91Ó°ÊÓ

The half-angle formulas are vital tools in trigonometry. They allow us to find the trigonometric values of half-angles when given the values for full angles.

Specifically, the half-angle formula for sine is:

\text{sin}\bigg(\frac{x}{2}\bigg) = \text{sqrt}\frac{1 - \text{cos}(x)}{2} \[ \text{sin} \bigg( \frac{x}{2} \bigg) = \sqrt{\frac{1 - \text{cos}(x)}{2}} \]

This formula is particularly useful when solving trigonometric problems involving angles not immediately found in standard trigonometric tables.

In the given problem, we used \text{cos}\(x\) to find \text{sin}\(\frac{x}{2}\). By substituting the given \text{cos}\(x\) value into the formula, we could solve for the exact value.

The Pythagorean identity is a fundamental relation in trigonometry representing the square of sine plus the square of cosine equals one. It’s often written as: \[ \text{sin}^2(x) + \text{cos}^2(x) = 1 \]

This identity is significant because it interrelates the sine and cosine functions, offering a way to find one if we know the other.

In the problem, we used the Pythagorean identity to determine \text{sin}\(x\) after knowing \text{cos}\(x\). Here’s the step-by-step for our context:

• Given \text{cos}\(x\) = -\frac{1}{4}\
• Substitute into the identity: \text{sin}^2(x) + \bigg(\frac{-1}{4}\bigg)^2 = 1\
• Then, \text{sin}^2(x) + \frac{1}{16} = 1\
• Solve for \text{sin}^2(x): \text{sin}^2(x) = 1 - \frac{1}{16} = \frac{15}{16}\
• Since the problem states x is in the 2nd quadrant, where sine is positive, \text{sin}(x) = \frac{\text{sqrt}(15)}{4}\

Understanding trigonometric values in various quadrants is key to solving many trigonometric problems. Each of the four quadrants has a specific sign identity for sine, cosine, and tangent functions.

The signs for trigonometric functions by quadrant are:

• 1st Quadrant: All functions are positive.
• 2nd Quadrant: Sine is positive; cosine and tangent are negative.
• 3rd Quadrant: Tangent is positive; sine and cosine are negative.
• 4th Quadrant: Cosine is positive; sine and tangent are negative.

In our problem, \text{cos}\(x\) = -\frac{1}{4}\ indicates that x is in the 2nd quadrant, as specified by \frac{\text{Ï€}}{2} \textless\ x \textless\ \text{Ï€}\. This helps in determining the sign of \text{sin}\(x\). Since sine is positive in the 2nd quadrant, \text{sin}\(x\) = \frac{\text{sqrt}(15)}{4}\ becomes positive.