigonometric function has different signs depending on the quadrant. For the exercise, knowing \(\sin \theta = \frac{1}{2}\) implies that \(\theta\) could be in the first or second quadrant.

• In the first quadrant, \(\cos \theta = \frac{\sqrt{3}}{2}\), \(\sin \theta = \frac{1}{2}\), hence \(\tan \theta = \frac{\sqrt{3}}{3} \)
• In the second quadrant, \(\cos \theta = -\frac{\sqrt{3}}{2}\), \(\sin \theta = \frac{1}{2}\), hence \(\tan \theta = -\frac{\sqrt{3}}{3} \)