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 = .

Step-by-step explanation:

Sine function is positive in I and II quadrant so values would be repeated.

To have inverse sine we have hence -pi/2 to pi/2

1.. To define the inverse sine function, we restrict the domain of sine to the interval__[-pi/2,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 =__pi/6____because sin_pi/6___ = _1/2____.

2. To define the inverse cosine function, we restrict the domain of cosine to the interval_[0,pi]____.

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 =__pi/3____because cos__pi/3___ = ___1/2___.

what is the question here

step-by-step explanation: