Question

question
express as a function of a different angle, 0° ≤ θ < 360°.
sin(339°)
answer attempt 2 out of 2
sin(□°)
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Explanation:

Step1: Identify the quadrant

339° is in the fourth quadrant (270° < 339° < 360°). The reference angle formula for fourth quadrant is \( 360^\circ - \theta \).

Step2: Find the reference angle

Calculate the reference angle: \( 360^\circ - 339^\circ = 21^\circ \).

Step3: Use sine's property in fourth quadrant

Sine is negative in the fourth quadrant, and \( \sin(360^\circ - \alpha)=-\sin(\alpha) \), but we need to express as a function of a different angle. Wait, actually, the co - terminal or reference angle relation. Wait, no, the question is to express \( \sin(339^\circ) \) as a function of a different angle. Since \( 339^\circ=360^\circ - 21^\circ \), and \( \sin(360^\circ - \alpha)=-\sin(\alpha) \), but we can also think about the angle in terms of its reference angle. Wait, the angle 339° can be related to - 21° (but - 21° is not in the range \( 0^\circ\leq\theta < 360^\circ \)). Wait, the reference angle is 21°, and 339° is equivalent to \( - 21^\circ+360^\circ \). But the sine function has the property \( \sin(\theta)=\sin(\theta + 360^\circ k) \) for integer k, and also \( \sin(360^\circ-\alpha)=-\sin(\alpha) \). But we need to express \( \sin(339^\circ) \) as a function of a different angle in the range \( 0^\circ\leq\theta < 360^\circ \). So \( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \), but if we consider the angle itself, wait, no, the question is to express it as a function of a different angle. Wait, maybe I made a mistake. Wait, 339° is in the fourth quadrant, and the angle can be related to 360 - 21 = 339, but we need a different angle. Wait, the reference angle is 21°, and the angle 339° is equal to - 21° + 360°, but since we need an angle in \( 0^\circ\leq\theta < 360^\circ \), the angle related to 339° is 360 - 21 = 339, no. Wait, no, the key is that \( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \), but we can also express it as \( \sin(339^\circ)=\sin(- 21^\circ) \), but - 21° is not in the range. Wait, the correct approach is: the angle 339° is 360° - 21°, so \( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \), but we need to write it as a function of a different angle. Wait, maybe the question is to use the reference angle or a co - terminal angle. Wait, the angle 339° can be represented as a function of 21° with a negative sign, but if we consider the angle in the range, the different angle is 360 - 21 = 339, no. Wait, I think I messed up. Let's start over.

The angle 339°: to find a different angle \( \theta \) in \( 0^\circ\leq\theta < 360^\circ \) such that \( \sin(339^\circ) \) can be expressed as a function of \( \theta \). We know that \( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \), and 21° is in the range \( 0^\circ\leq\theta < 360^\circ \). So we can express \( \sin(339^\circ) \) as \( -\sin(21^\circ) \), but the question says "express as a function of a DIFFERENT angle", so the angle is 21°, but with a negative sign? Wait, no, maybe the question is to use the angle's relation. Wait, the angle 339° is equal to 360° - 21°, so \( \sin(339^\circ)=\sin(360^\circ - 21^\circ) \), and by the sine subtraction formula \( \sin(A - B)=\sin A\cos B-\cos A\sin B \), but that's more complicated. The simpler way is to use the identity \( \sin(360^\circ - \alpha)=-\sin(\alpha) \), so \( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \). So the different angle is 21°, and we can write \( \sin(339^\circ)=-\sin(21^\circ) \), but the problem asks to express it as \( \sin(\square^\circ) \), wait, maybe I misread. Wait, the problem says "Express as a function of a DIFFERENT angle, \( 0^\circ\leq\theta < 360^\circ \)". So \( \sin(339^\circ)=\sin(360^\circ - 21^\circ) \), but \( \sin(360^\circ - 21^\circ)=-\sin(21^\circ) \), but if we consider the angle 339° is equivalent to - 21° + 360°, and \( \sin(-21^\circ + 360^\circ)=\sin(-21^\circ) \), but - 21° is not in the range. Wait, maybe the answer is \( \sin(339^\circ)=-\sin(21^\circ) \), but the box is for the angle inside the sine. Wait, no, maybe the question is to find the angle such that \( \sin(339^\circ) \) is equal to \( \sin(\theta) \) or related. Wait, no, the sine of 339° is equal to the sine of (339° - 360°)= - 21°, but - 21° is not in the range. The reference angle is 21°, and in the fourth quadrant, sine is negative. So \( \sin(339^\circ)=-\sin(21^\circ) \), so if we write it as a function of 21°, with a negative sign, but the problem has a box for the angle inside the sine. Wait, maybe there is a mistake in my approach. Let's check the angle 339°: 339° is 360° - 21°, so \( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \). So the different angle is 21°, and we can write \( \sin(339^\circ)=-\sin(21^\circ) \), but the problem's answer box is \( \sin(\square^\circ) \), maybe the question expects the angle with the sign considered? No, maybe I made a mistake. Wait, the angle 339° is in the fourth quadrant, and the angle can be represented as 360° - 21°, so \( \sin(339^\circ)=\sin(360^\circ - 21^\circ) \), and since \( \sin(360^\circ - \alpha)=-\sin(\alpha) \), but if we just consider the angle, the different angle is 21°, and we can write \( \sin(339^\circ)=-\sin(21^\circ) \), but the box is for the angle inside the sine. Wait, maybe the question is to find the angle \( \theta \) such that \( \sin(339^\circ)=\sin(\theta) \) with \( \theta eq339^\circ \) in the range. But \( \sin(\theta)=\sin(339^\circ) \) implies \( \theta = 339^\circ+360^\circ k \) or \( \theta=180^\circ - 339^\circ+360^\circ k \). For \( k = 0 \), \( 180^\circ - 339^\circ=- 159^\circ \) (not in range), for \( k = 1 \), \( - 159^\circ+360^\circ = 201^\circ \). Wait, \( \sin(339^\circ)=\sin(201^\circ) \)? No, \( \sin(201^\circ)=\sin(180^\circ + 21^\circ)=-\sin(21^\circ) \), and \( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \), so \( \sin(339^\circ)=\sin(201^\circ) \). Wait, 201° is in the third quadrant (180° < 201° < 270°). Let's check: \( \sin(201^\circ)=\sin(180^\circ + 21^\circ)=-\sin(21^\circ) \), \( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \), so \( \sin(339^\circ)=\sin(201^\circ) \). So the different angle is 201°. Wait, that's a better approach. Let's verify:

\( \sin(201^\circ)=\sin(180^\circ + 21^\circ)=-\sin(21^\circ) \)

\( \sin(339^\circ)=\sin(360^\circ - 21^\circ)=-\sin(21^\circ) \)

So \( \sin(339^\circ)=\sin(201^\circ) \)

So the angle is 201°.

Answer:

\( \sin(201^\circ) \) (Wait, no, earlier we thought 21°, but now 201°. Wait, let's check the sine values. \( \sin(339^\circ)\approx\sin(339)=\sin(-21)\approx - 0.3584 \), \( \sin(201^\circ)=\sin(180 + 21)=-\sin(21)\approx - 0.3584 \), \( \sin(21^\circ)\approx0.3584 \). So \( \sin(339^\circ)=-\sin(21^\circ)=\sin(201^\circ) \). So the correct angle is 201°. So the answer is 201.