# Thread: ST and CT by vectors

1. ## ST and CT by vectors

Using vectors prove that

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

2. Hello ns1954
Originally Posted by ns1954
Using vectors prove that

a) $\displaystyle \sin(\alpha-\beta)= \sin \alpha\cos \beta-\cos \alpha\sin \beta$
b) $\displaystyle \cos (\alpha-\beta)=\cos \alpha\cos \beta + \sin \alpha\sin \beta$
First, note that the dot product of vectors $\displaystyle x_1\,\textbf i+y_1\,\textbf j$ and $\displaystyle x_2\,\textbf i+y_2\,\textbf j$ is:
$\displaystyle (x_1\,\textbf i+y_1\,\textbf j).(x_2\,\textbf i+y_2\,\textbf j)= x_1x_2 +y_1y_2$
$\displaystyle =|x_1\,\textbf i+y_1\,\textbf j|\,|x_2\,\textbf i+y_2\,\textbf j|\cos \theta$, where $\displaystyle \theta$ 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 $\displaystyle A (\cos \alpha, \sin\alpha), B (\cos\beta, \sin\beta)$. The radii $\displaystyle OA, OB$ make angles $\displaystyle \alpha, \beta$ with the positive $\displaystyle x$-axis, measured anticlockwise. So $\displaystyle \angle BOA = \alpha - \beta$, measured anticlockwise from $\displaystyle OB$ to $\displaystyle OA$.

Using vector notation, we have:
$\displaystyle \textbf{OA} = \cos\alpha\,\textbf i+\sin\alpha\,\textbf j$

$\displaystyle \textbf{OB} = \cos\beta\,\textbf i+\sin\beta\,\textbf j$
Then, using the above result:
$\displaystyle \textbf{OA}.\textbf{OB} = \cos\alpha\cos\beta+\sin\alpha\sin\beta$
$\displaystyle =|OA||OB|\cos(\alpha - \beta)$

$\displaystyle =\cos(\alpha - \beta)$, since $\displaystyle |OA| = |OB| = 1$.
$\displaystyle \Rightarrow \cos(\alpha-\beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta$
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Then the proof of the first result. Note that:
$\displaystyle \sin(\alpha-\beta)=\cos\Big(\pi/2-(\alpha -\beta)\Big)$, using the result just proved
$\displaystyle =\cos(\pi/2+\beta - \alpha)$
Then, with the point $\displaystyle A$ defined as above, consider the point $\displaystyle B, (-\sin\beta, \cos\beta)$ - see the attached diagram. Then the radius $\displaystyle OB$ makes an angle $\displaystyle (\pi/2+\beta)$ with the $\displaystyle x$-axis. Therefore $\displaystyle \angle AOB = \pi/2+\beta - \alpha$.

Then, as before:
$\displaystyle \textbf{OA}.\textbf{OB} = \cos\alpha(-\sin\beta)+\sin\alpha\cos\beta$
$\displaystyle =|OA|.|OB|\cos(\pi/2+\beta - \alpha)$

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