If \(\rm tan\:x=-\dfrac{3}{4}\) and x is in the second quadrant, then what is the value of sin x ⋅ cos x?
Understanding the Problem: Finding sin x cos x
The problem asks us to find the value of the product of \(\sin x\) and \(\cos x\) given that \(\tan x = -\dfrac{3}{4}\) and the angle \(x\) is located in the second quadrant. To solve this, we first need to find the individual values of \(\sin x\) and \(\cos x\) using the given information about \(\tan x\) and the quadrant.
Using Trigonometric Identities to Find sin x and cos x
We are given \(\tan x = -\dfrac{3}{4}\). We know the identity that relates \(\tan x\) and \(\sec x\):
\(\sec^2 x = 1 + \tan^2 x\)
Substitute the given value of \(\tan x\):
\(\sec^2 x = 1 + \left(-\dfrac{3}{4}\right)^2\)
\(\sec^2 x = 1 + \dfrac{9}{16}\)
To add these, we find a common denominator:
\(\sec^2 x = \dfrac{16}{16} + \dfrac{9}{16}\)
\(\sec^2 x = \dfrac{16 + 9}{16}\)
\(\sec^2 x = \dfrac{25}{16}\)
Now, take the square root of both sides to find \(\sec x\):
\(\sec x = \pm \sqrt{\dfrac{25}{16}}\)
\(\sec x = \pm \dfrac{5}{4}\)
Determining the Sign of sec x and cos x in the Second Quadrant
The problem states that angle \(x\) is in the second quadrant. We need to consider the signs of trigonometric functions in different quadrants.
| Quadrant | sin x | cos x | tan x | sec x | csc x | cot x |
|---|---|---|---|---|---|---|
| I (0° to 90°) | + | + | + | + | + | + |
| II (90° to 180°) | + | - | - | - | + | - |
| III (180° to 270°) | - | - | + | - | - | + |
| IV (270° to 360°) | - | + | - | + | - | - |
In the second quadrant, the cosine function (\(\cos x\)) is negative. Since \(\sec x = \dfrac{1}{\cos x}\), the sign of \(\sec x\) is the same as the sign of \(\cos x\). Therefore, in the second quadrant, \(\sec x\) is negative.
So, we must choose the negative value for \(\sec x\):
\(\sec x = -\dfrac{5}{4}\)
Now we can find \(\cos x\) using the reciprocal relationship:
\(\cos x = \dfrac{1}{\sec x} = \dfrac{1}{-\dfrac{5}{4}} = -\dfrac{4}{5}\)
Finding sin x
We know \(\tan x = \dfrac{\sin x}{\cos x}\). We can rearrange this to find \(\sin x\):
\(\sin x = \tan x \cdot \cos x\)
Substitute the given value of \(\tan x\) and the calculated value of \(\cos x\):
\(\sin x = \left(-\dfrac{3}{4}\right) \cdot \left(-\dfrac{4}{5}\right)\)
\(\sin x = \dfrac{(-3) \cdot (-4)}{4 \cdot 5}\)
\(\sin x = \dfrac{12}{20}\)
Simplify the fraction:
\(\sin x = \dfrac{3}{5}\)
Let's check the sign of \(\sin x\) in the second quadrant from the table above. In the second quadrant, \(\sin x\) is positive. Our calculated value \(\sin x = \dfrac{3}{5}\) is positive, which matches the expected sign.
Calculating the Product sin x ⋅ cos x
Now that we have the values for \(\sin x\) and \(\cos x\), we can calculate their product:
\(\sin x \cdot \cos x = \left(\dfrac{3}{5}\right) \cdot \left(-\dfrac{4}{5}\right)\)
\(\sin x \cdot \cos x = \dfrac{3 \cdot (-4)}{5 \cdot 5}\)
\(\sin x \cdot \cos x = \dfrac{-12}{25}\)
\(\sin x \cdot \cos x = -\dfrac{12}{25}\)
Thus, the value of \(\sin x \cdot \cos x\) is \(-\dfrac{12}{25}\).
Summary of Steps
- Used the identity \(\sec^2 x = 1 + \tan^2 x\) to find \(\sec x\).
- Determined the sign of \(\sec x\) based on the quadrant of \(x\) (second quadrant implies \(\cos x\) and \(\sec x\) are negative).
- Calculated \(\cos x\) using the reciprocal relationship \(\cos x = 1/\sec x\).
- Calculated \(\sin x\) using the relationship \(\sin x = \tan x \cdot \cos x\).
- Verified the sign of \(\sin x\) based on the quadrant of \(x\) (second quadrant implies \(\sin x\) is positive).
- Calculated the product \(\sin x \cdot \cos x\).
Revision Table: Key Values and Identities
| Given Information | Identities Used | Calculated Values | Final Product |
|---|---|---|---|
| \(\tan x = -\dfrac{3}{4}\) | \(\sec^2 x = 1 + \tan^2 x\) | \(\sec x = -\dfrac{5}{4}\) | \(\sin x \cdot \cos x = -\dfrac{12}{25}\) |
| \(x\) is in the second quadrant | \(\cos x = \dfrac{1}{\sec x}\) | \(\cos x = -\dfrac{4}{5}\) | |
| \(\sin x = \tan x \cdot \cos x\) | \(\sin x = \dfrac{3}{5}\) |
Additional Information: Trigonometric Functions in Quadrants
Understanding the signs of trigonometric functions in each quadrant is crucial for solving many trigonometry problems. The coordinate plane is divided into four quadrants based on the signs of the x and y coordinates. For a point (x, y) on the terminal side of an angle \(\theta\) in standard position, with distance \(r\) from the origin (\(r = \sqrt{x^2 + y^2}\) and \(r > 0\)):
- \(\sin \theta = y/r\)
- \(\cos \theta = x/r\)
- \(\tan \theta = y/x\)
In the second quadrant, x-coordinates are negative, and y-coordinates are positive. Therefore:
- \(\sin x = \text{positive/positive} = \text{positive}\)
- \(\cos x = \text{negative/positive} = \text{negative}\)
- \(\tan x = \text{positive/negative} = \text{negative}\)
The reciprocal functions (\(\csc x\), \(\sec x\), \(\cot x\)) have the same signs as their corresponding functions (\(\sin x\), \(\cos x\), \(\tan x\)).
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