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Let $x = -16\pi/3$. Part a: Determine the reference angle of $x$. Part b: Find the exact values of $\sin x$, $\tan x$, and $\sec x$ in simplest form.

A) a) $5\pi/3$, b) $\sin x = -\sqrt{3}/2, \tan x = \sqrt{3}/3, \sec x = -2$
B) a) $\pi/3$, b) $\sin x = \sqrt{3}/2, \tan x = -\sqrt{3}/3, \sec x = -2$
C) a) $5\pi/3$, b) $\sin x = -\sqrt{3}/2, \tan x = -\sqrt{3}/3, \sec x = -2$
D) a) $\pi/3$, b) $\sin x = \sqrt{3}/2, \tan x = \sqrt{3}/3, \sec x = -2$

User Mumin Ka
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2 Answers

5 votes

Final answer:

The reference angle of
$x = -16\pi/3$ is
$\pi/3$. The exact values of
$\sin x$,
$\tan x$, and
$\sec x$ are
$\sin x = √(3)/2$,
$\tan x = -√(3)/3$, and
$\sec x = -2$.

Explanation:

To find the reference angle of
$x = -16\pi/3$, we need to determine the angle in the first quadrant that has the same trigonometric values as
$x$. Since the reference angle is always positive and less than or equal to
$\pi/2$, we can find it by finding the equivalent angle within the range of
$0$ to
$\pi/2$.

In this case, if we add
$2\pi$ repeatedly to
$x = -16\pi/3$, we get
$x = -10\pi/3,
-4\pi/3,
2\pi/3$, which is equivalent to $\pi/3$. Therefore, the reference angle of
$x = -16\pi/3$ is
$\pi/3$.

Now, let's find the exact values of
$\sin x$,
$\tan x$, and
$\sec x$ using the reference angle
$\pi/3$.

Since
$\sin$ is positive in the first and second quadrants, and negative in the third and fourth quadrants, the value of $\sin x$ is negative.

Using the reference angle
$\pi/3$, we know that the point on the unit circle corresponding to
$\pi/3$is
$(1/2, √(3)/2)$. Therefore,
$\sin x = -√(3)/2$.

Since
$\tan$ is positive in the first and third quadrants, and negative in the second and fourth quadrants, the value of
$\tan x$ is negative.

Using the reference angle
$\pi/3$, we know that the point on the unit circle corresponding to
$\pi/3$ is
$(1/2, √(3)/2)$. Therefore,
$\tan x = -√(3)/3$.

Finally, the value of
$\sec x$ can be found using the reciprocal of
$\cos$which is
$(1)/(\cos x)$. Since
$\cos x$ is negative in the second and third quadrants, the value of
$\sec x$ is negative.

Using the reference angle
$\pi/3$, we know that the point on the unit circle corresponding to
$\pi/3$ is
$(1/2, √(3)/2)$. Therefore,
$\sec x = -2$.

In conclusion, the exact values of
$\sin x$,
$\tan x$, and
$\sec x$ for
$x = -16\pi/3$ in simplest form are $\sin x =
-√(3)/2$, $\tan x = -√(3)/3$, and
$\sec x = -2$.

User Junya
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8.9k points
6 votes

Final Answer:

a)
$5\pi/3$, b) $\sin x = -√(3)/2, \tan x = -√(3)/3, \sec x = -2$

Explanation:

For part a), the reference angle of $x = -16\pi/3$ is $5\pi/3$. This is determined by finding the positive acute angle formed between the terminal side of the angle and the x-axis when considering the negative coterminal angle.

For part b), evaluating the trigonometric functions:

-
$\sin x = -√(3)/2$: This is obtained from the unit circle corresponding to the sine value at $5\pi/3$.

-
$\tan x = -√(3)/3$: Obtained by dividing
$\sin x$ by $\cos x$ ($\tan x = \sin x / \cos x$).

- $\sec x = -2$: This is the reciprocal of $\cos x$, which is $-1/2$ on the unit circle, giving
$\sec x = -2$.

These values are derived from the unit circle, understanding the relationship between trigonometric functions and the coordinates on the circle at the given angle $5\pi/3$, ensuring the signs are correctly represented for each function.

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User MiBrock
by
8.4k points

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