If sin x = 0.6 and x ∈ (0, π⁄2), find sin 2x + cos 2x.

Correct — D. x lies in (0, π⁄2), so cos x is positive.

cos x = √(1 − 0.6²) = √(1 − 0.36) = √0.64 = 0.8

sin 2x = 2 sin x cos x = 2 × 0.6 × 0.8 = 0.96
cos 2x = 1 − 2 sin²x = 1 − 2 × 0.36 = 1 − 0.72 = 0.28

sin 2x + cos 2x = 0.96 + 0.28 = 1.24 → option (d)

Why the others are wrong

• (a)
  2,68 — 2,68, read as 2.68, is too big: sin θ + cos θ never exceeds √2 ≈ 1.41. It is 0.96 + 1.72, from the wrong-signed cos 2x = 1 + 2 sin²x.
• (b)
  0.52 — 0.52 is smaller than sin 2x = 0.96 on its own. Since cos 2x = 0.28 is positive, the sum has to be more than 0.96.
• (c)
  0.68 — 0.68 is 0.96 − 0.28: it takes cos 2x with the wrong sign, as sin²x − cos²x = −0.28. The identity is cos²x − sin²x = 0.64 − 0.36 = 0.28.

The double-angle identities turn the ratios of x into the ratios of 2x:

sin 2x = 2 sin x cos x
cos 2x = cos²x − sin²x = 1 − 2 sin²x = 2 cos²x − 1

sin x = 0.6 = 3⁄5 is the 3-4-5 right triangle, so cos x = 4⁄5 = 0.8 in the first quadrant. Knowing the triangle skips the square root.

Key facts

• sin 2x = 2 sin x cos x.
• cos 2x = cos²x − sin²x = 1 − 2 sin²x = 2 cos²x − 1.
• In the first quadrant, sin x = 3⁄5 gives cos x = 4⁄5 and tan x = 3⁄4.
• sin θ + cos θ lies between −√2 and √2 for every θ.

Common traps

• Using cos 2x = sin²x − cos²x, which flips its sign and gives 0.68.
• Dropping the 2 in sin 2x = 2 sin x cos x, which gives 0.48 instead of 0.96.
• Taking cos x = −0.8: the interval (0, π⁄2) in the stem rules out the negative root.

No directly related past PYQ was found.