Question
Two adjacent sides of a parallelogram PQRS are given by $\overrightarrow{PQ} = \widehat{j} + \widehat{k}$ and $\overrightarrow{PS} = \widehat{i} - \widehat{j}$. 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 $\sin^{2}\left( \frac{5\alpha}{2} \right) - \sin^{2}\left( \frac{\alpha}{2} \right)$ is equal to :
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Solution
Sol. [Key Concept: Angle between the sides from $cos\theta$; use $\sin^{2}A - \sin^{2}B = sin(A + B)sin(A - B)$.]
Let angle between $\overrightarrow{PQ}$ & $\overrightarrow{PS}$ is $\theta$
$$\begin{matrix} & cos\theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PS}}{|\overrightarrow{PQ}||\overrightarrow{PS}|} \\ & cos\theta = \frac{0 - 1 + 0}{\sqrt{2}\sqrt{2}} \\ & cos\theta = \frac{- 1}{2} \\ & \theta = \frac{2\pi}{3} \\ & \alpha = \frac{2\pi}{3} - \frac{\pi}{2}\text{~}\text{So}\text{~}\alpha = \frac{\pi}{6} \\ & \sin^{2}\frac{5\alpha}{2} - \sin^{2}\frac{\alpha}{2} = sin\left( \frac{5\alpha}{2} + \frac{\alpha}{2} \right)sin\left( \frac{5\alpha}{2} - \frac{\alpha}{2} \right) \\ & \ = sin\frac{6\alpha}{2} \times sin\frac{4\alpha}{2} \\ & \ = sin3\alpha \times sin2\alpha \\ & \ = sin\frac{\pi}{2} \times sin\frac{\pi}{3} \\ & \ = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \end{matrix}$$
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