Double Angle Formulas Quiz: Solving Equations and Evaluating Values Using Double-Angle Formulas

  • 11th Grade
Reviewed by Editorial Team
The ProProfs editorial team is comprised of experienced subject matter experts. They've collectively created over 10,000 quizzes and lessons, serving over 100 million users. Our team includes in-house content moderators and subject matter experts, as well as a global network of rigorously trained contributors. All adhere to our comprehensive editorial guidelines, ensuring the delivery of high-quality content.
Learn about Our Editorial Process
| By Thames
T
Thames
Community Contributor
Quizzes Created: 7682 | Total Attempts: 9,547,133
| Questions: 20 | Updated: Dec 16, 2025
Please wait...
Question 1 / 20
0 %
0/100
Score 0/100
1) Select all identities that are correct.

Explanation

The tangent identity has denominator 1 − tan^2θ. The other listed forms are correct.

Submit
Please wait...
About This Quiz
Double Angle Formulas Quiz: Solving Equations And Evaluating Values Using Double-angle Formulas - Quiz

How do double-angle formulas help simplify and solve trigonometric expressions? In this quiz, you’ll explore how identities like sin(2𝜃),
cos(2𝜃), and tan(2𝜃) emerge from foundational trig relationships. You’ll practice evaluating expressions, solving equations, and choosing the most efficient form of each identity. Step by step, you’ll build confidence applying double-angle formulas... see moreto problems that require precision, algebraic reasoning, and a strong grasp of trigonometric structure.
see less

2)
You may optionally provide this to label your report, leaderboard, or certificate.
2) Solve on [0, 2π): sin(2θ) = 0.

Explanation

sin(2θ)=0 ⇒ 2θ = kπ ⇒ θ = kπ/2. In [0,2π) the distinct solutions are θ = 0, π/2, π, 3π/2.

Submit
3) If sinθ = 3/5 and cosθ = 4/5 with θ in Quadrant I, find sin(2θ).

Explanation

Use sin(2θ) = 2sinθ cosθ = 2·(3/5)·(4/5) = 24/25. Quadrant I confirms positive sign.

Submit
4) If sinθ = √3/2 and θ in Quadrant II, then cos(2θ) < 0.

Explanation

cos(2θ)=1−2sin^2θ=1−2·(3/4)=1−3/2=−1/2

Submit
5) If cosθ = 2/3 with θ in Quadrant I, find cos(2θ).

Explanation

cos(2θ)=2cos^2θ−1=2·(4/9)−1=8/9−1=−1/9.

Submit
6) Evaluate exactly: cos(2·30°)

Explanation

2·30° = 60°. cos60° = 1/2.

Submit
7) Solve on [−π, π]: sin(2θ) = √2/2.

Explanation

sin(2θ)=√2/2 ⇒ 2θ = π/4 + 2kπ or 3π/4 + 2kπ. Divide by 2 to obtain θ = π/8 + kπ and θ = 3π/8 + kπ. In [−π, π], these are π/8, 5π/8, −3π/8, −7π/8.

Submit
8) Select all correct evaluations.

Explanation

Evaluate: 2·45°=90° ⇒ sin90°=1, cos90°=0, tan90° undefined. Also 2·60°=120° ⇒ cos120°=−1/2. 2·30°=60° ⇒ sin60°=√3/2.

Submit
9) Given sinθ = −5/13 and θ in Quadrant IV, find tan(2θ).

Explanation

Quadrant IV: sinθ0. cosθ=12/13 ⇒ tanθ=−5/12. Then tan(2θ)=2(−5/12)/(1−(25/144))=(−10/12)/(119/144)=(−5/6)·(144/119)=−120/119.

Submit
10) Select all correct values of θ in [0, 2π) that solve cos(2θ) = 0.

Explanation

cos(2θ)=0 ⇒ 2θ = π/2 + kπ ⇒ θ = π/4 + kπ/2. In [0,2π) these are θ = π/4, 3π/4, 5π/4, 7π/4.

Submit
11) Simplify to a sine-only expression: cos(2θ) =

Explanation

From sin^2θ + cos^2θ = 1, replace cos^2θ by 1 − sin^2θ in cos^2θ − sin^2θ to get 1 − 2sin^2θ.

Submit
12) Select all solutions on [0, 2π) to cos(2θ) = −1.

Explanation

cos(2θ)=−1 ⇒ 2θ = π + 2kπ ⇒ θ = π/2 + kπ. In [0,2π) these are θ = π/2 and 3π/2.

Submit
13) Evaluate exactly: tan(2·30°)

Explanation

2·30° = 60°. tan60° = √3.

Submit
14) Tan(2θ) = 2tanθ/(1 − tan^2θ) whenever 1 − tan^2θ ≠ 0.

Explanation

Standard double-angle identity for tangent derived from tan(a+b). Denominator restriction prevents division by zero.

Submit
15) Solve for θ in [0, 2π): cos(2θ) = 1/2.

Explanation

cos(2θ)=1/2 ⇒ 2θ = ±π/3 + 2kπ ⇒ θ = π/6 + kπ and 5π/6 + kπ. In [0,2π) this gives θ = π/6, 5π/6, 7π/6, 11π/6.

Submit
16) Given tanθ = −3/4 and θ in Quadrant IV, find tan(2θ).

Explanation

tan(2θ)=2tanθ/(1−tan^2θ)=2(−3/4)/(1−9/16)=(−3/2)/(7/16)=(−3/2)·(16/7)=−24/7.

Submit
17) If tanθ = 1, then tan(2θ) is undefined.

Explanation

tan(2θ) = 2tanθ/(1 − tan^2θ). For tanθ=1, denominator 1 − 1 = 0, so undefined.

Submit
18) Which is NOT an equivalent form of cos(2θ)?

Explanation

cos(2θ) has three standard forms: cos^2θ − sin^2θ, 1 − 2sin^2θ, and 2cos^2θ − 1. The expression 2sin^2θ − 1 is not equal to cos(2θ); the sine-only form is 1 − 2sin^2θ.

Submit
19) If cosθ = −3/5 and θ is in Quadrant II, find cos(2θ).

Explanation

cos(2θ) = 2cos^2θ − 1 = 2·(9/25) − 1 = 18/25 − 1 = −7/25. Quadrant of θ does not affect cos^2θ, so result is negative.

Submit
20) If cosθ = 0, then sin(2θ) = 0.

Explanation

sin(2θ)=2sinθ cosθ. If cosθ=0, the product is zero regardless of sinθ (when defined).

Submit
×
Saved
Thank you for your feedback!
View My Results
Cancel
  • All
    All (20)
  • Unanswered
    Unanswered ()
  • Answered
    Answered ()
Select all identities that are correct.
Solve on [0, 2π): sin(2θ) = 0.
If sinθ = 3/5 and cosθ = 4/5 with θ in Quadrant I, find sin(2θ).
If sinθ = √3/2 and θ in Quadrant II, then cos(2θ) < 0.
If cosθ = 2/3 with θ in Quadrant I, find cos(2θ).
Evaluate exactly: cos(2·30°)
Solve on [−π, π]: sin(2θ) = √2/2.
Select all correct evaluations.
Given sinθ = −5/13 and θ in Quadrant IV, find tan(2θ).
Select all correct values of θ in [0, 2π) that solve cos(2θ) = 0.
Simplify to a sine-only expression: cos(2θ) =
Select all solutions on [0, 2π) to cos(2θ) = −1.
Evaluate exactly: tan(2·30°)
Tan(2θ) = 2tanθ/(1 − tan^2θ) whenever 1 − tan^2θ ≠ 0.
Solve for θ in [0, 2π): cos(2θ) = 1/2.
Given tanθ = −3/4 and θ in Quadrant IV, find tan(2θ).
If tanθ = 1, then tan(2θ) is undefined.
Which is NOT an equivalent form of cos(2θ)?
If cosθ = −3/5 and θ is in Quadrant II, find cos(2θ).
If cosθ = 0, then sin(2θ) = 0.
Alert!

Advertisement