Find the value of

Correct — B. The stem figure reads 3 sin15° − 4 sin³15°, which is exactly the right-hand side of the triple-angle identity for sine.

Rule: sin 3θ = 3 sin θ − 4 sin³θ

Put θ = 15°, so 3θ = 45°:
3 sin15° − 4 sin³15° = sin 45°
sin 45° = 1⁄√2 → option (b)

Check it numerically: sin15° ≈ 0.2588, so 3(0.2588) − 4(0.2588)³ = 0.7765 − 0.0693 = 0.7071, and 1⁄√2 = 0.7071.

Why the others are wrong

• (a)
  √2 ≈ 1.414 is greater than 1, and the whole expression collapses to sin 45°. No sine of any angle exceeds 1, so this can be struck out before touching the identity. √2 is cosec 45°, the reciprocal of the keyed value.
• (c)
  2 fails the same bound: the expression equals sin 45°, and a sine lies between −1 and 1. The 3 and the 4 are the identity's fixed coefficients, not numbers to combine arithmetically.
• (d)
  1⁄2 is sin 30° — the value the expression would take if the identity carried θ to 2θ. It is sin 3θ, so 15° becomes 45°, not 30°.

Three sine terms in one expression with the coefficients 3 and 4 is the signature of the triple-angle identity. It is not a formula to derive under exam pressure — it is one to recognise.

sin 3θ = 3 sin θ − 4 sin³θ comes from expanding sin(2θ + θ) and replacing cos²θ by 1 − sin²θ, which is why only sines survive.

Once you see it, the awkward 15° stops mattering: the identity carries the angle to 45°, whose sine is a standard value.

Key facts

• sin 3θ = 3 sin θ − 4 sin³θ.
• cos 3θ = 4 cos³θ − 3 cos θ — note the coefficients swap places and the sign pattern flips.
• sin 45° = cos 45° = 1⁄√2 ≈ 0.7071.
• Every sine and cosine lies between −1 and 1, so any option above 1 is impossible for an expression that reduces to a single sine.

Common traps

• Expanding sin15° as (√6 − √2)⁄4 and grinding through the cube instead of spotting the identity.
• Matching the 3 and 4 to the cosine version and answering as though the angle were being cubed.
• Rationalising 1⁄√2 to √2⁄2 and then failing to match it against the option images.

No directly related past PYQ was found.