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  • If p(x) = [cos x sin x] then show that P(x).P(y) = P(x + y) = P(y).P(x)

If P(x)= \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} then show that P(x).P(y) = P(x + y) = P(y).P(x)

Answers (1)

We are given that, 

P(x)= \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}

P(y)= \begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix}

\begin{array}{l} \therefore P(x) \cdot P(y)=\left[\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]\left[\begin{array}{cc} \cos y & \sin y \\ -\sin y & \cos y \end{array}\right] \\ \Rightarrow P(x) \cdot P(y)=\left[\begin{array}{cc} \cos x \cos y+\sin x(-\sin y) & \cos x \sin y+\sin x \cos y \\ -\sin x \cos y-\sin y(\cos x) & -\sin x \sin y+\cos x \cos y \end{array}\right] \end{array}

We know that-

\\ \\\cos x \cos y + \sin x \sin y = \cos (x - y) \\\cos x \sin y + \sin x \cos y = \sin (x + y) \\and \cos x \cos y - \sin x \sin y = \cos (x + y)

\Rightarrow P(x) \cdot P(y)=\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]
By comparing with equation 1 we can say that:
\\\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]=\mathrm{P}(x+y) \\\therefore P(x) \cdot P(y)=P(x+y)
Similarly, we can show for P(y) \cdot P(x)
\\mathrm{P}(\mathrm{y}) \cdot \mathrm{P}(\mathrm{x})=\left[\begin{array}{cc}\cos \mathrm{y} & \sin \mathrm{y} \\ -\sin \mathrm{y} & \cos \mathrm{y}\end{array}\right]\left[\begin{array}{cc}\cos \mathrm{x} & \sin \mathrm{x} \\ -\sin \mathrm{x} & \cos \mathrm{x}\end{array}\right]

By the rule of matrix multiplication, we have -

\\ P(y) \cdot P(x)=\left[\begin{array}{cc}\cos y \cos x+\sin y(-\sin x) & \cos y \sin x+\sin y \cos x \\ -\sin y \cos x-\sin x \cos y & -\sin y \sin x+\cos x \cos y\end{array}\right] \\\\\Rightarrow P(y) \cdot P(x)=\left[\begin{array}{cc}\cos x \cos y-\sin x \sin y & \sin x \cos y+\cos x \sin y \\ -(\sin x \cos y+\cos x \sin y) & \cos x \cos y-\sin x \sin y\end{array}\right] \\\\\Rightarrow P(y) \cdot P(x)=\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]_{-...(3)}\\ \therefore From equation 2 and 3, we have,\\ P(x) \cdot P(y)=P(y) \cdot P(x)=P(x+y)

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