Gelfands Trigonometry $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$

Question

Trying Prove the identity $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ using the figure provided in Gelfands trigonometry.

What I have so far

$\sin(\alpha - \beta) = \frac{CD}{AC} = \frac{PQ}{AC} = \frac{BQ}{AC} - \frac{BP}{AC}$

$\sin(\alpha) = \frac{BQ}{AB} \implies AB\sin(\alpha) = BQ$

$\sin(\alpha - \beta) = \frac{AB\sin(\alpha)}{AC} - \frac{BP}{AC}$

$\frac{AB}{AC} = \frac{1}{\cos(\beta)}$ #corrected

$\sin(\alpha - \beta) = \frac{\sin(\alpha)}{\cos(\beta)} - \frac{BP}{AC}$ # corrected

Im stuck on what to do with $\frac{BP}{AC}$. I've seen the posts here about the derivation of $\sin(\alpha + \beta)$ from the same diagram and I understand that proof perfectly well, but I am stuck on this one.

Answer 1

The given diagram unnecessarily complicates what should be an otherwise simple proof. I shall use this one, which has a right angle at $B$ instead:

The angles are same as in the original diagram, i.e. $\angle BAC=\beta,\angle BAD=\alpha$ and $\angle CAD=\alpha-\beta$. We now take $$\sin(\alpha-\beta)=\frac{CD}{AC}=\frac{BQ-BP}{AC}$$

$$\implies\sin(\alpha-\beta)=\frac{BQ}{AB}\cdot\frac{AB}{AC}-\frac{BP}{BC}\cdot\frac{BC}{AC}$$

The only thing left to notice is that $\angle PBC=\alpha$ and thus:

$$\bbox[5px,border:2px solid #C0A000]{\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta}$$

The formula for $\cos(\alpha-\beta)$ follows in a similar fashion.

Answer 2

The key to this problem is to express all sides in terms of the sines and cosines of $\alpha$ and $\beta$. Doing a bit of angle-chasing, then it can be shown that $\angle PBC=\alpha-\beta$. Therefore, we can express $BP$ as

\begin{align*} BP & =BC\cos(\alpha-\beta)\\ & =AB\sin\beta\cos(\alpha-\beta) \end{align*}

Where the last line was obtained since $BC=AB\sin\beta$. Therefore, our expression becomes

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

Now, we need to find an expression for $\cos(\alpha-\beta)$ in terms of the sines and cosines of $\alpha$ and $\beta$. Referring back to the diagram, then it can be shown that

\begin{align*} \cos(\alpha-\beta) & =\frac {AQ+QD}{AC}\\ & =\frac {AB}{AC}\cos\alpha+\frac {AB}{AC}\sin\beta\sin(\alpha-\beta) \end{align*}

Thus, we get that

\begin{align*} \sin(\alpha-\beta) & =\frac {\sin\alpha}{\cos\beta}-\tan\beta\cos(\alpha-\beta)\\\cos(\alpha-\beta) & =\frac {\cos\alpha}{\cos\beta}+\tan\beta\sin(\alpha-\beta) \end{align*}

Substituting the equation for $\cos(\alpha-\beta)$ into $\sin(\alpha-\beta)$ and using the identity $1+\tan^2\theta=\sec^2\theta$, then you arrive at your result. A similar method can be used to prove the expansion for $\cos(\alpha-\beta)$.

Answer 3

The OP question was based on figure 1 Gelflands figure 1 angle concatenation methods but the top answer is based on the figure 3 method. The OP's question is valid and everywhere the solution for sin(α−β) falls back to figure 3.

So here's answering with the op original diagram i.e figure 1 solution solution to figure 1

Using the original figure 1, you can see how the assumption of a unit radius on the hypotenuse AB contributes to the following equations

$\sin α = \sin(α−β) \cos β + \cos(α−β) \sin β$

and,

$\cos α = \cos(α−β) \cos β - \sin(α−β) \sin β$

substituting,

$\cos(α−β) = (1/\cos β )(\cos α + \sin(α−β)\ sin β )$

in the previous equation gives

$\sin(α−β) \cos β = \sin α - \sin β((1/\cos β )(\cos α + \sin(α−β) \sin β ))$

Simplifying, we get

$$ \sin(α−β) (\cos β + \sin2β/\cos β) = \sin α - \sin β \cos α/\cos β \\ \implies \sin(α−β) = \sin α \cos β - \cos α \sin β$$

Answer 4

Another perspective on this question: how we can also get the sine and cosine addition formulae from this construction.

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Gelfand's method is very clever: splitting a ratio of unrelated lengths into multiplying two ratios, each of which is a trigonometric ratio. However, I do not find this very natural to attempt at first glance.

More natural would be to use the line segments $BQ$ and $AD$, and their parallel segments $CD$ and $PC$. Using this we can observe that:

$$AQ + QD = AD \implies AQ + PC = AD$$ $$BP + PQ = BQ \implies BP + CD = BQ$$

Now as $AD$ is parallel to $PC$, $\angle PCA = \alpha - \beta \implies \angle PBC = \alpha - \beta$, so $\Delta ACD \sim \angle \Delta BCP$. Thus:

$$AB \cos \alpha + BC \sin(\alpha - \beta) = AC \cos(\alpha - \beta) \tag{1}$$ $$BC \cos(\alpha - \beta) + AC \sin(\alpha - \beta) = AB \sin \alpha \tag{2}$$

and as $BC = AB \sin \beta, AC = AB \cos \beta$:

$$\cos \alpha + \sin \beta \sin(\alpha - \beta) = \cos \beta \cos(\alpha - \beta) \tag{3}$$ $$\sin \beta \cos(\alpha - \beta) + \cos \beta \sin(\alpha - \beta) = \sin \alpha \tag{4}$$

These formulae look awfully close to what we want to prove. So indeed, relabelling $\alpha - \beta = \alpha', \beta = \beta'$, we have $\alpha = \alpha' + \beta'$ and making $\cos \alpha$ the subject in equation $(3)$:

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

which yield the cosine and sine addition formulae.

Answer 5

Try using the whole angle $\alpha$ as in the proof for the sin and cos addition formulas:

$$\sin \alpha = \frac{BQ}{AB} = \frac{BP}{AB} + \frac{CD}{AB}$$ $$=\frac{BP}{BC} \frac{BC}{AB} + \frac{CD}{AC} \frac{AC}{AB}$$ $$=\cos(\alpha - \beta) \sin \beta + \sin(\alpha - \beta) \cos \beta$$

and also:

$$\cos \alpha = \frac{AQ}{AB} = \frac{AD}{AB} - \frac{PC}{AB}$$ $$= \frac{AD}{AC} \frac{AC}{AB} - \frac{PC}{BC} \frac{BC}{AB}$$ $$= \cos(\alpha - \beta) \cos \beta - \sin(\alpha - \beta) \sin \beta$$

Now let $\cos(\alpha - \beta) = x, \sin(\alpha - \beta) = y$, and we have some simultaneous equations:

$$x \cos \beta - y \sin \beta = \cos \alpha$$ $$x \sin \beta + y \cos \beta = \sin \alpha$$

This means that:

$\sin \beta(x \cos \beta - y \sin \beta) = \cos \alpha \sin \beta \tag{1}$

$\cos \beta(x \sin \beta + y \cos \beta) = \sin \alpha \cos \beta \tag{2}.$

Thus $(1) - (2)$ gives:

$$-y \sin^2 \beta - y \cos^2 \beta = \cos \alpha \sin \beta - \sin \alpha \cos \beta, -y = \cos \alpha \sin \beta - \sin \alpha \cos \beta.$$

Remembering what $y$ is, we have the identity for $\sin(\alpha - \beta)$.

Solving similarly for $\cos(\alpha - \beta)$ gives the desired identity.