Proofs of some trigonometric identities

Remember all those trigonometric identities in the front cover of your calculus book that were too hard to memorize and you didn’t have to anyway? Not the simple ones like \(\)\(\sin^2 x + \cos^2 x = 1\) or \(\tan^2 x + 1 = \sec^2 x\). I mean the angle-addition and -subtraction formulas and the like. We proved them pretty easily in my Advanced Calculus class at Western Michigan University, starting with some assumed knowledge of vector calculus. I don’t know what other ways there are to prove all of them, but this way starts with \(\cos (\alpha – \beta) \) and derives all of them from there.

\( \boldsymbol{ 1. \cos(\alpha – \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta } \)

First, imagine two unit vectors with their tails at the origin and their heads on the unit circle. Vector \(\vec{a}\) makes an angle of \(\alpha\) with the horizontal axis and vector \(\vec{b}\) makes an angle of \(\beta\). Thus, the angle between them is \(\alpha – \beta\). And each vector written in component form is \(\vec{a} = \langle \cos \alpha, \sin \alpha \rangle \) and \(\vec{b} = \langle \cos \beta, \sin \beta \rangle \). Recall that their dot product is
$$
\begin{eqnarray}
\langle \cos \alpha, \sin \alpha \rangle \cdot \langle \cos \beta, \sin \beta \rangle &=& \| \vec{a} \| \| \vec{b} \| \cos(\alpha – \beta) \nonumber \\
\cos \alpha \cos \beta + \sin \alpha \sin \beta &=& 1 \cdot 1 \cdot \cos(\alpha – \beta) \nonumber \\
\end{eqnarray}
$$

\( \boldsymbol{ 2. \cos (\alpha + \beta) = \cos \alpha \cos \beta – \sin \alpha \sin \beta\ } \)

The next one is easy because we can just replace \(\beta\) with \(-\beta\):
$$
\begin{eqnarray}
\cos(\alpha + \beta) &=& \cos(\alpha – (-\beta)) \nonumber \\
&=& \cos \alpha \cos (-\beta) + \sin \alpha \sin(-\beta) \nonumber \\
&=& \cos \alpha \cos \beta + \sin \alpha (-\sin \beta) \nonumber \\
&=& \cos \alpha \cos \beta – \sin \alpha \sin \beta \nonumber \\
\end{eqnarray}
$$

That one relies on the fact that \(\cos (-\alpha) = \cos \alpha\) and \(\sin (-\alpha) = -\sin \alpha\), so…I guess you have to know that. The way we “proved” them is to simply draw an angle into the fourth quadrant that was the same magnitude as \(+\alpha\) and observe that the cosine (horizontal distance) was the same and the sine (vertical distance) was the same in magnitude but opposite in sign. I don’t know what other, more rigorous ways there are to prove that \( \cos(-\alpha) = \cos \alpha\) and \(\sin(-\alpha) = -\sin \alpha\), but there are only so many things you can spend time proving in a semester of Analysis.

Before we do the equivalent \(\sin\) identities, it is easiest to do the following two:

\( \boldsymbol{ 3. \cos \left(\frac{\pi}{2} – \alpha \right) = \sin \alpha } \)

$$
\begin{eqnarray}
\cos \left(\frac{\pi}{2} – \alpha \right) &=& \cos \frac{\pi}{2} \cos \alpha + \sin \frac{\pi}{2} \sin \alpha \nonumber \\
&=& 0 \cdot \cos \alpha + 1 \cdot \sin \alpha \nonumber \\
&=& \sin \alpha
\end{eqnarray}
$$

\( \boldsymbol{ 4. \sin \left(\frac{\pi}{2} – \alpha \right) = \cos \alpha} \)

From #3, since the sine of an angle equals the cosine of \(\frac{\pi}{2} \) minus that angle, we can easily transform \(\sin \left(\frac{\pi}{2} – \alpha \right)\):
$$
\begin{eqnarray}
\sin \left(\frac{\pi}{2} – \alpha \right) &=& \cos \left(\frac{\pi}{2} – \left(\frac{\pi}{2} – \alpha \right) \right) \nonumber \\
&=& \cos \left(\frac{\pi}{2} – \frac{\pi}{2} + \alpha \right) \nonumber \\
&=& \cos \alpha
\end{eqnarray}
$$

Now we can do the \(\sin\) angle-addition and angle-subtraction identities:

\( \boldsymbol{ 5. \sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha} \)

$$
\begin{eqnarray}
\sin(\alpha + \beta) &=& \cos \left(\frac{\pi}{2} – (\alpha + \beta) \right) \nonumber \\
&=& \cos \left( \left(\frac{\pi}{2} – \alpha \right) – \beta \right) \nonumber \\
&=& \cos \left(\frac{\pi}{2} – \alpha \right) \cos \beta + \sin \left(\frac{\pi}{2} – \alpha \right) \sin \beta \nonumber \\
&=& \sin \alpha \cos \beta + \cos \alpha \sin \beta \nonumber \\
\end{eqnarray}
$$

\( \boldsymbol{ 6. \sin(\alpha – \beta) = \sin \alpha \cos \beta – \sin \beta \cos \alpha} \)

$$
\begin{eqnarray}
\sin(\alpha – \beta) &=& \cos \left(\frac{\pi}{2} – (\alpha – \beta) \right) \nonumber \\
&=& \cos \left( \left(\frac{\pi}{2} – \alpha \right) + \beta \right) \nonumber \\
&=& \cos \left(\frac{\pi}{2} – \alpha \right) \cos \beta – \sin \left(\frac{\pi}{2} – \alpha \right) \sin \beta \nonumber \\
&=& \sin \alpha \cos \beta – \cos \alpha \sin \beta
\end{eqnarray}
$$

And now we can do the double-angle identities:

\( \boldsymbol{ 7. \cos 2\alpha = \cos^2 \alpha – \sin^2 \alpha} \)
$$
\begin{eqnarray}
\cos 2 \alpha &=& \cos(\alpha + \alpha) \nonumber \\
&=& \cos \alpha \cos \alpha – \sin \alpha \sin \alpha \nonumber \\
&=& \cos^2 \alpha – \sin^2 \alpha
\end{eqnarray}
$$

\( \boldsymbol{ 8. \sin 2\alpha = 2\sin \alpha \cos \alpha} \)
$$
\begin{eqnarray}
\sin 2\alpha &=& \sin(\alpha + \alpha) \nonumber \\
&=& \sin \alpha \cos \alpha + \sin \alpha \cos \alpha \nonumber \\
&=& 2\sin \alpha \cos \alpha
\end{eqnarray}
$$

\( \boldsymbol{ 9. \cos \alpha – \cos \beta = -2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha – \beta}{2}\right)} \)
$$
\begin{split}
\cos \alpha – \cos \beta = & ~ \cos \left(\frac{\alpha + \beta}{2} + \frac{\alpha – \beta}{2}\right) – \cos \left(\frac{\alpha + \beta}{2} – \frac{\alpha – \beta}{2}\right) \nonumber \\
= & ~ \cos\left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right) – \sin \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha – \beta}{2}\right) – \\ & \left[\cos \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right) + \sin \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha – \beta}{2}\right) \right] \nonumber \\
= & ~ -2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha – \beta}{2}\right)
\end{split}
$$

\( \boldsymbol{ 10. \cos \alpha + \cos \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha – \beta}{2}\right)} \)
$$
\begin{split}
\cos \alpha + \cos \beta = & ~ \cos \left(\frac{\alpha + \beta}{2} + \frac{\alpha – \beta}{2}\right) + \cos \left(\frac{\alpha + \beta}{2} – \frac{\alpha – \beta}{2}\right) \nonumber \\
= & ~ \cos\left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right) – \sin \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha – \beta}{2}\right) + \\ & \cos \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right) + \sin \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha – \beta}{2}\right) \nonumber \\
= & ~ 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha – \beta}{2}\right)
\end{split}
$$

\( \boldsymbol{ 11. \sin \alpha – \sin \beta = 2 \sin \left(\frac{\alpha – \beta}{2}\right) \cos \left(\frac{\alpha + \beta}{2}\right)} \)
$$
\begin{split}
\sin \alpha – \sin \beta = & ~ \sin \left(\frac{\alpha + \beta}{2} + \frac{\alpha – \beta}{2}\right) – \sin \left(\frac{\alpha + \beta}{2} – \frac{\alpha – \beta}{2}\right) \\
= & ~ \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right) + \sin \left(\frac{\alpha – \beta}{2} \right) \cos \left(\frac{\alpha + \beta}{2}\right) – \\
& ~ \left[ \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right) – \sin \left(\frac{\alpha – \beta}{2}\right) \cos \left(\frac{\alpha + \beta}{2}\right) \right] \\
= & ~ 2 \sin \left(\frac{\alpha – \beta}{2}\right) \cos \left(\frac{\alpha + \beta}{2}\right)
\end{split}
$$

\( \boldsymbol{ 12. \sin \alpha + \sin \beta = 2 \sin \left(\frac{\alpha – \beta}{2}\right) \cos \left(\frac{\alpha + \beta}{2}\right)} \)
$$
\begin{split}
\sin \alpha + \sin \beta = & ~ \sin \left(\frac{\alpha + \beta}{2} + \frac{\alpha – \beta}{2}\right) + \sin \left(\frac{\alpha + \beta}{2} – \frac{\alpha – \beta}{2}\right) \\
= & ~ \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right) + \sin \left(\frac{\alpha – \beta}{2} \right) \cos \left(\frac{\alpha + \beta}{2}\right) + \\
& ~ \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right) – \sin \left(\frac{\alpha – \beta}{2}\right) \cos \left(\frac{\alpha + \beta}{2}\right) \\
= & ~ 2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha – \beta}{2}\right)
\end{split}
$$

This entry was posted in Math. Bookmark the permalink.

Leave a Reply

Your email address will not be published. Required fields are marked *