What is the value of \(\sin\left(2\text{n}\pi+\frac{5\pi}{6}\right)\sin\left(2\text{n}\pi−\frac{5\pi}{6}\right)\) where n ∈ Z ?
Evaluating the Trigonometric Expression \(\sin(2\text{n}\pi+\frac{5\pi}{6})\sin(2\text{n}\pi−\frac{5\pi}{6})\)
The problem asks us to find the value of the expression \(\sin\left(2\text{n}\pi+\frac{5\pi}{6}\right)\sin\left(2\text{n}\pi−\frac{5\pi}{6}\right)\), where \(n\) is an integer (n ∈ Z).
To solve this, we can use the property of the sine function that states \(\sin(x + 2k\pi) = \sin(x)\) and \(\sin(x - 2k\pi) = \sin(x)\) for any integer \(k\). In our expression, we have \(2n\pi\) being added or subtracted, and since \(n\) is an integer, \(2n\pi\) is a multiple of the period of the sine function (\(2\pi\)).
Step-by-Step Calculation
Let's consider the two parts of the expression separately:
- The first term is \(\sin\left(2\text{n}\pi+\frac{5\pi}{6}\right)\). Using the periodicity property \(\sin(x + 2k\pi) = \sin(x)\), we can simplify this as: \[\sin\left(2\text{n}\pi+\frac{5\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right)\]
- The second term is \(\sin\left(2\text{n}\pi−\frac{5\pi}{6}\right)\). Using the periodicity property \(\sin(x - 2k\pi) = \sin(x)\), and also knowing that \(\sin(-x) = -\sin(x)\), we can simplify this. Note that \(\sin(2n\pi - x) = \sin(-x) = -\sin(x)\). So, \[\sin\left(2\text{n}\pi−\frac{5\pi}{6}\right) = -\sin\left(\frac{5\pi}{6}\right)\]
Now, substitute these simplified terms back into the original expression:
\[\sin\left(2\text{n}\pi+\frac{5\pi}{6}\right)\sin\left(2\text{n}\pi−\frac{5\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) \times \left(-\sin\left(\frac{5\pi}{6}\right)\right)\] \[= -\left(\sin\left(\frac{5\pi}{6}\right)\right)^2\]Evaluating \(\sin\left(\frac{5\pi}{6}\right)\)
To find the value of \(\sin\left(\frac{5\pi}{6}\right)\), we can use the angle identity \(\sin(\pi - \theta) = \sin(\theta)\). We can write \(\frac{5\pi}{6}\) as \(\pi - \frac{\pi}{6}\).
\[\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right)\]The value of \(\sin\left(\frac{\pi}{6}\right)\) is known to be \(\frac{1}{2}\).
\[\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\]Final Calculation
Now, substitute the value of \(\sin\left(\frac{5\pi}{6}\right)\) back into the simplified expression:
\[-\left(\sin\left(\frac{5\pi}{6}\right)\right)^2 = -\left(\frac{1}{2}\right)^2\] \[= -\left(\frac{1^2}{2^2}\right) = -\left(\frac{1}{4}\right)\] \[= -\frac{1}{4}\]Thus, the value of the given expression is \(-\frac{1}{4}\).
Let's verify this against the given options.
| Option | Value | Matches Calculation? |
|---|---|---|
| 1 | \(-\frac{1}{4}\) | Yes |
| 2 | \(-\frac{3}{4}\) | No |
| 3 | \(\frac{1}{4}\) | No |
| 4 | \(\frac{3}{4}\) | No |
The calculated value \(-\frac{1}{4}\) matches Option 1.
Revision Table: Key Trigonometry Concepts
| Concept | Formula / Property | Example |
|---|---|---|
| Periodicity of Sine | \(\sin(x + 2k\pi) = \sin(x)\) \(\sin(x - 2k\pi) = \sin(x)\) (for integer \(k\)) |
\(\sin(2\pi + x) = \sin(x)\) \(\sin(4\pi - x) = \sin(-x)\) |
| Sine of Negative Angle | \(\sin(-x) = -\sin(x)\) | \(\sin(-\frac{\pi}{4}) = -\sin(\frac{\pi}{4})\) |
| Sine in Second Quadrant | \(\sin(\pi - \theta) = \sin(\theta)\) | \(\sin(\frac{5\pi}{6}) = \sin(\pi - \frac{\pi}{6}) = \sin(\frac{\pi}{6})\) |
| Standard Angle Value | \(\sin(\frac{\pi}{6}) = \frac{1}{2}\) | Used to evaluate \(\sin(\frac{5\pi}{6})\) |
Additional Information: Trigonometric Identities and Periodicity
Understanding the periodicity of trigonometric functions is crucial for simplifying expressions like the one in this question. The sine function has a period of \(2\pi\), which means its values repeat every \(2\pi\) interval.
For any angle \(\theta\) and any integer \(n\), the following holds true:
- \(\sin(\theta + 2n\pi) = \sin(\theta)\)
- \(\cos(\theta + 2n\pi) = \cos(\theta)\)
- \(\tan(\theta + n\pi) = \tan(\theta)\) (Note: tangent has a period of \(\pi\))
The identity \(\sin(-x) = -\sin(x)\) indicates that the sine function is an odd function. This property is also frequently used in simplifying trigonometric expressions.
Evaluating trigonometric functions for angles like \(\frac{5\pi}{6}\) involves relating them to standard angles (\(\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}\), etc.) using quadrant rules or angle identities. The angle \(\frac{5\pi}{6}\) lies in the second quadrant, where the sine function is positive. \(\frac{5\pi}{6}\) is \(150^\circ\), and it makes an angle of \(30^\circ\) with the negative x-axis, or is \(\pi - \frac{\pi}{6}\).
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