If A = 15°, then what is the value of (11√2 cos 3A + 10 sin 2A) ⁄ (7√3 sin 4A)?

Correct — A. The stem is an image: with A = 15°, find (11√2 cos 3A + 10 sin 2A) ⁄ (7√3 sin 4A). Substitute and it is the standard-angle table.

3A = 45°, 2A = 30°, 4A = 60°

Numerator: 11√2 · cos 45° + 10 · sin 30°
= 11√2 · 1⁄√2 + 10 · 1⁄2
= 11 + 5 = 16

Denominator: 7√3 · sin 60° = 7√3 · √3⁄2 = 21⁄2 = 10.5

16 ÷ 10.5 = 32⁄21 → option (a), which prints the fraction 32 over 21.

Why the others are wrong

• (b)
  Option (b) prints 7⁄11. That is the two coefficients 7 and 11 read straight off the expression, with cos 45°, sin 30° and sin 60° never evaluated at all.
• (c)
  Option (c) prints 11⁄7, the same coefficient ratio the other way up. It drops the 10 sin 2A term, which adds 5, and it ignores that √3 sin 60° = 3⁄2 rather than 1.
• (d)
  Option (d) prints 21⁄32, the right value inverted. It is what you get by dividing the denominator by the numerator, 10.5 ÷ 16, instead of the other way round.

A trigonometric value question of this shape reduces to the standard-angle table: 0°, 30°, 45°, 60°, 90°.

The setter hides those angles behind multiples — 3A, 2A, 4A — so that one substitution, A = 15°, turns every argument into a table entry.

The coefficients are chosen to cancel. √2 cos 45° = 1 and √3 sin 60° = 3⁄2 are what the 11√2 and the 7√3 are there for, and they are what leaves whole numbers behind.

Key facts

• sin 30° = 1⁄2, cos 45° = 1⁄√2, sin 60° = √3⁄2.
• √2 cos 45° = 1, so a √2 written in front of cos 45° exists to cancel.
• √3 sin 60° = 3⁄2, so 7√3 sin 60° = 21⁄2 = 10.5.

Common traps

• Substituting A = 15° into one term and leaving the others as 3A or 4A.
• Reading cos 45° as √3⁄2, a value that belongs to sin 60°.
• Inverting at the last step, since 21⁄32 is sitting on the option list waiting for it.

No directly related past PYQ was found.