Using vectors, prove the identity cos(α-β)=cosαcosβ+sinαsinβ
Using vectors, prove the identity $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$.
Explanation of Method
Vector proof of the cosine subtraction formula.
Step 1: Two unit vectors
Take the points $P$ and $Q$ on the unit circle at the angles $\alpha$ and $\beta$ from the positive $x$-axis: \[\vec p=\overrightarrow{OP}=(\cos\alpha,\sin\alpha),\qquad \vec q=\overrightarrow{OQ}=(\cos\beta,\sin\beta),\qquad |\vec p|=|\vec q|=1.\]
- $\vec p=(\cos\alpha,\sin\alpha)$
- $\vec q=(\cos\beta,\sin\beta)$
Step 2: Scalar product — definition
Rotating $\vec q$ by $\alpha-\beta$ gives $\vec p$. The angle $\varphi\in[0^\circ,180^\circ]$ between the vectors therefore satisfies $\cos\varphi=\cos(\alpha-\beta)$, because cosine is even and $360^\circ$-periodic. Hence
\[ \vec p\cdot\vec q=|\vec p||\vec q|\cos\varphi=\cos(\alpha-\beta). \qquad (\mathrm I) \]
Step 3: Scalar product — coordinates
\[ \vec p\cdot\vec q=\cos\alpha\cos\beta+\sin\alpha\sin\beta. \qquad (\mathrm{II}) \]
Step 4: Conclusion
Both expressions equal $\vec p\cdot\vec q$, and the argument works for all real $\alpha,\beta$: $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$.
Step 5: Consequences
| Substitution | Identity |
|---|---|
| $\beta\to-\beta$ | $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ |
| $\beta=\alpha$ | $1=\cos^2\alpha+\sin^2\alpha$ |
| $\sin x=\cos(90^\circ-x)$ | $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$ |
| then $\beta\to-\beta$ | $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$ |
For the sine formula: \[\sin(\alpha-\beta)=\cos\big((90^\circ-\alpha)+\beta\big)=\cos(90^\circ-\alpha)\cos\beta-\sin(90^\circ-\alpha)\sin\beta=\sin\alpha\cos\beta-\cos\alpha\sin\beta.\]
Result
\[\boxed{\,\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\,}\]
Remark
Note. The geometric angle between $\vec p$ and $\vec q$ is $|\alpha-\beta|$ reduced to $[0^\circ,180^\circ]$, not always $\alpha-\beta$ itself. It has the same cosine, so the proof holds for all angles.