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JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 16

Two adjacent sides of a parallelogram PQRS are given by PQ→=j^+k^\overrightarrow{PQ}=\hat j+\hat k and PS→=i^−j^\overrightarrow{PS}=\hat i-\hat j. If the side PS is rotated about the point P by an acute angle α\alpha in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then sin⁡2(5α2)−sin⁡2(α2)\sin^2\left(\dfrac{5\alpha}2\right)-\sin^2\left(\dfrac\alpha2\right) is equal to :

Choose one option
Answer and worked solution

Official answer: B

Angle between the sides. cos⁡∠QPS=(0,1,1)⋅(1,−1,0)22=−12\cos\angle QPS=\dfrac{(0,1,1)\cdot(1,-1,0)}{\sqrt2\sqrt2}=-\frac12, i.e. 120°. Rotating PS by an acute angle to make it 90° with PQ needs α=30∘\alpha=30^\circ.

Value. sin⁡2A−sin⁡2B=sin⁡(A+B)sin⁡(A−B)\sin^2A-\sin^2B=\sin(A+B)\sin(A-B), so
sin⁡275∘−sin⁡215∘=sin⁡90∘sin⁡60∘=32.\sin^275^\circ-\sin^215^\circ=\sin90^\circ\sin60^\circ=\frac{\sqrt3}2.

(B)

Topics:Dot productTrigonometric ratios and identities
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