Even Odd Identities Quiz: Fundamental Even Odd Definitions

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1) Classify cosθ by parity.

Even

Odd

Neither

Depends on θ

cos(−θ)=cosθ for all θ, so cosine is even by definition of even functions.

Explanation

cos(−θ)=cosθ for all θ, so cosine is even by definition of even functions.

About This Quiz

Even Odd Identities Quiz: Fundamental Even Odd Definitions - Quiz

What determines whether a trigonometric function is even or odd? In this quiz, you’ll explore how symmetry in the coordinate plane shapes the behavior of trig graphs and their algebraic rules. You’ll analyze reflections across axes, observe sign changes, and connect these visual patterns to symbolic definitions. With each question,... see moreyou’ll build stronger intuition for how functions respond to negative inputs and why these identities become essential tools in simplifying trig expressions and verifying equations. see less

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2) Tan(−θ)=−tanθ.

True

False

tanθ=sinθ/cosθ with sin odd and cos even. Then tan(−θ)=sin(−θ)/cos(−θ)=(−sinθ)/(cosθ)=−tanθ.

Explanation

tanθ=sinθ/cosθ with sin odd and cos even. Then tan(−θ)=sin(−θ)/cos(−θ)=(−sinθ)/(cosθ)=−tanθ.

3) Select all functions that are odd.

Sinθ

Cosθ

Tanθ

Cotθ

Cscθ

sin, tan, cot, and csc are odd: f(−θ)=−f(θ). cos and sec are even: f(−θ)=f(θ).

Explanation

sin, tan, cot, and csc are odd: f(−θ)=−f(θ). cos and sec are even: f(−θ)=f(θ).

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4) If f is even and g is odd, then f·g is odd.

True

False

Let f(−x)=f(x) and g(−x)=−g(x). Then (fg)(−x)=f(−x)g(−x)=f(x)(−g(x))=−f(x)g(x), so fg is odd.

Explanation

Let f(−x)=f(x) and g(−x)=−g(x). Then (fg)(−x)=f(−x)g(−x)=f(x)(−g(x))=−f(x)g(x), so fg is odd.

5) The quotient of two odd functions (where defined) is even.

True

False

If f and g are odd, (f/g)(−x)=f(−x)/g(−x)=(−f(x))/(−g(x))=f(x)/g(x). Thus the quotient is even when the denominator is nonzero.

Explanation

If f and g are odd, (f/g)(−x)=f(−x)/g(−x)=(−f(x))/(−g(x))=f(x)/g(x). Thus the quotient is even when the denominator is nonzero.

6) Evaluate tan(−θ)/tanθ (when defined).

-1

Tan^2θ

Cot^2θ

Since tan(−θ)=−tanθ, the ratio tan(−θ)/tanθ=(−tanθ)/tanθ=−1 whenever tanθ≠0.

Explanation

Since tan(−θ)=−tanθ, the ratio tan(−θ)/tanθ=(−tanθ)/tanθ=−1 whenever tanθ≠0.

7) Simplify sin(−θ).

−sinθ

Sinθ

Cosθ

−cosθ

Sine is odd: sin(−θ)=−sinθ because reflecting θ across the x-axis flips the y-coordinate on the unit circle.

Explanation

Sine is odd: sin(−θ)=−sinθ because reflecting θ across the x-axis flips the y-coordinate on the unit circle.

8) Select all identities that are always true.

Cos(−θ)=cosθ

Sec(−θ)=secθ

Tan(−θ)=tanθ

Sin(−θ)=sinθ

Cot(−θ)=cotθ

cos and sec are even so they are unchanged under −θ. tan, sin, and cot are odd, so they change sign under −θ.

Explanation

cos and sec are even so they are unchanged under −θ. tan, sin, and cot are odd, so they change sign under −θ.

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9) Sec(−θ)=−secθ.

True

False

secθ=1/cosθ and cosine is even. So sec(−θ)=1/cos(−θ)=1/cosθ=secθ, not −secθ.

Explanation

secθ=1/cosθ and cosine is even. So sec(−θ)=1/cos(−θ)=1/cosθ=secθ, not −secθ.

10) Simplify cos(−θ).

Cosine is even: cos(−θ)=cosθ because x-coordinates on the unit circle are unchanged under reflection across the x-axis.

Explanation

Cosine is even: cos(−θ)=cosθ because x-coordinates on the unit circle are unchanged under reflection across the x-axis.

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11) Simplify sin(−θ)·cos(−θ).

−sinθ·cosθ

Sinθ·cosθ

−sinθ/ cosθ

Sinθ/ cosθ

sin is odd and cos is even. So sin(−θ)·cos(−θ)=(−sinθ)(cosθ)=−(sinθ·cosθ).

Explanation

sin is odd and cos is even. So sin(−θ)·cos(−θ)=(−sinθ)(cosθ)=−(sinθ·cosθ).

12) Which expression is even in θ?

Sin^2θ

Sinθ

Tanθ

Cotθ

sin is odd, but squaring removes the sign: sin^2(−θ)=(sin(−θ))^2=(−sinθ)^2=sin^2θ, so it is even.

Explanation

sin is odd, but squaring removes the sign: sin^2(−θ)=(sin(−θ))^2=(−sinθ)^2=sin^2θ, so it is even.

13) Simplify cot(−θ).

cotθ=cosθ/sinθ with cos even and sin odd, so cot(−θ)=cos(−θ)/sin(−θ)=cosθ/(−sinθ)=−cotθ.

Explanation

cotθ=cosθ/sinθ with cos even and sin odd, so cot(−θ)=cos(−θ)/sin(−θ)=cosθ/(−sinθ)=−cotθ.

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14) Select all statements that are always true.

Cos(−θ)=cosθ

Sin^3(−θ)=−sin^3θ

Tan^2(−θ)=−tan^2θ

Sinθ·cosθ is even

Csc(−θ)=−cscθ

A: cos is even. B: odd power of an odd function stays odd, so sin^3(−θ)=−sin^3θ. C: tan^2 is even, so the given negative is false. D: sin is odd and cos is even, so the product is odd, not even. E: csc is odd.

Explanation

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15) Select all true statements using exact values at 30°.

Sin(−30°)=−1/2

Cos(−30°)=√3/2

Tan(−30°)=−1/√3

Sec(−30°)=sec(30°)

Csc(−30°)=csc(30°)

At 30°: sin30°=1/2, cos30°=√3/2, tan30°=1/√3. Using parity: sin is odd (becomes −1/2), cos is even (stays √3/2), tan is odd (becomes −1/√3), sec is even (unchanged). csc is odd so it changes sign.

Explanation

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