SSC CGL 2024 Tier-I · 17 September 2024, Shift 2 · Quantitative Aptitude · Question 15 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

If sin A + cos A = 1⁄(2√2), then find the value of sin⁴A + cos⁴A.

Answer

Why

Correct — B.

The stem is printed as an image: if sin A + cos A = 1 ⁄ (2√2), find sin⁴A + cos⁴A.

Square the given equation:
(sin A + cos A)² = (1 ⁄ (2√2))² = 1 ⁄ 8

Expand the left side, using sin²A + cos²A = 1:
1 + 2 sin A cos A = 1 ⁄ 8
2 sin A cos A = 1 ⁄ 8 − 1 = −7 ⁄ 8
sin A cos A = −7 ⁄ 16

Now sin⁴A + cos⁴A = (sin²A + cos²A)² − 2 sin²A cos²A:
= 1 − 2 × (−7 ⁄ 16)² = 1 − 2 × 49 ⁄ 256
= 1 − 49 ⁄ 128 = 79 ⁄ 128 → option (b).

Why the others are wrong

(a)
Option (a) shows 5 ⁄ 13. No step here produces a denominator of 13: squaring 1 ⁄ (2√2) gives 1 ⁄ 8, and the final division is by 128.
(c)
Option (c) shows 79 ⁄ 126, one digit from the key. The denominator arises as 2 × 49 ⁄ 256 = 49 ⁄ 128, so it has to be a power of two — and 126 is not.
(d)
Option (d) shows 7 ⁄ 13. The 7 is a real piece of the working, from 2 sin A cos A = −7 ⁄ 8, but the answer needs 1 − 2(sin A cos A)², not the 7 lifted out on its own.

Concept

Treat sin A and cos A as two numbers with a known sum and an unknown product, and this stops being trigonometry.

Squaring the sum gives sin²A + cos²A + 2 sin A cos A, and the first two terms collapse to 1. That single move converts the given sum into the product sin A cos A.

The target then rewrites in the same two quantities: sin⁴A + cos⁴A = (sin²A + cos²A)² − 2 sin²A cos²A = 1 − 2(sin A cos A)².

Every SSC item of this family is the same two lines: square the sum to get the product, then feed the product into the identity.

Both the stem and the four options are printed as images in the response sheet, so the option list must be read off the pictures: (a) 5 ⁄ 13, (b) 79 ⁄ 128, (c) 79 ⁄ 126, (d) 7 ⁄ 13. Note also that sin A cos A comes out negative, so A is not a first-quadrant angle — the identity squares it, so the answer is unaffected.

Key facts

(sin A + cos A)² = 1 + 2 sin A cos A, because sin²A + cos²A = 1.
sin⁴A + cos⁴A = 1 − 2 sin²A cos²A.
Here (1 ⁄ (2√2))² = 1 ⁄ 8, giving sin A cos A = −7 ⁄ 16 and sin⁴A + cos⁴A = 79 ⁄ 128.
sin A + cos A can only lie between −√2 and √2, and 1 ⁄ (2√2) ≈ 0.354 sits inside that range.

Study next

Deriving sin⁶A + cos⁶A = 1 − 3 sin²A cos²A from the same substitutionItems that give sin A − cos A instead, where squaring yields 1 − 2 sin A cos ASquaring surd fractions such as 1 ⁄ (2√2) without rationalising first

Common traps

Computing 1 ⁄ 8 − 1 as +7 ⁄ 8 and losing the minus sign.
Writing sin⁴A + cos⁴A = (sin²A + cos²A)² and forgetting the −2 sin²A cos²A correction, which returns 1.
Squaring 1 ⁄ (2√2) as 1 ⁄ (2 × 2) = 1 ⁄ 4 instead of 1 ⁄ (4 × 2) = 1 ⁄ 8.

SSC prints these trigonometric stems as pictures, so open the image at full size before writing anything down — the whole item turns on reading 2√2 rather than 2√3.

The same squaring-to-collect-sin²-plus-cos² move drives 17 Sep 2024, 12:30, Quant Q.6, where a = x cos θ + y sin θ and b = x sin θ − y cos θ and you are asked for a² + b².

Related PYQs

No directly related past PYQ was found.