Question

1. To define the inverse sine function, we restrict the domain of sine to the interval______. On this interval the sine function is one-to-one, and its inverse function sin−1 is defined by sin−1(x) = y ⇔ sin____ = ____. For example, sin−1 1/2 =______because sin____ = _____. 2. To define the inverse cosine function, we restrict the domain of cosine to the interval_____. On this interval the cosine function is one-to-one and its inverse function cos−1 is defined by cos−1(x) = y ⇔ cos_____ = ______. For example, cos−1 1 2 =______because cos_____ = ______.

Answer 1

The completed statement are presented as follows;

1. To define the inverse sine function, we restrict the domain of sine to the interval  $-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$. On this interval the sine function is one-to-one, and its inverse function sin⁻¹ is defined by sin⁻¹(x) = y ⇔ sin(y) = x. For example, sin⁻¹(1/2) = π/6 because sin(π/6) = 1/2

2.  To define the inverse cosine function, we restrict the domain of sine to the interval 0 ≤ x ≤ π. On this interval the cosine function is one-to-one, and its inverse function cos⁻¹ is defined by cos⁻¹(x) = y ⇔ cos(y) = x. For example, cos⁻¹(1/2) = π/3 because cos(π/3) = 1/2

The reason the above angles and trigonometric function values are correct is as follows:

1. The sine function is a periodic (repeating) function expressed using the parent formula as y = sin(x)

The period is the length or time in which a cycle of a periodic function is completed, after which an identical repetition of the cycle is started

The y-values of the sine function have a range of -1 ≤ y ≤ 1. A domain over which each x-value maps to exactly one y-value is the domain $-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$

Within the above domain, it can be said that each x-value of the sine function has only one y-value, therefore;

• To define the inverse sine function, we restrict the domain of sine to the interval  $-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$

The inverse of the sine function is found as follows;

f(x) = y = sin(x)

f⁻¹(x) = sin⁻¹(y) = arcsin(sin(x)) = x

x = sin⁻¹(y)

∴ f⁻¹(x) is y = sin⁻¹(x) which gives;

sin(y) = x, by the definition of invers function

Which gives;

The inverse function of x = sin(y) is y = sin⁻¹(x)

Therefore, we have;

• On this interval the sine function is one-to-one, and its inverse function sin⁻¹ is defined by sin⁻¹(x) = y ⇔ sin(y) = x

Using examples we have;

sin(π/6) = 1/2 therefore, sin⁻¹(1/2) = π/6, which gives;

• For example, sin⁻¹(1/2) = π/6 because sin(π/6) = 1/2

2. The cosine function is also a periodic (repeating) function expressed using the parent formula as y = cos(x)

The domain of the cosine function where each input maps unto exactly one output value and there is a one to one relationship is 0 ≤ x ≤ π, in which an inverse function can be defined

Therefore;

• To define the inverse cosine function, we restrict the domain of sine to the interval 0 ≤ x ≤ π.

The cosine function inverse  is found as follows;

f(x) = y = cos(x)

f⁻¹(x) = cos⁻¹(y) = x

x = cos⁻¹(y)

Representing the above equation as a function of y gives

∴ y = cos⁻¹(x) which gives;

cos(y) = x

Therefore, from x = cos(y), we get, y = cos⁻¹(x)

Which gives;

• On this interval the cosine function is one-to-one, and its inverse function cos⁻¹ is defined by cos⁻¹(x) = y ⇔ cos(y) = x

Using examples we have;

cos(π/3) = 1/2 therefore, cos⁻¹(1/2) = π/3, which gives;

• For example, cos⁻¹(1/2) = π/3 because cos(π/3) = 1/2

Learn more about the inverse of the sine and cosine functions here:

https://brainly.com/question/17091413

https://brainly.com/question/12015707