Common Core Standard HSF-TF.A.4 Questions

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Grade 11 Trigonometry CCSS: HSF-TF.A.4

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What is the value of

$sin(-theta) ?$

$-x_1/y_1$
$x_1/y_1$
$-y_1/x_1$
$-y_1$

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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The period of a periodic function is the smallest value of

$T$ such that

$f(x+T) = f(x)$ for all values of

$x$ in the domain of

$f$ . Using the information in the previous questions, which of the following gives the best reasoning as to why the period of the sine function is

$2pi ?$

Since values of $T$ less than $2pi$ weren't valid, $2pi$ must work.
Using $T=2pi$ and trying a few values of $theta$ , such as $(3pi)/7$ and $(8pi)/11$ , the equation $sin(theta+2pi) = sin(theta)$ is valid. Therefore, it is true for all values of $theta$ .
For $T=2pi$ , $theta + T$ and $theta$ are coterminal angles for any value of $theta$ . Therefore, referring to the unit circle, $sin(theta)=b$ and $sin(theta+2pi)=b$ , and therefore $sin(theta+2pi)=sin(theta)$ .
There are no other values of $T$ such that $sin(x+T) = sin(x)$ .

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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What conclusion from the previous questions can be reached?

$sin(theta) = y_1$ and $sin(-theta) = -x_1$ for $0 <= theta <= pi/2$ .
$sin(theta) = y_1/x_1 = 1/sin(-theta)$ for $0 <= theta <= pi/2$ .
$sin(theta) = y_1/x_1 = -sin(-theta)$ for $0 <= theta <= pi/2$ .
$sin(theta) = y_1 = -sin(-theta)$ for $0 <= theta <= pi/2$ .

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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The period of the cosine function is also

$2pi$ , and showing this is true is similar to the process used for the sine function. However, the period of the tangent function is

$pi$ . Which of the following equations correctly shows why the period of tangent is

$pi?$

$tan(theta + pi) = sin(theta+pi)/cos(theta+pi) = (-sin(theta))/(-cos(theta)) = sin(theta)/cos(theta) = tan(theta)$
$tan(theta + pi) = sin(theta+pi)/cos(theta+pi) = (sin(theta)+pi)/(cos(theta)+pi) = sin(theta)/cos(theta) = tan(theta)$
$tan(theta + pi) = sin(theta+pi)/cos(theta+pi) = (sin(theta)+sin(pi))/(cos(theta)+cos(pi)) = (sin(theta)+0)/(cos(theta)+0) = tan(theta)$
$tan(theta + pi) = tan(theta) + tan(pi) = tan(theta)$

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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Let

$pi/2 < alpha <= pi$ be an angle in standard position, whose terminal arm intersects circle A at

$(-x_2, y_2)$ . Can the same conclusion be reached concerning

$sin(alpha)$ as was reached for

$sin(theta)$ , with the same reasoning? Why?

Yes, since neither conclusion depends upon the x-values (it doesn't matter if they're positive or negative).
Yes, since $sin(alpha) = sin(theta)$ .
No, because the reasoning applied is only valid for angles in the first quadrant.
No, because the sine function is only positive in the first quadrant.

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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Looking at

$pi < beta <= (3pi)/2$ and

$(3pi)/2 < gamma <= 2pi$ , where both angles are in standard position, and their terminal arms intersect circle A at

$(-x_3, -y_3)$ and

$(x_4, -y_4)$ respectively, which of the following explains why the sine of an angle equals the negative sine of the negative of that angle for angles between

$pi$ and

$2pi ?$

Since sine is periodic, whatever is true for the sine function for values between $0$ and $pi$ must also be true for values between $pi$ and $2pi$ .
Using the same reasoning as before, for angle $beta$ it follows that $sin(beta) = -y_3 = -sin(-beta)$ and for angle $gamma$ that $sin(gamma) = -y_4 = -sin(-gamma)$ .
As seen before, the sine of an angle in one quadrant is always the opposite of the sine of an angle in an adjacent quadrant. Thus $sin(beta) = -sin(-beta)$ and $sin(gamma) = - sin(-gamma)$ .
Looking at angles between $pi$ and $2pi$ , the sine of these angles is always negative. Therefore, $sin(beta) = - sin(-beta)$ and $sin(gamma) = - sin(-gamma)$ .

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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Which of the following reasons allows one to conclude that

$sin(Theta)$ is odd?

It has been shown that $sin(-omega) = - sin(omega)$ for $0 <= omega <= 2pi$ . Since sine is periodic with a period of $2pi$ , this result must be true for $omega in RR$ .
It has been shown that $sin(-omega) = -sin(omega)$ for $0 <= omega <= 2pi$ . Since $-1 <= sin(omega) <= 1$ is true not only for values between 0 and $2pi$ , but for all real values of $omega$ , it follows that $sin(-omega) = -sin(omega)$ for $omega in RR$ .
Since the values of sine from the first quadrant are simply repeated in the other quadrants (except for being sometimes negative), all that one needed to confirm was that $sin(-omega)=-sin(omega)$ for $0 <= omega <= pi/2$ , and then it must be true for $omega in RR$ .
Using circle A, it has been seen that, for $0 <= omega <= 2pi$ , the sine of $omega$ is always greater than or equal to zero, while the sine of $-omega$ is always less than or equal to zero. Therefore, since sine is periodic with period $2pi$ , this result must be true for $omega in RR$ .

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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Which of the following statements best explains the symmetry of

$cos(Theta) ?$ Assume that

$Theta$ can be interpreted as an angle in standard position, whose terminal arm intersects circle A at the point

$(X_1,Y_1)$ , where

$X_1, Y_1$ are real numbers.

Since we know that sine and cosine are the same function, except that cosine is different by a phase shift of $pi/2$ , therefore the same properties must apply to both functions. Thus, cosine is an odd function.
Using the same reasoning as was applied to the sine function in the previous questions, it can be seen that that $cos(Theta) = X_1$ and $cos(-Theta) = - X_1$ . Therefore $cos(Theta) = -cos(-Theta)$ , showing that cosine is odd.
For any given angle $Theta$ , the value of $cos(Theta)$ can be interpreted as $X_1$ . Using similar reasoning as was used in previous questions, it can be seen that $cos(-Theta) = X_1$ , where $X_1$ will always be of the same sign. Hence, $cos(Theta) = cos(-Theta)$ , showing that the cosine function is even.
Looking at the unit circle, the sine of any angle $Theta$ will be $Y_1$ and the cosine of the angle will be $X_1$ . The properties of the sine function can therefore be seen in the properties of the cosine function, if we simply reflect the coordinate axis about the line $y=x$ . Therefore, an odd function will become an even function, and thus cosine is an even function.

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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Using the same reasoning as was done with the sine function, it can be shown that the tangent function is odd. However, there are values for which the tangent function is undefined (its domain is not all real numbers). How is this possible, given that the unit circle has no undefined values on its circumference?

Since the tangent function has a period of $pi$ instead of $2pi$ , it is actually associated with the unit semicircle centered at the origin. Therefore, because of the discontinuity of it being a semicircle instead of a full circle, the undefined values arise.
The values of the tangent function do not occur on circle A. Referring to a right triangle, the tangent of either acute angle is the opposite side divided by the adjacent side. Since this value does not include the hypotenuse, the points on circle A are actually only an approximation of the tangent function values. Therefore, although the tangent function has undefined values, these do not arise in the approximated values of circle A.
Since the values of $tan(Theta)$ can be represented as $Y_1/X_1$ , the values where $X_1=0$ are undefined and correspond to the domain restrictions of the tangent function.
Since the values of $tan(Theta)$ can be represented as $X_1/Y_1$ , the values where $Y_1=0$ are undefined and correspond to the domain restrictions of the tangent function.

Grade 11 Trigonometry CCSS: HSF-TF.A.4

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For this investigation of trigonometric functions and their symmetry, the unit circle was used (circle A). Why was the unit circle used, and not a circle with a different radius (not equal to one unit)?

Because the absolute value of the maximum and minimum values of the sine and cosine functions are 1.
The circumference of a circle is $C = 2 pi r$ . If the radius is not equal to 1, the circumference would be greater than or less than $2pi$ . Since the period of sine and cosine is $2pi$ , this would mean that the values of these functions on the circle would not fit.
It simplifies the math, allowing the coordinates of the points on the unit circle to be equal to the sine and cosine of angles in standard position.
There is no particular reason, it is simply convention to use the unit circle. All the math and reasoning would have been exactly the same using a circle centered at the origin with a different radius.