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Verify that $$ \left[\begin{matrix}\cos32 & -\sin32\\ \sin32 & \cos32 \end{matrix}\right]\cdot\left[\begin{matrix}\cos40 & -\sin40\\ \sin40 & \cos40 \end{matrix}\right]=\left[\begin{matrix}\cos72 & -\sin72\\ \sin72 & \cos72 \end{matrix}\right] $$ and explain why this result could have been expected.

I have verified that the product matrix is correct using the matrix operation techniques, but I am not sure as to why the result is expected.

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The matrix

$$R\left(\theta\right)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$

when acted on a vector, rotates the vector by an angle $\theta$. Thus it is expected that a rotation by $\theta_{1}$, followed by a rotation by $\theta_{2}$, will be equivalent to a rotation by $\theta_{1}+\theta_{2}$. In matrix notation this is translated into

$$R\left(\theta_{1}\right)R\left(\theta_{2}\right)=R\left(\theta_{1}+\theta_{2}\right)$$

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  • $\begingroup$ I don't quite understand what is meant by $\theta_1$ and $\theta_2$? $\endgroup$ – geo_freak Dec 17 '17 at 19:05
  • $\begingroup$ @geo_freak Just two angles. $\endgroup$ – eranreches Dec 17 '17 at 19:07
  • $\begingroup$ [+1] Thorough explanation. In your last sentence, you need a translation to explain a rotation ... :) $\endgroup$ – Jean Marie Dec 17 '17 at 21:08
  • $\begingroup$ Maybe the OP could have a look at (math.stackexchange.com/questions/363652/…) $\endgroup$ – Jean Marie Dec 17 '17 at 21:11
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In addition to the comment above, I think it deserves a little explaination why: $R\left(\theta_{1}\right)R\left(\theta_{2}\right)=R\left(\theta_{1}+\theta_{2}\right)$

By using matrix multiplication we get:

$$R\left(\theta_{1}\right)R\left(\theta_{2}\right)=\begin{pmatrix}\cos\theta_{1}&-\sin\theta_{1}\\\sin\theta_{1}&\cos\theta_{1}\end{pmatrix}\cdot\begin{pmatrix}\cos\theta_{2}&-\sin\theta_{2}\\\sin\theta_{2}&\cos\theta_{2}\end{pmatrix} = \begin{pmatrix}\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2} & -\left(\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}\right)\\ \sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2} & -\sin\theta_{1}\sin\theta_{2}+\cos\theta_{1}\cos\theta_{2} \end{pmatrix}$$

We will use the trigonometric identities:

$$\left(1\right) \cos\left(\theta_{1}+\theta_{2}\right)=\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2}\\ \left(2\right) \sin\left(\theta_{1}+\theta_{2}\right)=\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}$$ And we will recieve: $$R\left(\theta_{1}\right)R\left(\theta_{2}\right)=\begin{pmatrix}\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2} & -\left(\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}\right)\\ \sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2} & -\sin\theta_{1}\sin\theta_{2}+\cos\theta_{1}\cos\theta_{2} \end{pmatrix}=\begin{pmatrix}\cos\left(\theta_{1}+\theta_{2}\right) & \sin\left(\theta_{1}+\theta_{2}\right)\\ \sin\left(\theta_{1}+\theta_{2}\right) & \cos\left(\theta_{1}+\theta_{2}\right) \end{pmatrix} = R\left(\theta_{1}+\theta_{2}\right)$$ As said in the comment above.

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  • $\begingroup$ I think this has been done by the OP. What is lacking him/her (see the last sentence of the question) is to know in which way the result could have been anticipated. There, a geometrical understanding is necessary (see the answer of @erenreches). $\endgroup$ – Jean Marie Dec 17 '17 at 21:15
  • $\begingroup$ It might be obvious for him, but for future reference, other people might come and look for the same answer and be puzzled why $R\left(\theta_{1}\right)+R\left(\theta_{2}\right)=R\left(\theta_{1}+\theta_{2}\right)$. So it won't hurt to show the algebraic reasoning behind the rotation. $\endgroup$ – theshopen Dec 17 '17 at 21:23
  • $\begingroup$ I accept your argument. Nevertheless, his question was elsewhere. $\endgroup$ – Jean Marie Dec 17 '17 at 21:26

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