The Function \[$ F \$\] Is Given By \[$ F(t) = \sin^2 T - 1 \$\]. For How Many Values Of \[$ T \$\] Does \[$ F(t) = 0 \$\]?A. None B. One C. Two D. Infinitely Many
Introduction
In mathematics, a function is a relation between a set of inputs, called the domain, and a set of possible outputs, called the range. The function { f $}$ given by { f(t) = \sin^2 t - 1 $}$ is a trigonometric function that involves the sine of an angle. In this article, we will explore the roots of this function, which are the values of { t $}$ that make { f(t) = 0 $}$.
Understanding the Function
The function { f(t) = \sin^2 t - 1 $}$ can be rewritten as { f(t) = (\sin t)^2 - 1 $}$. This is a quadratic function in terms of { \sin t $}$. To find the roots of this function, we need to set it equal to zero and solve for { t $}$.
Setting the Function Equal to Zero
Setting the function equal to zero, we get:
{ \sin^2 t - 1 = 0 $}$
This can be rewritten as:
{ (\sin t)^2 = 1 $}$
Solving for sin(t)
To solve for { \sin t $}$, we can take the square root of both sides of the equation:
{ \sin t = \pm 1 $}$
Finding the Values of t
The sine function has a period of { 2\pi $}$, which means that it repeats every { 2\pi $}$ radians. Therefore, we can find the values of { t $}$ that satisfy the equation { \sin t = \pm 1 $}$ by considering the following intervals:
- { 0 \leq t < 2\pi $}$
- { 2\pi \leq t < 4\pi $}$
- { 4\pi \leq t < 6\pi $}$
- { 6\pi \leq t < 8\pi $}$
Solving for t in Each Interval
In the interval { 0 \leq t < 2\pi $}$, the sine function is positive, so we have:
{ \sin t = 1 $}$
This gives us:
{ t = \frac{\pi}{2} $}$
In the interval { 0 \leq t < 2\pi $}$, the sine function is negative, so we have:
{ \sin t = -1 $}$
This gives us:
{ t = \frac{3\pi}{2} $}$
In the interval { 2\pi \leq t < 4\pi $}$, the sine function is positive, so we have:
{ \sin t = 1 $}$
This gives us:
{ t = \frac{5\pi}{2} $}$
In the interval { 2\pi \leq t < 4\pi $}$, the sine function is negative, so we have:
{ \sin t = -1 $}$
This gives us:
{ t = \frac{7\pi}{2} $}$
In the interval { 4\pi \leq t < 6\pi $}$, the sine function is positive, so we have:
{ \sin t = 1 $}$
This gives us:
{ t = \frac{9\pi}{2} $}$
In the interval { 4\pi \leq t < 6\pi $}$, the sine function is negative, so we have:
{ \sin t = -1 $}$
This gives us:
{ t = \frac{11\pi}{2} $}$
In the interval { 6\pi \leq t < 8\pi $}$, the sine function is positive, so we have:
{ \sin t = 1 $}$
This gives us:
{ t = \frac{13\pi}{2} $}$
In the interval { 6\pi \leq t < 8\pi $}$, the sine function is negative, so we have:
{ \sin t = -1 $}$
This gives us:
{ t = \frac{15\pi}{2} $}$
Conclusion
In conclusion, we have found that the function { f(t) = \sin^2 t - 1 $}$ has infinitely many roots, which are the values of { t $}$ that make { f(t) = 0 $}$. These roots are given by:
{ t = \frac{\pi}{2} + k\pi $}$
where { k $}$ is an integer.
Final Answer
Q: What is the function f(t) given by f(t) = sin^2 t - 1?
A: The function f(t) is a trigonometric function that involves the sine of an angle. It can be rewritten as f(t) = (sin t)^2 - 1.
Q: What is the purpose of this article?
A: The purpose of this article is to explore the roots of the function f(t), which are the values of t that make f(t) = 0.
Q: How do you find the roots of the function f(t)?
A: To find the roots of the function f(t), we need to set it equal to zero and solve for t. This involves taking the square root of both sides of the equation and solving for sin t.
Q: What are the possible values of sin t?
A: The possible values of sin t are ±1.
Q: How do you find the values of t that satisfy the equation sin t = ±1?
A: To find the values of t that satisfy the equation sin t = ±1, we need to consider the intervals 0 ≤ t < 2π, 2π ≤ t < 4π, 4π ≤ t < 6π, and 6π ≤ t < 8π.
Q: What are the values of t that satisfy the equation sin t = 1?
A: The values of t that satisfy the equation sin t = 1 are t = π/2, 5π/2, 9π/2, and 13π/2.
Q: What are the values of t that satisfy the equation sin t = -1?
A: The values of t that satisfy the equation sin t = -1 are t = 3Ï€/2, 7Ï€/2, 11Ï€/2, and 15Ï€/2.
Q: How many roots does the function f(t) have?
A: The function f(t) has infinitely many roots, which are the values of t that make f(t) = 0.
Q: What is the general formula for the roots of the function f(t)?
A: The general formula for the roots of the function f(t) is t = π/2 + kπ, where k is an integer.
Q: What is the significance of the roots of the function f(t)?
A: The roots of the function f(t) are important because they help us understand the behavior of the function and its applications in various fields such as physics, engineering, and mathematics.
Q: Can you provide an example of how the roots of the function f(t) are used in real-world applications?
A: Yes, the roots of the function f(t) are used in the design of electrical circuits, where the function f(t) represents the voltage across a capacitor. The roots of the function f(t) help us determine the frequency at which the capacitor will resonate.
Q: What are some common mistakes to avoid when finding the roots of the function f(t)?
A: Some common mistakes to avoid when finding the roots of the function f(t) include:
- Not considering all possible intervals for the values of t
- Not taking the square root of both sides of the equation
- Not solving for sin t correctly
- Not considering the periodic nature of the sine function
Q: How can you verify the roots of the function f(t)?
A: You can verify the roots of the function f(t) by plugging the values of t back into the original equation and checking if f(t) = 0. You can also use a graphing calculator or a computer algebra system to visualize the function and its roots.