Question

which trigonometric function is an odd function and why? the sine function is odd because it is represented by the y-coordinate of the points on the unit circle, and therefore \\(\sin(-x) = -\sin(x)\\). the sine function is odd because it is represented by the y-coordinate of the points on the unit circle, and therefore \\(\sin(x) = \sin(-x)\\). the cosine function is odd because it is represented by the x-coordinate of the points on the unit circle, and therefore \\(\cos(x) = \cos(-x)\\). the cosine function is odd because it is represented by the x-coordinate of the points on the unit circle, and therefore \\(\cos(-x) = -\cos(x)\\).

Explanation:

To determine which trigonometric function is odd, we recall the definition of an odd function: a function \( f(x) \) is odd if \( f(-x)=-f(x) \).

• For the sine function: On the unit circle, \( \sin(x) \) corresponds to the \( y \)-coordinate of the point \((\cos(x),\sin(x))\). For \( -x \), the point on the unit circle is \((\cos(-x),\sin(-x)) = (\cos(x),-\sin(x))\) (since cosine is even, \( \cos(-x)=\cos(x) \), and sine reflects over the \( x \)-axis, so \( \sin(-x)=-\sin(x) \)). Thus, \( \sin(-x)=-\sin(x) \), satisfying the odd function definition.
• For the second option about sine: \( \sin(x)=\sin(-x) \) is the property of an even function, not odd, so this is incorrect.
• For cosine: \( \cos(x) \) corresponds to the \( x \)-coordinate on the unit circle. \( \cos(-x)=\cos(x) \) (even function property), so \( \cos(x)=\cos(-x) \) means cosine is even, not odd. The fourth option's claim about cosine being odd with \( \cos(-x)=-\cos(x) \) is false.

So the correct statement is the first one.

Answer:

The sine function is odd because it is represented by the \( y \)-coordinate of the points on the unit circle, and therefore \( \sin(-x) = -\sin(x) \).