HI, how can solved this question? What sort of eqn should i form? Right angle triangle eqn? Thanks
I copied your question and the graph into the following figure, please see the attachment. consider the given matrix system, you are asking to prove that: $\displaystyle x_2=x_1\cos\theta + y_1\sin\theta$ and $\displaystyle y_2 = -x_1\sin\theta + y_1\cos\theta$.
According to your description and the attached figure, we know that $\displaystyle x_1=OT,\;y_1=PT$ and $\displaystyle x_2=PR=OQ,\;y_2=PQ$. So what we need to show is $\displaystyle OQ=(OT)\cos\theta + (PT)\sin\theta$ and $\displaystyle PQ = -(OT)\sin\theta + (PT)\cos\theta$. Next I am going to show these two identities actually hold, all we need is just a few manipulations of the line segments and trig. definitions. Please consider the two right triangles $\displaystyle \triangle OST$ and $\displaystyle \triangle PSQ$.
First, we have
$\displaystyle PQ=(PS)\cos\theta=(PT-ST)\cos\theta=\left[PT-(OS)\sin\theta\right]\cos\theta$
$\displaystyle =\left[PT-\left(\frac{OT}{\cos\theta}\right)\sin\theta\right]\cos\theta= -(OT)\sin\theta + (PT)\cos\theta$
Then, consider
$\displaystyle OQ=OS+SQ=\frac{OT}{\cos\theta}+(PS)\sin\theta=\fra c{OT}{\cos\theta}+(PT-ST)\sin\theta$
$\displaystyle =\frac{OT}{\cos\theta}+(PT-(OT)\tan\theta)\sin\theta=\frac{OT}{\cos\theta}+(P T)\sin\theta-(OT)\frac{\sin^2\theta}{\cos\theta}$
$\displaystyle =(OT)\cos\theta + (PT)\sin\theta$
This completes the proof.
Roy