ST and CT by vectors

**ns1954** · February 3rd 2010, 05:46 AM

Using vectors prove that

a) $\sin(\alpha-\beta)= \sin \alpha\cos \beta-\cos \alpha\sin \beta$
b) $\cos (\alpha-\beta)=\cos \alpha\cos \beta + \sin \alpha\sin \beta$

**Grandad** · February 3rd 2010, 07:03 AM

Hello ns1954

Originally Posted by ns1954

Using vectors prove that

a) $\sin(\alpha-\beta)= \sin \alpha\cos \beta-\cos \alpha\sin \beta$
b) $\cos (\alpha-\beta)=\cos \alpha\cos \beta + \sin \alpha\sin \beta$

First, note that the dot product of vectors $x_1\,\textbf i+y_1\,\textbf j$ and $x_2\,\textbf i+y_2\,\textbf j$ is:

$(x_1\,\textbf i+y_1\,\textbf j).(x_2\,\textbf i+y_2\,\textbf j)= x_1x_2 +y_1y_2$

$=|x_1\,\textbf i+y_1\,\textbf j|\,|x_2\,\textbf i+y_2\,\textbf j|\cos \theta$ , where $\theta$ is the angle between the vectors

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Next, we'll prove the second result first, because it's more straightforward.

Consider the points on the unit circle $A (\cos \alpha, \sin\alpha), B (\cos\beta, \sin\beta)$ . The radii $OA, OB$ make angles $\alpha, \beta$ with the positive $x$ -axis, measured anticlockwise. So $\angle BOA = \alpha - \beta$ , measured anticlockwise from $OB$ to $OA$ .

Using vector notation, we have:

$\textbf{OA} = \cos\alpha\,\textbf i+\sin\alpha\,\textbf j$

$\textbf{OB} = \cos\beta\,\textbf i+\sin\beta\,\textbf j$

Then, using the above result:

$\textbf{OA}.\textbf{OB} = \cos\alpha\cos\beta+\sin\alpha\sin\beta$

$=|OA||OB|\cos(\alpha - \beta)$

$=\cos(\alpha - \beta)$ , since $|OA| = |OB| = 1$ .

$\Rightarrow \cos(\alpha-\beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta$

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Then the proof of the first result. Note that:

$\sin(\alpha-\beta)=\cos\Big(\pi/2-(\alpha -\beta)\Big)$ , using the result just proved

$=\cos(\pi/2+\beta - \alpha)$

Then, with the point $A$ defined as above, consider the point $B, (-\sin\beta, \cos\beta)$ - see the attached diagram. Then the radius $OB$ makes an angle $(\pi/2+\beta)$ with the $x$ -axis. Therefore $\angle AOB = \pi/2+\beta - \alpha$ .

Then, as before: