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Trigonometric identities
Eugene Kritchevski, Vanier College, last updated on March 6, 2016
Proof of basic sum and difference formulas
In this note, we look at the underlying geometrical principle behind the trigonometric identites such as
$$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$
$$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$
$$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$
$$\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(b)$$
We will see that these formulas follow from a simple geometrical principle. If a segment $[AB]$ in the plane is rotated, then the resulting segment $[A'B']$ has the same length.
Given $\alpha$ and $\beta$, consider the correspoinding points $A(\cos(\alpha),\sin(\alpha))$ and $B(\cos(\beta),\sin(\beta))$ on the unit circle. Let us work out a useful formula for the square of the distance from $A$ to $B$:
$$
\begin{equation*}
\begin{split}
|AB|^2&=\of{\cos(\alpha)-\cos(\beta)}^2+\of{\sin(\alpha)-\sin(\beta)}^2\\
&=\cos^2(\alpha)-2\cos(\alpha)\cos(\beta) +\cos^2(\beta)+\sin^2(\alpha)-2\sin(\alpha)\sin(\beta)+\sin^2(\beta)\\
&=\cos^2(\alpha) + \sin^2(\alpha)+ \cos^2(\beta) + \sin^2(\beta) -2\cos(\alpha)\cos(\beta)-2\sin(\alpha)\sin(\beta) \\
&=2 -2\cos(\alpha)\cos(\beta)-2\sin(\alpha)\sin(\beta) \\
&=2 -2[\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)] \\
\end{split}
\end{equation*}
$$
Given $\theta$, consider the points $A'(\cos(\alpha+\theta),\sin(\alpha+\theta))$ and $B'(\cos(\beta+\theta),\sin(\beta+\theta))$ on the unit circle. Then $A'$ and $B'$ result from rotating $A$ and $B$ by an angle $\theta$ around the origin. If $\theta>0$, then the rotation is counterclockwise. If $\theta < 0$, the rotaion is clockwise. If $\theta=0$, then $A=A'$ and $B=B'$.
The segment $[A'B']$ results from rotating the segment $[AB]$ an so
$$|AB|=|A'B'|.$$
Now
$$|AB|^2=2-2[\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)]$$
and
$$|A'B'|^2=2-2[\cos(\alpha+\theta)\cos(\beta+\theta)+\sin(\alpha+\theta)\sin(\beta+\theta)]$$
Since $|AB|^2=|A'B'|^2$, we conclude that
$$(*)\qquad \cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)=\cos(\alpha+\theta)\cos(\beta+\theta)+\sin(\alpha+\theta)\sin(\beta+\theta)\qquad \textrm{ for any choice of }\alpha,\beta,\theta$$
Many trigonometric identities follow from $(*)$ obtained by taking special combinations of $\alpha,\beta$ and $\theta$ in $(*)$.
With $\theta=-\beta$, we get
$$\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)=\cos(\alpha-\beta)\underbrace{\cos(\beta-\beta)}_{1}+\sin(\alpha-\beta)\underbrace{\sin(\beta-\beta)}_{0}=\cos(\alpha-\beta)$$
and so
$$(1)\qquad\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)\qquad(\textrm{ for any } \alpha,\beta)$$
Setting $\alpha=0$ in $(1)$, we get
$$(2)\qquad\cos(-\beta)=\cos(\beta) \qquad(\textrm{ for any } \beta)$$
Taking $\alpha=\pi/2$ in $(1)$ gives
$$(3)\qquad\cos(\pi/2-\beta)=\sin(\beta)\qquad(\textrm{ for any } \beta)$$
and replacing $\beta$ with $\pi/2-\beta$, we get
$$(3')\qquad \cos(\beta)=\sin(\pi/2-\beta)\qquad(\textrm{ for any } \beta)$$
Taking $\alpha=-\pi/2$ in $(1)$ gives $\cos(-\pi/2-\beta)=-\sin(\beta)$.
Also using $(2)$ and $(3)$, we have $\sin(-\beta)=\cos(\pi/2+\beta)=\cos(-\pi/2-\beta)$. It follows that
$$(4)\qquad\sin(-\beta)=-\sin(\beta) \qquad(\textrm{ for any } \beta)$$
Using $(1)$, $(2)$ and $(4)$ gives
$$(5)\qquad\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\qquad(\textrm{ for any } \alpha,\beta)$$
The identities for $\sin(\alpha\pm\beta)$ can be found by combining the previous identities:
$$\sin(\alpha+\beta)=\cos((\alpha+\beta)-\pi/2)=\cos(\alpha+(\beta)-\pi/2))=\cos(\alpha)\cos(\beta-\pi/2)-\sin(\alpha)\sin(\beta-\pi/2)$$ $$=\cos(\alpha)\sin(\beta)+\sin(\alpha)\sin(\pi/2-\beta)=\cos(\alpha)\sin(\beta)+\sin(\alpha)\cos(\beta)$$
so
$$(6)\qquad\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)\qquad(\textrm{ for any } \alpha,\beta)$$
Now replacing $\beta$ with $-\beta$ gives
$$(7)\qquad\sin(\alpha-\beta)=\sin(\alpha)\cos(\beta)-\sin(\beta)\cos(\alpha)\qquad(\textrm{ for any } \alpha,\beta)$$