Unit Circle Interpretation Quiz: Fundamental Shape And Unit Circle Interpretation

1) Tan(−θ) = −tanθ, so y = tanθ is an odd function symmetric about the origin.

True

False

sin is odd and cos is even, so tan(−θ) = sin(−θ)/cos(−θ) = (−sinθ)/(cosθ) = −tanθ. This is the definition of an odd function.

Explanation

sin is odd and cos is even, so tan(−θ) = sin(−θ)/cos(−θ) = (−sinθ)/(cosθ) = −tanθ. This is the definition of an odd function.

2) Tanθ = sinθ/cosθ is defined exactly when cosθ ≠ 0.

True

False

By definition tanθ = sinθ/cosθ. Division by zero is not allowed, so tanθ is defined precisely at angles with cosθ ≠ 0.

Explanation

By definition tanθ = sinθ/cosθ. Division by zero is not allowed, so tanθ is defined precisely at angles with cosθ ≠ 0.

3) One full basic cycle of y = tanθ is most naturally graphed on which interval?

−π/2 to π/2

0 to 2π

−π to π

−π/4 to 3π/4

Between consecutive asymptotes at −π/2 and π/2, the graph runs through one complete increasing branch crossing the origin. The function repeats every π.

Explanation

Between consecutive asymptotes at −π/2 and π/2, the graph runs through one complete increasing branch crossing the origin. The function repeats every π.

4) Select all θ in [0, 2π) where tanθ is undefined.

π/2

π

3π/2

0

π/4

tanθ = sinθ/cosθ is undefined when cosθ = 0. On [0, 2π) that occurs at θ = π/2 and 3π/2.

Explanation

tanθ = sinθ/cosθ is undefined when cosθ = 0. On [0, 2π) that occurs at θ = π/2 and 3π/2.

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5) What is the fundamental period of y = tanθ?

π

2π

π/2

No period

Using tanθ = sinθ/cosθ and angle addition, tan(θ+π) = tanθ for all θ where defined. Hence the smallest positive period is π.

Explanation

Using tanθ = sinθ/cosθ and angle addition, tan(θ+π) = tanθ for all θ where defined. Hence the smallest positive period is π.

6) State the sign of tanθ in Quadrant III.

In Quadrant III, both sinθ and cosθ are negative, so tanθ = sinθ/cosθ is positive.

Explanation

In Quadrant III, both sinθ and cosθ are negative, so tanθ = sinθ/cosθ is positive.

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7) Solve tanθ = √3 on [0, 2π). Select all solutions.

π/3

2π/3

4π/3

5π/3

π/6

tanθ = √3 at reference angle π/3. Tangent is positive in Quadrants I and III, giving θ = π/3 and 4π/3 on [0, 2π).

Explanation

tanθ = √3 at reference angle π/3. Tangent is positive in Quadrants I and III, giving θ = π/3 and 4π/3 on [0, 2π).

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8) The range of y = tanθ is all real numbers.

True

False

As θ approaches the vertical asymptotes, tanθ tends to ±∞, and by continuity on each interval it takes every real value exactly once per period.

Explanation

As θ approaches the vertical asymptotes, tanθ tends to ±∞, and by continuity on each interval it takes every real value exactly once per period.

9) Which of the following angles is a zero of y = tanθ on [0, 2π)?

π/6

π/4

π

π/3

tanθ = 0 when sinθ = 0 and cosθ ≠ 0. On [0, 2π), this occurs at θ = 0 and θ = π. Among the options, only θ = π satisfies sinπ = 0 and cosπ = −1 ≠ 0. The other options give tan(π/6) = 1/√3, tan(π/4) = 1, and tan(π/3) = √3, none of which equal zero.

Explanation

tanθ = 0 when sinθ = 0 and cosθ ≠ 0. On [0, 2π), this occurs at θ = 0 and θ = π. Among the options, only θ = π satisfies sinπ = 0 and cosπ = −1 ≠ 0. The other options give tan(π/6) = 1/√3, tan(π/4) = 1, and tan(π/3) = √3, none of which equal zero.

10) Evaluate tan(135°).

At 135°, sin135° = √2/2 and cos135° = −√2/2, so tan135° = (√2/2)/(−√2/2) = −1.

Explanation

At 135°, sin135° = √2/2 and cos135° = −√2/2, so tan135° = (√2/2)/(−√2/2) = −1.

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11) Between any two consecutive vertical asymptotes, the graph of y = tanθ is strictly increasing.

True

False

On each interval (−π/2 + kπ, π/2 + kπ), tanθ has derivative sec^2θ > 0, so it increases strictly across the interval.

Explanation

On each interval (−π/2 + kπ, π/2 + kπ), tanθ has derivative sec^2θ > 0, so it increases strictly across the interval.

12) Evaluate tan(45°).

On the unit circle, sin45° = √2/2 and cos45° = √2/2, so tan45° = (√2/2)/(√2/2) = 1.

Explanation

On the unit circle, sin45° = √2/2 and cos45° = √2/2, so tan45° = (√2/2)/(√2/2) = 1.

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13) If (x, y) = (cosθ, sinθ) = (√3/2, 1/2), what is tanθ?

√3/2

√3

1/2

1/√3

tanθ = y/x = (1/2)/(√3/2) = 1/√3.

Explanation

tanθ = y/x = (1/2)/(√3/2) = 1/√3.

14) On [0, 2π), in which intervals is tanθ > 0?

(0, π/2)

(π/2, π)

(π, 3π/2)

(3π/2, 2π)

{π/2}

Since tanθ = sinθ/cosθ, tanθ is positive when sinθ and cosθ have the same sign: Quadrants I and III, i.e., (0, π/2) and (π, 3π/2).

Explanation

Since tanθ = sinθ/cosθ, tanθ is positive when sinθ and cosθ have the same sign: Quadrants I and III, i.e., (0, π/2) and (π, 3π/2).

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15) Which statement about the tangent graph is true?

It has zeros at θ = kπ for all integers k.

It has zeros at θ = π/2 + kπ.

It is even about the y-axis.

Its period is 2π.

tanθ = 0 when sinθ = 0 and cosθ ≠ 0, i.e., θ = kπ. It is odd and has period π, not 2π.

Explanation

tanθ = 0 when sinθ = 0 and cosθ ≠ 0, i.e., θ = kπ. It is odd and has period π, not 2π.

16) For θ with cosθ ≠ 0, the slope of the line from the origin to (cosθ, sinθ) equals

Cosθ/sinθ

Sinθ/cosθ

Secθ/cscθ − 1

Sinθ·cosθ

Slope = rise/run = y/x = sinθ/cosθ = tanθ when cosθ ≠ 0.

Explanation

Slope = rise/run = y/x = sinθ/cosθ = tanθ when cosθ ≠ 0.

17) Select all expressions equivalent to tanθ (where defined).

Y/x

Sinθ/cosθ

Cosθ/sinθ

Secθ/cscθ

−sinθ/cosθ

On the unit circle, tanθ = y/x = sinθ/cosθ; also secθ/cscθ = (1/cosθ)/(1/sinθ) = sinθ/cosθ = tanθ. The expressions cosθ/sinθ = cotθ and −sinθ/cosθ = −tanθ.

Explanation

On the unit circle, tanθ = y/x = sinθ/cosθ; also secθ/cscθ = (1/cosθ)/(1/sinθ) = sinθ/cosθ = tanθ. The expressions cosθ/sinθ = cotθ and −sinθ/cosθ = −tanθ.

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18) The graph of y = tanθ passes through the origin.

True

False

tan0 = 0, so the point (0, 0) lies on y = tanθ.

Explanation

tan0 = 0, so the point (0, 0) lies on y = tanθ.

19) Write the general formula for the vertical asymptotes of y = tanθ.

tanθ is undefined when cosθ = 0, which happens at θ = π/2 + kπ, k ∈ ℤ.

Explanation

tanθ is undefined when cosθ = 0, which happens at θ = π/2 + kπ, k ∈ ℤ.

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20) Select all angles in [−π/2, π/2) where tanθ > 1.

−π/4

0

π/6

π/3

3π/8

On [−π/2, π/2), tanθ increases from −∞ to +∞ and crosses 1 at θ = π/4. Thus tanθ > 1 for θ in (π/4, π/2): examples include π/3 and 3π/8. The others give tanθ ≤ 1.

Explanation

On [−π/2, π/2), tanθ increases from −∞ to +∞ and crosses 1 at θ = π/4. Thus tanθ > 1 for θ in (π/4, π/2): examples include π/3 and 3π/8. The others give tanθ ≤ 1.

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