Question 101 markMCQSection A

  1. If cos⁡A=12\cos A = \frac{1}{2}, then the value of sin⁡2A+2cos⁡2A\sin^2 A + 2\cos^2 A is : (a) 3/2 (b) 5/4 (c) -1 (d) 1/2

Introduction to Trigonometry
Evaluating trigonometric expressions

Options

(A)32\frac{3}{2}
(B)54\frac{5}{4}
(C)−1-1
(D)12\frac{1}{2}

Official answer

(b) 54\frac{5}{4} — sin⁡2A+2cos⁡2A=1+cos⁡2A=1+14=54\sin^2 A + 2\cos^2 A = 1 + \cos^2 A = 1 + \frac{1}{4} = \frac{5}{4}.

Answer keywords:trigonometric identitysin²A + cos²A = 1cos A = 1/2A = 60°value 5/4evaluating expression

Marking scheme

  1. 1.1 mark: correct option (b) 54\frac{5}{4}.
  2. 2.Internal justification: 1+cos⁡2A=1+14=541 + \cos^2 A = 1 + \frac{1}{4} = \frac{5}{4} (or direct A=60°A = 60° substitution).

Hint

Rewrite sin⁡2A+2cos⁡2A\sin^2 A + 2\cos^2 A as (sin⁡2A+cos⁡2A)+cos⁡2A=1+cos⁡2A(\sin^2 A + \cos^2 A) + \cos^2 A = 1 + \cos^2 A, then substitute cos⁡2A=14\cos^2 A = \frac{1}{4}.

Quick oral answer

Group it as sin²A plus cos²A, which is 1, plus one more cos²A; since cos A is one-half, cos²A is one-fourth, giving 1 plus one-fourth, that is five-fourths.

Common mistakes

  1. 1.Getting 32\frac{3}{2} by writing the expression as 1+2cos⁡2A1 + 2\cos^2 A instead of 1+cos⁡2A1 + \cos^2 A (double-counting the cos² term).
  2. 2.Choosing −1-1 despite the expression being a sum of squares that can never be negative.
  3. 3.Forgetting to square: using cos⁡A=12\cos A = \frac{1}{2} directly in place of cos⁡2A=14\cos^2 A = \frac{1}{4}.

Analysis and explanation

Rewriting sin²A + 2cos²A as (sin²A + cos²A) + cos²A = 1 + cos²A instantly gives 1 + 1/4 = 5/4 when cos A = 1/2.


Concept

  • Identity sin²A + cos²A = 1 lets the expression be regrouped as 1 + cos²A.
  • Given cos A = 1/2 ⇒ cos²A = 1/4.

Key steps

  • sin⁡2A+2cos⁡2A=1+cos⁡2A=1+14=54\sin^2 A + 2\cos^2 A = 1 + \cos^2 A = 1 + \frac{1}{4} = \frac{5}{4} → option (b).
  • Cross-check: cos⁡A=12⇒A=60°⇒sin⁡A=32\cos A = \frac{1}{2} \Rightarrow A = 60° \Rightarrow \sin A = \frac{\sqrt{3}}{2}; sin⁡2A+2cos⁡2A=34+2(14)=54\sin^2 A + 2\cos^2 A = \frac{3}{4} + 2\left(\frac{1}{4}\right) = \frac{5}{4} ✓.

Common mistakes

  • (a) 3/2: comes from wrongly writing 1 + 2cos²A instead of 1 + cos²A (double-counting one cos²A term).
  • (c) −1: impossible, since the expression is a sum of squares and must be positive — a sign-error trap.
  • (d) 1/2: results from dropping the sin²A term entirely.

Real-world

  • The grouping trick (splitting 2cos²A into cos²A + cos²A so one part combines with sin²A) is a standard time-saver in trigonometry exams.

Interesting facts

cos 60° = 1/2 is one of the standard angle values every Class 10 student memorises; 60° is also the interior angle of an equilateral triangle.

Expressions like this are the algebra behind alternating-current and wave power calculations, where averaging sin² and cos² terms is routine in physics and engineering.

Question details

Q10 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Introduction to Trigonometry
Topic
Evaluating trigonometric expressions
Bloom's level
apply
Syllabus topic
Unit 5: Trigonometry — Trigonometric Ratios of Specific Angles and the Identity sin²A + cos²A = 1
Author
CBSE
Reviewed by
Padhantu Academic Team

Spotted a mistake or something unclear?

Tell us and we will fix it.

Frequently asked questions

What is the fastest way to solve this?

Split 2cos⁡2A2\cos^2 A into cos⁡2A+cos⁡2A\cos^2 A + \cos^2 A, group sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1, leaving 1+cos⁡2A1 + \cos^2 A. With cos⁡2A=14\cos^2 A = \frac{1}{4}, the answer is 54\frac{5}{4}.

Can I just find the angle A instead?

Yes. cos⁡A=12\cos A = \frac{1}{2} gives A=60°A = 60°, so sin⁡A=32\sin A = \frac{\sqrt{3}}{2}. Then sin⁡2A+2cos⁡2A=34+2(14)=54\sin^2 A + 2\cos^2 A = \frac{3}{4} + 2\left(\frac{1}{4}\right) = \frac{5}{4} — the same answer.

Why can the answer never be −1?

The expression is a sum of squared quantities multiplied by positive numbers, so it is always positive. Option (c) is a trap.