HSF.TF.A.4 Tutor Intake — Symmetry and Periodicity of Trig Functions

Cosine is even because $\cos(-x) = a = \cos(x)$; the x-coordinates are the same.

Cosine is odd because the point for $-x$ is a reflection of the point for $x$.

Cosine is even because cosine values are always positive.

Cosine is odd because $\cos(-x) = -a = -\cos(x)$; the reflection negates the value.

Sine is even because both points share the x-coordinate $a$.

Sine is odd because $\sin(-x) = -b = -\sin(x)$; the y-coordinates are opposites.

Sine is odd because sine values are negative for negative angles.

Sine is even because $\sin(-x) = b = \sin(x)$; the y-coordinate is unchanged.

Tangent is even because $\tan(-x) = \dfrac{\cos(-x)}{\sin(-x)} = \dfrac{\cos x}{-\sin x} = -\tan(x)$,
so $\tan(-x) = -\tan(x)$ — confirming even symmetry.

Tangent is even because both sine and cosine are defined for all angles.

Tangent is odd because $\tan(-x) = \dfrac{\sin(-x)}{\cos(-x)} = \dfrac{-\sin x}{\cos x} = -\tan(x)$.

Tangent is neither even nor odd because it has vertical asymptotes.

The claim is correct — $\sin(0 + \pi) = \sin(0)$ confirms the period is $\pi$.

The claim is incorrect. Period requires $f(x + T) = f(x)$ for ALL $x$.
Counterexample: $\sin\!\left(\tfrac{\pi}{2} + \pi\right) = \sin\!\left(\tfrac{3\pi}{2}\right) = -1 \neq 1 = \sin\!\left(\tfrac{\pi}{2}\right)$.

The claim is incorrect because $\pi$ is not where sine equals zero.

The claim is correct because the sine graph visually repeats after $\pi$.

$\dfrac{1}{2}$

$-\dfrac{1}{2}$

$\dfrac{\sqrt{3}}{2}$

$-\dfrac{\sqrt{3}}{2}$