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

Answer:

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___.

Answer 2

Answer:

1) [-pi/2 , pi/2]

2 & 3) sin(y) = x

4) pi/6

5 & 6) pi/6,½

1) [0,pi]

2 & 3) cos(y) = x

4) pi/3

5 & 6) pi/3, ½