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# Trig Functions & Domain Restrictions (Grades 11-12)

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## Trig Functions & Domain Restrictions

1.
Which of the following best explains why the domain of a trigonometric function needs to be restricted in order for the inverse to be a function?
1. Trigonometric functions are periodic, meaning that there are multiple input values for a single output value. Therefore, for any inverse trigonometric function there will be multiple output values for a given input value.
2. The domain of trigonometric functions is all real numbers, or all real numbers with some exceptions as in the case of the tangent function. This will necessarily lead to issues with the inverse, since there are infinite input values.
3. The range of trigonometric functions is usually restricted to the values between -1 and 1. Therefore the domain must be restricted as well.
4. Trigonometric functions are based on the ratio of the sides of right triangles. As such, in order to ensure that the inverse function can also be applied to these ratios, the input of trigonometric functions must be restricted to an interval less than $pi$, such as $-pi/2 < theta < pi/2$.
2.
The inverse of the function $f(x) = sin(x)$ is not a function, unless its domain is restricted. Which of the following domain restrictions would ensure that the inverse of $f(x)$ is a function? Choose all correct answers.
1. $-pi <= x < pi$
2. $(-3pi)/2 <= x <= (-pi)/2$
3. $0 <= x <= pi$
4. $(3pi)/2 < x < (5pi)/2$
3.
The inverse of $f(x) = cos(x)$ is not a function, unless the domain of $f(x)$ is restricted. Which of the following domain restrictions would ensure that $f^(-1)(x)$ is a function? Choose all correct answers.
1. $-pi <= x <= 0$
2. $-2pi < x < (-3pi)/2$
3. $(-pi)/2 <= x < pi/2$
4. $pi/2 <= x <= (3pi)/2$
4.
The inverse of $f(x) = tan(x)$ is not a function, unless its domain is restricted. Which of the following domain restrictions would ensure that $f^(-1)(x)$ is a function? There may be more than one correct answer.
1. $0 < x < pi$
2. $(-pi)/2 < x < pi/2$
3. $-pi<= x <= 0$
4. $(-3pi)/2 < x < -pi$
5.
What is the standard domain restriction for $f(x) = sin(x)$, in order to ensure that its inverse is a function?
1. $0 <= x <= pi$
2. $(-3pi)/2 <= x <= (-pi)/2$
3. $(-pi)/2 <= x <= pi/2$
4. $-pi <= x <= 0$
6.
What is the standard domain restriction of $f(x) = cos(x)$ so that its inverse is a function?
1. $pi/2 <= x <= pi$
2. $(-pi)/2 <= x <= pi/2$
3. $0 <= x <= pi$
4. $-pi <= x <= 0$
7.
What is the standard domain restriction of $f(x) = tan(x)$ so that its inverse is a function?
1. $pi/2 < x < pi$
2. $-pi < x < pi$
3. $0 < x < pi$
4. $-pi/2 < x < pi/2$
8.
When restricting the domain of $f(x) = sin(x)$, which of the following would ensure that the inverse $f^(-1)(x)$ is a function? There may be more than one correct answer.
1. On the chosen restricted domain, $f(x)$ is strictly increasing.
2. On the chosen restricted domain, $f(x)$ is strictly decreasing.
3. On the chosen restricted domain, there are no x-intercepts.
4. On the chosen restricted domain, there are no y-intercepts.
9.
Let $f(theta)=cos(theta)$. The angle $theta$ can be thought of as an angle in standard position whose terminal arm intersects the unit circle (centered at the origin) at the point $(x,y)$. Therefore, $f(theta) = cos(theta) = x$. Which of the following best explains why the standard domain restriction of the cosine function, $0 <= theta <= pi$, ensures that the inverse of cosine is a function?
1. Because it restricts the values to a semicircle. Any such restriction would ensure that $f^(-1)(theta)$ is a function.
2. This restriction of $theta$ ensures that there are no points $(x,y)$ in the third or fourth quadrants. Any restriction with points in the third and fourth quadrants would mean that $f(theta)$ is not one-to-one.
3. The values of $x$ are strictly decreasing, and therefore do not repeat. This means that $f(theta)$ is one-to-one for this restricted domain.
4. The values of $x$ are bounded by -1 and 1. Therefore the function $f(theta)$ must be one-to-one.
10.
For $f(x)=tan(x)$, why is the standard domain restriction $-pi/2 < x < pi/2$ instead of $0 <= x < pi ?$
1. On the interval $[0,pi)$, the tangent function is not strictly increasing, and so the inverse will not be a function.
2. It is easier to not have a discontinuity in the inverse function.
3. An inverse function cannot have an asymptote.
4. The standard domain restriction used for the tangent function is the only way to ensure that 0 is included in the range of the inverse function.
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