Problem 46 - Contemporary Precalculus

Question

If \(x\) and \(y\) are in the second quadrant, \(\sin x=1 / 3,\) and \(\cos y=-3 / 4,\) find the exact value of \(\sin (x+y)\) \(\cos (x+y), \tan (x+y),\) and find the quadrant in which \(x+y\) lies.

Step-by-Step Solution

Answer

Answer: The angle \(x+y\) lies in the fourth quadrant. The exact values of the trigonometric functions are: \(\sin(x+y) = -\frac{1}{4} - \frac{\sqrt{14}}{6}\), \(\cos(x+y) = \frac{\sqrt{2}}{2} - \frac{\sqrt{7}}{12}\), and \(\tan(x+y) = \frac{-\frac{1}{4} - \frac{\sqrt{14}}{6}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{7}}{12}}\).

Step 1: Calculate values of \(\cos x\) and \(\sin y\)

Since both \(x\) and \(y\) are in the second quadrant, we know that: \(\sin x > 0\), \(\cos x < 0\), \(\sin y > 0\), \(\cos y < 0\). We have \(\sin x = \frac{1}{3}\), and because, \(\sin^2 x + \cos^2 x = 1\), we can solve for \(\cos x\). \(\cos^2 x = 1 - \sin^2 x = 1 - (\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}\) Since \(\cos x < 0\) in the second quadrant, \(\cos x = -\frac{2\sqrt{2}}{3}\) Similarly, we have \(\cos y = -\frac{3}{4}\), and because, \(\sin^2 y + \cos^2 y = 1\), we can solve for \(\sin y\). \(\sin^2 y = 1 - \cos^2 y = 1 - (-\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}\) Since \(\sin y > 0\) in the second quadrant, \(\sin y = \frac{\sqrt{7}}{4}\)

Step 2: Calculate \(\sin (x+y)\) and \(\cos(x+y)\) using Sum of Angles formulas

Using the Sum of Angles formulas: \(\sin (x+y) = \sin x \cos y + \cos x \sin y\) \(\sin (x+y) = (\frac{1}{3})(-\frac{3}{4}) + (-\frac{2\sqrt{2}}{3})(\frac{\sqrt{7}}{4})\) \(\sin (x+y) = -\frac{1}{4} - \frac{\sqrt{14}}{6}\) \(\cos (x+y) = \cos x \cos y - \sin x \sin y\) \(\cos (x+y) = (-\frac{2\sqrt{2}}{3})(-\frac{3}{4}) - (\frac{1}{3})(\frac{\sqrt{7}}{4})\) \(\cos (x+y) = \frac{\sqrt{2}}{2} - \frac{\sqrt{7}}{12}\)

Step 3: Calculate \(\tan (x+y)\)

Now, let's use the quotient identity to find the value of \(\tan (x+y)\): \(\tan (x+y) = \frac{\sin (x+y)}{\cos (x+y)} = \frac{-\frac{1}{4} - \frac{\sqrt{14}}{6}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{7}}{12}}\)

Step 4: Determine the quadrant for \(x+y\)

To determine the quadrant for \(x+y\), we need to analyze the sine, cosine, and tangent: \(\sin (x+y) = -\frac{1}{4} - \frac{\sqrt{14}}{6} < 0\), so sine is negative. \(\cos (x+y) = \frac{\sqrt{2}}{2} - \frac{\sqrt{7}}{12} > 0\), so cosine is positive. \(\tan (x+y) = \frac{-\frac{1}{4} - \frac{\sqrt{14}}{6}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{7}}{12}} < 0\), so tangent is negative. Since sine is negative and cosine is positive, \(x+y\) must lie in the fourth quadrant.

Key Concepts

Sum of Angles

Understanding how to calculate the sum of angles is essential when dealing with trigonometric functions. This concept is particularly useful when you want to find the trigonometric values of an angle that is a combination of two known angles. In this exercise, we used the sum of angles formulas to determine \( \sin(x+y) \) and \( \cos(x+y) \). These formulas are:

• The sine of the sum of two angles: \( \sin(a+b) = \sin a \cos b + \cos a \sin b \)
• The cosine of the sum of two angles: \( \cos(a+b) = \cos a \cos b - \sin a \sin b \)

By plugging in the known values for \( \sin x, \cos y, \cos x, \) and \( \sin y \), we derived the exact values for \( \sin(x+y) \) and \( \cos(x+y) \). These are crucial for understanding more about the combined angle, such as its tangent value and which quadrant it lies in.

Second Quadrant

In trigonometry, knowing the quadrant where your angle resides is vital because it determines the sign of your trigonometric functions. Angles in the second quadrant have some specific characteristics:

• For angles in the second quadrant: \( 90^\circ < x < 180^\circ \)
• The sine function is positive, while the cosine function is negative.
• The values for \( x \) and \( y \) in our problem both lie in this quadrant, which is why \( \sin x \) and \( \sin y \) are positive, but \( \cos x \) and \( \cos y \) are negative.

This sign knowledge helps in applying trigonometric identities correctly. It forms the base for calculating more complex expressions involving both angles like \( \sin(x+y) \) and \( \cos(x+y) \). Understanding the constraints of the second quadrant enables us to predict the behavior of these trigonometric functions as part of the angle sum.

Exact Values

The idea of finding exact values in trigonometry is to compute values without rounding or approximating. This is crucial for more accurate mathematical solutions. In the given problem, we dealt with the exact values associated with sine and cosine functions using known values and Pythagorean identities. Here's how this ties into our exercise:

• We started with \( \sin x = \frac{1}{3} \) and calculated \( \cos x \) using the identity \( \sin^2 x + \cos^2 x = 1 \).
• Conversely, for \( \cos y = -\frac{3}{4}, \) we calculated \( \sin y \).
• These calculations yielded \( \cos x = -\frac{2\sqrt{2}}{3} \) and \( \sin y = \frac{\sqrt{7}}{4}\), ensuring we worked with precise value.

Using these exact numbers, we were able to plug them into the sum of angles formulas and accurately determine \( \sin(x+y) \) and \( \cos(x+y) \), which further allowed us to find \( \tan(x+y)\). Maintaining exact values minimizes errors across more extensive computations.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all angles. These identities are potent tools in solving trigonometric equations as they allow the transformation and simplification of expressions. Major identities used in this problem include:

• The Pythagorean Identity: \( \sin^2 x + \cos^2 x = 1 \), which we used to find unknown sine or cosine values from known counterparts.
• The quotient identity to find the tangent: \( \tan(x+y) = \frac{\sin(x+y)}{\cos(x+y)} \)

These identities help maintain accuracy while simplifying complex trigonometric problems. By understanding and applying these identities, we could determine the sine, cosine, and tangent of the sum of the two given angles clearly and precisely. They enable us to manipulate functions in a manner that makes finding relationships and solving equations much more manageable. These identities are thus fundamental tools in trigonometry.