Two adjacent sides of a parallelogram PQRS are given by PQ = j + k and PS = i - j. If the side PS is rotated about the point P by an acute angle α in…

Let $\theta$θ be the angle between the vectors $\vec{PQ}$PQ​ and $\vec{PS}$PS.

$$\cos \theta = \dfrac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}| |\vec{PS}|}$$cosθ=∣PQ​∣∣PS∣PQ​⋅PS​

$$\cos \theta = \dfrac{(\hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j})}{\sqrt{1^2 + 1^2} \sqrt{1^2 + (-1)^2}}$$cosθ=12+12​12+(−1)2​(j^​+k^)⋅(i^−j^​)​

$$\cos \theta = \dfrac{-1}{\sqrt{2} \times \sqrt{2}} = -\dfrac{1}{2}$$cosθ=2​×2​−1​=−21​

$$\theta = 120^{\circ}$$θ=120∘

The side PS is rotated by an acute angle $\alpha$α in the plane of the parallelogram to become perpendicular to PQ. The new angle between the sides is $90^{\circ}$90∘.

$$\alpha = 120^{\circ} - 90^{\circ} = 30^{\circ}$$α=120∘−90∘=30∘

The given expression is $\sin^2\left(\dfrac{5\alpha}{2}\right) - \sin^2\left(\dfrac{\alpha}{2}\right)$sin2(25α​)−sin2(2α​).

Using the trigonometric identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$sin2A−sin2B=sin(A+B)sin(A−B):

$$\sin^2\left(\dfrac{5\alpha}{2}\right) - \sin^2\left(\dfrac{\alpha}{2}\right) = \sin\left(\dfrac{5\alpha}{2} + \dfrac{\alpha}{2}\right) \sin\left(\dfrac{5\alpha}{2} - \dfrac{\alpha}{2}\right)$$sin2(25α​)−sin2(2α​)=sin(25α​+2α​)sin(25α​−2α​)

$$= \sin(3\alpha) \sin(2\alpha)$$=sin(3α)sin(2α)

Substituting $\alpha = 30^{\circ}$α=30∘:

$$= \sin(90^{\circ}) \sin(60^{\circ})$$=sin(90∘)sin(60∘)

$$= 1 \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}$$=1×23​​=23​​

Answer: $\dfrac{\sqrt{3}}{2}$23​​