Problem 6 - Precalculus with Calculus Previews

Question

If \(\cos t=\frac{3}{10}\), find all possible values of \(\sin t\)

Step-by-Step Solution

Answer

\( \sin t = \pm \frac{\sqrt{91}}{10} \)

Step 1: Recall the Pythagorean identity

The Pythagorean identity in trigonometry states that for any angle \( t \), \( \sin^2 t + \cos^2 t = 1 \). We will use this identity to find the value of \( \sin t \) given that \( \cos t = \frac{3}{10} \).

Step 2: Substitute the known value

Substitute \( \cos t = \frac{3}{10} \) into the Pythagorean identity: \( \sin^2 t + \left(\frac{3}{10}\right)^2 = 1 \), which simplifies to \( \sin^2 t + \frac{9}{100} = 1 \).

Step 3: Solve for \( \sin^2 t \)

Rearrange the equation to solve for \( \sin^2 t \): \( \sin^2 t = 1 - \frac{9}{100} = \frac{100}{100} - \frac{9}{100} = \frac{91}{100} \).

Step 4: Determine \( \sin t \)

To find \( \sin t \), take the square root of both sides of the equation: \( \sin t = \pm \sqrt{\frac{91}{100}} \). Calculate the square root: \( \sin t = \pm \frac{\sqrt{91}}{10} \).

Step 5: Consider both possible solutions

Since sine can be positive or negative depending on the quadrant, both \( \sin t = \frac{\sqrt{91}}{10} \) and \( \sin t = -\frac{\sqrt{91}}{10} \) are valid solutions.

Key Concepts

Pythagorean identity

In trigonometry, the Pythagorean identity is a fundamental concept that relates the square of the sine and cosine functions. It states:

• \( \sin^2 t + \cos^2 t = 1 \),

where \( t \) represents an angle in a right triangle. This identity is derived from the Pythagorean theorem, which asserts that the square of the hypotenuse equals the sum of the squares of the other two sides in a right triangle.

The Pythagorean identity is crucial in trigonometry as it allows us to find one trigonometric function if we know the value of the other. For instance, if you know \( \cos t \), as we do in this problem, you can easily determine \( \sin t \) by rearranging the identity:

• \( \sin^2 t = 1 - \cos^2 t \).

Sine function

The sine function, denoted as \( \sin t \), is one of the primary trigonometric functions. It expresses the ratio of the opposite side to the hypotenuse in a right triangle. The sine of an angle varies between -1 and 1 for all possible angles. This range becomes crucial when determining possible values since \( \sin^2 t \) simplifies to a real number within these limits.

• In this exercise, we calculated \( \sin^2 t = \frac{91}{100} \),
• we then determined \( \sin t = \pm \frac{\sqrt{91}}{10} \).

The use of the positive or negative root indicates that the sine value depends heavily on which quadrant the angle \( t \) resides.

Cosine function

The cosine function, \( \cos t \), complements the sine function and measures the ratio of the adjacent side to the hypotenuse in a right triangle. Cosine also ranges between -1 and 1. For the angle \( t \), its cosine is given to us as \( \frac{3}{10} \) in the problem. This particular exercise showcases a typical use of the cosine function to find corresponding sine values.

• By applying the Pythagorean identity, we substituted \( \cos t = \frac{3}{10} \)
• into \( \sin^2 t + \cos^2 t = 1 \), leading to \( \sin^2 t = 1 - \frac{9}{100} \).

This manipulation allows solving for \( \sin t \), highlighting the interdependent nature of sine and cosine.

Quadrants in trigonometry

In trigonometry, the angle's position - known as the quadrant - plays a significant role in determining the sign of trigonometric functions like sine and cosine. There are four quadrants in the Cartesian plane:

• Quadrant I: Both sine and cosine are positive.
• Quadrant II: Sine is positive, cosine is negative.
• Quadrant III: Both sine and cosine are negative.
• Quadrant IV: Sine is negative, cosine is positive.

Knowing the cosine value helps us determine possible angles \( t \) might reside in. With \( \cos t = \frac{3}{10} \), which is positive, \( t \) could be in the first or fourth quadrant. The respective sine value could thus be either positive or negative, depending on the specific quadrant, resulting in two potential solutions: \( \sin t = \frac{\sqrt{91}}{10} \) or \( \sin t = -\frac{\sqrt{91}}{10} \). Understanding these properties is essential to correctly solving trigonometric problems.