Finding the Value of tan (-1125°)

This problem asks us to find the value of the tangent function for a negative angle, specifically −1125°. We can solve this using properties of trigonometric functions and angle reduction formulas.

Understanding Tangent of a Negative Angle

The tangent function is an odd function. This means that for any angle \(\theta\), the tangent of the negative angle is the negative of the tangent of the positive angle. Mathematically, this is expressed as:

\[ \tan(-\theta) = -\tan(\theta) \]

Using this property, we can rewrite the given problem:

\[ \tan(-1125^\circ) = -\tan(1125^\circ) \]

Now, we need to find the value of \(\tan(1125^\circ)\).

Reducing Large Angles for Tangent

Trigonometric functions have a periodicity of 360° (or \(2\pi\) radians). This means that adding or subtracting any multiple of 360° to an angle does not change the value of its trigonometric function. For the tangent function specifically, it also has a period of 180° (or \(\pi\) radians), meaning \(\tan(\theta + 180^\circ n) = \tan(\theta)\) for any integer \(n\). However, it's standard practice to reduce angles by subtracting multiples of 360° to find an equivalent angle between 0° and 360° (or 0° and 180° for tangent's primary period).

To reduce the angle 1125°, we find how many times 360° fits into 1125°. We can do this by dividing 1125 by 360:

\[ \frac{1125}{360} \approx 3.125 \]

This tells us that 1125° is more than 3 full cycles of 360°. We subtract 3 times 360° from 1125°:

\[ 1125^\circ - 3 \times 360^\circ = 1125^\circ - 1080^\circ = 45^\circ \]

So, the angle 1125° is coterminal with 45°. This means:

\[ \tan(1125^\circ) = \tan(45^\circ) \]

Evaluating tan (45°)

The tangent of 45° is a standard trigonometric value that is often memorized or derived from the properties of a 45-45-90 right triangle. In such a triangle, the opposite side and the adjacent side to the 45° angle are equal.

\[ \tan(45^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{side}}{\text{side}} = 1 \]

So, \(\tan(45^\circ) = 1\).

Final Calculation of tan (-1125°)

Now we can substitute the value of \(\tan(1125^\circ)\) back into our expression from the first step:

\[ \tan(-1125^\circ) = -\tan(1125^\circ) = -\tan(45^\circ) \]

Since we found that \(\tan(45^\circ) = 1\), we have:

\[ \tan(-1125^\circ) = -(1) = -1 \]

Therefore, the value of \(\tan(-1125^\circ)\) is −1.

Summary of Steps:

1. Use the property \(\tan(-\theta) = -\tan(\theta)\) to handle the negative angle.
2. Reduce the large positive angle by subtracting multiples of 360°.
3. Evaluate the tangent of the resulting standard angle (45°).
4. Apply the negative sign from step 1 to get the final answer.

Revision Table: Key Trigonometric Values

Additional Information on Trigonometric Properties

Understanding the properties of trigonometric functions for negative angles and large angles is crucial for solving many problems. Here are some key identities:

• Even and Odd Functions:
  • Cosine and Secant are even functions: \(\cos(-\theta) = \cos(\theta)\), \(\sec(-\theta) = \sec(\theta)\).
  • Sine, Cosecant, Tangent, and Cotangent are odd functions: \(\sin(-\theta) = -\sin(\theta)\), \(\csc(-\theta) = -\csc(\theta)\), \(\tan(-\theta) = -\tan(\theta)\), \(\cot(-\theta) = -\cot(\theta)\).
• Periodicity:
  • Sine, Cosine, Secant, Cosecant have a period of 360° (or \(2\pi\)): \(f(\theta + 360^\circ n) = f(\theta)\).
  • Tangent and Cotangent have a period of 180° (or \(\pi\)): \(f(\theta + 180^\circ n) = f(\theta)\). While you can use the 180° period for tangent/cotangent reduction, using 360° reduction first is a general method applicable to all trig functions.

These properties allow us to evaluate trigonometric functions for any angle by relating it back to an angle between 0° and 360°, or even between 0° and 90° using quadrantal analysis and reference angles.