JEE Main 2022 — 25 June, Evening Shift. Previous Year Question.
Problem
The value of $2\sin(12°)-\sin(72°)$ is:
Key insight. $72°$ isn’t one of the “standard” angles with a simple sine value, but it splits neatly as $60°+12°$ — an angle whose sine and cosine are known. Expanding it that way turns the whole expression into a combination of $\sin12°$ and $\cos12°$, which can then be recombined into a single sine of a new, more useful angle.
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Approach
Expand $\sin(72°)$ as $\sin(60°+12°)$ using the angle-addition formula, substitute the known values of $\sin60°$ and $\cos60°$, and simplify. The resulting expression turns out to be a constant multiple of $(\cos30°\sin12°-\sin30°\cos12°)$, which collapses into a single sine of a difference angle — landing on the well-known value $\sin18°$.
Solution
Step 1 — Expand sin(72°)
$$\sin(72°) = \sin(60°+12°) = \sin60°\cos12°+\cos60°\sin12°$$
$$= \frac{\sqrt3}{2}\cos12° + \frac{1}{2}\sin12°$$
Step 2 — Substitute into the original expression
$$2\sin12° – \sin72° = 2\sin12° – \frac{\sqrt3}{2}\cos12° – \frac{1}{2}\sin12°$$
$$= \frac{3}{2}\sin12° – \frac{\sqrt3}{2}\cos12°$$
Step 3 — Factor out √3
$$\frac{3}{2}\sin12° – \frac{\sqrt3}{2}\cos12° = \sqrt3\left(\frac{\sqrt3}{2}\sin12° – \frac{1}{2}\cos12°\right)$$
(since $\dfrac{3}{2\sqrt3} = \dfrac{\sqrt3}{2}$)
Step 4 — Recognise cos30° and sin30°
$$\sqrt3\left(\cos30°\sin12° – \sin30°\cos12°\right)$$
This matches the expansion of $\sin(A-B)$ with $A=12°$, $B=30°$:
$$= \sqrt3\sin(12°-30°) = \sqrt3\sin(-18°) = -\sqrt3\sin18°$$
Step 5 — Substitute the known value of sin 18°
Using the standard result $\sin18° = \dfrac{\sqrt5-1}{4}$:
$$-\sqrt3\cdot\frac{\sqrt5-1}{4} = \frac{\sqrt3(1-\sqrt5)}{4}$$
Answer
$$\frac{\sqrt3(1-\sqrt5)}{4}$$
Common mistakes
- Trying to split $72°$ as $2\times36°$ instead of $60°+12°$. This is a valid identity too, but it requires knowing $\sin36°$ and $\cos36°$ in surd form and leads to a much messier simplification than using the already-present $12°$.
- Forgetting the standard value of $\sin18°$. Recognising $\sin(-18°)=-\sin18°$ and substituting $\sin18°=\dfrac{\sqrt5-1}{4}$ is the step that finally turns the surds into the matching answer choice — without it, the expression stays in an unrecognisable intermediate form.
Practise next
- Find the value of $2\cos(12°)-\cos(48°)$, using the same expand-and-recombine method with $48°=60°-12°$.
Show answer
$\sqrt3\cos42^\circ\approx1.2872$. Expand the second term with $48^\circ=60^\circ-12^\circ$:
$$\cos48^\circ=\tfrac12\cos12^\circ+\tfrac{\sqrt3}{2}\sin12^\circ,$$
so the expression is $\tfrac32\cos12^\circ-\tfrac{\sqrt3}{2}\sin12^\circ=\sqrt3\left(\cos30^\circ\cos12^\circ-\sin30^\circ\sin12^\circ\right)=\sqrt3\cos42^\circ$.
Worth noting the contrast with the problem above, which recombines to $-\sqrt3\sin18^\circ$ and then simplifies further because $\sin18^\circ=\tfrac{\sqrt5-1}{4}$ is a special value. $42^\circ$ is not, so $\sqrt3\cos42^\circ$ is as far as this one goes.
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