If sin² A − cos² A =1⁄2, find the value of cos² A.
- (a)1⁄2
- (b)1⁄4
- (c)3⁄4
- (d)2⁄3
Answer
Why
Correct — B. Replace sin²A using sin²A = 1 − cos²A, so the equation has one unknown.
(1 − cos²A) − cos²A = 1⁄2
1 − 2cos²A = 1⁄2
2cos²A = 1 − 1⁄2 = 1⁄2
cos²A = 1⁄4 → option (b)
Check: sin²A = 3⁄4, and 3⁄4 − 1⁄4 = 1⁄2, as required. A = 60° is one angle that fits.
Why the others are wrong
- (a)1⁄2 — 1⁄2 is the right-hand side copied across. If cos²A were 1⁄2, sin²A would be 1⁄2 as well, and their difference would be 0.
- (c)3⁄4 — 3⁄4 is sin²A, not cos²A. Put cos²A = 3⁄4 and the left side becomes 1⁄4 − 3⁄4 = −1⁄2, the wrong sign.
- (d)2⁄3 — 2⁄3 leaves sin²A = 1⁄3, so sin²A − cos²A = 1⁄3 − 2⁄3 = −1⁄3, wrong in both size and sign.
Concept
The identity sin²A + cos²A = 1 lets you trade one square for the other. Any linear equation in sin²A and cos²A then has a single unknown.
The left side is also a double-angle form: cos²A − sin²A = cos 2A. So the equation says cos 2A = −1⁄2.
Then cos²A = (1 + cos 2A) ⁄ 2 = (1 − 1⁄2) ⁄ 2 = 1⁄4, the same answer by a second route.
The question does not say A is acute, and it does not need to. Every angle with cos 2A = −1⁄2 has cos²A = 1⁄4.
Key facts
- sin²A + cos²A = 1 for every angle A.
- cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A.
- sin 60° = √3⁄2 and cos 60° = 1⁄2, so sin²60° = 3⁄4 and cos²60° = 1⁄4.
Study next
Common traps
- Finding sin²A = 3⁄4 first and answering it, when the question asks for cos²A.
- Flipping the sign: writing 2cos²A − 1 = 1⁄2 instead of 1 − 2cos²A = 1⁄2 gives cos²A = 3⁄4.
The same substitution solves 10 sin²θ + 6 cos²θ = 7 at 24 Sep 2024, 16:00, Quant Q.25: replacing cos²θ gives 6 + 4 sin²θ = 7, so sin²θ = 1⁄4 and θ = 30°.
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