Two adjacent sides of a parallelogram PQRS are given by PQ=j+k and PS=i-j.

Question

Two adjacent sides of a parallelogram PQRS are given by $\overset{	o \leftarrow }{PQ}=\hat{j}+\hat{k}$ and $\overset{	o \leftarrow }{PS}=\hat{i}-\hat{j}$. If the side PS is rotated about the point P by an acute angle $\alpha $ in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then ${\sin }^{2}(\frac{5\alpha }{2})-{\sin }^{2}(\frac{\alpha }{2})$ is equal to:

Accepted Answer

Let $\theta$ be the angle between $\vec{PQ}$ and $\vec{PS}$: $\cos\theta = \dfrac{\vec{PQ}\cdot\vec{PS}}{|\vec{PQ}||\vec{PS}|} = \dfrac{0+(-1)+0}{\sqrt2\cdot\sqrt2} = -\dfrac12 \Rightarrow \theta = \dfrac{2\pi}{3}$.

Rotating $PS$ by acute angle $\alpha$ to make it perpendicular to $PQ$ requires $\theta - \alpha = \dfrac\pi2$ (since rotating toward $PQ$ reduces the angle from $\frac{2\pi}3$ to $\frac\pi2$), giving $\alpha = \dfrac{2\pi}3 - \dfrac\pi2 = \dfrac\pi6$.

$$\sin^2\frac{5\alpha}2 - \sin^2\frac\alpha2 = \sin\left(\frac{5\alpha}2+\frac\alpha2\right)\sin\left(\frac{5\alpha}2-\frac\alpha2\right) = \sin(3\alpha)\sin(2\alpha)$$

With $\alpha=\pi/6$: $=\sin\frac\pi2\cdot\sin\frac\pi3 = 1\cdot\frac{\sqrt3}2 = \frac{\sqrt3}{2}$

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