I derived the expansion of $\cos (\alpha + \beta)$ as follows, please take a look :

Figure - 1

Here, $\angle AOB = \alpha$, $\angle BOC = \beta$, $\angle AOC = (\alpha + \beta)$, $a = \cos \alpha$, $b = \sin \alpha$, $m = \cos \beta$, $n = \sin \beta$, $x = \cos (\alpha + \beta)$ and $y = \sin (\alpha + \beta)$.
Something to be noticed : $a^2 + b^2 = m^2 + n^2 = x^2 + y^2 = 1$ as they all are sines and cosines of $\alpha$, $\beta$ and $(\alpha + \beta)$ respectively. These equations will be used further in the derivation.

Now, from the above diagram, we can find the length of chord $BC$ using the distance formula which states that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ on the Cartesian Plane is : $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. So, we get : $$BC = \sqrt{(x-a)^2+(y-b)^2} = \sqrt{x^2+a^2-2ax+y^2+b^2-2by} $$$$ = \sqrt{(a^2+b^2)+(x^2+y^2)-2ax-2by} = \sqrt{2-2ax-2by}$$ Now, what I do is rotate the axes by an angle of $\alpha$ anticlockwise to obtain this :

Figure - 2

[The dotted lines represent the old axes whereas the red lines denote the new axes. The new X-Axis coincides with the line segment $OB$].
Now, the length of chord $BC$ will remain the same irrespective of the position of the coordinate axes.
And, the point that was previously $B(a,b)$ is now $(0,1)$ and the point that was previously $C(x,y)$ is now $(m,n)$ because $\beta$ becomes an angle in standard position, so the point where the terminal side of $\beta$ will intersect with the unit circle will be $(\cos \beta, \sin \beta)$, which as I mentioned earlier, can be written as $(m,n)$ because $m = cos \beta$ and $n = \sin \beta$.
So, in this "new axes plane", the length of what previously used to be $BC$ will now be : $$\sqrt{(m-1)^2+(n-0)^2} = \sqrt{m^2+1-2m+n^2} = \sqrt{(m^2+n^2)+1-2m} = \sqrt{2-2m}$$

Now, from the two values of the length of $BC$ that we have obtained by aligning the coordinate axes at two different positions are : $\sqrt{2-2ax-2by}$ and $\sqrt{2-2m}$. We can equate these two values to arrive at the value of $x$, which will give us the value of $\cos (\alpha + \beta)$. $$\sqrt{2-2ax-2by} = \sqrt{2-2m}$$. By squaring both sides, we obtain : $$2-2ax-2by = 2-2m$$ Now, we can cancel the term $2$ on both the sides to obtain this : $$-2ax-2by = -2m$$ Dividing both sides by $(-2)$, we will obtain : $$ax+by=m$$ $$ax+b\sqrt{1-x^2}=m$$ $$b\sqrt{1-x^2} = m-ax$$ By squaring both the sides, we will obtain : $$b^2(1-x^2) = (m-ax)^2 = m^2+a^2x^2-2max$$ $$\therefore b^2-b^2x^2 = m^2+a^2x^2-2max$$ Now, we can convert this equation to the standard quadratic equation form $(Ax^2+Bx+C)$ and obtain this : $$(a^2+b^2)x^2+(-2ma)x+(m^2-b^2)=0$$ Here, $A = a^2+b^2$ which is equal to $1$, $B = -2ma$ and $C = m^2-b^2$
Let $D$ denote the discriminant. $$\therefore D = B^2-4AC = B^2-4C = 4m^2a^2 - 4(m^2-b^2) = 4m^2a^2 - 4m^2 + 4b^2$$ $$\therefore D = 4(m^2a^2-m^2+b^2) = 4(m^2a^2-m^2+1-a^2) $$$$ = 4(a^2(m^2-1)-1(m^2-a^2)) = 4(m^2-1)(a^2-1)$$ $$\therefore D = 4(-n^2)(-b^2) = 4n^2b^2 $$$$ \implies \sqrt{D} = 2nb$$. Now, by applying the quadratic formula, $$x = \dfrac{-B \pm \sqrt{D}}{2A} = \dfrac{2ma \pm 2nb}{2} = ma \pm nb$$ From this, we get our final result, $$\cos(\alpha + \beta) = \cos \alpha \cos \beta \pm \sin \alpha \sin \beta$$

Now, the only problem seems to be the $\pm$ symbol. How do I get rid of that and determine which symbol to actually put in its place?

Also, please let me know about your thoughts on the derivation.

  • 2
    $\begingroup$ Set $\beta=-\alpha$ to obtain: $1=\cos^2\alpha\mp\sin^2\alpha$. What sign do you choose? :) $\endgroup$ – user Jun 3 '20 at 19:00
  • $\begingroup$ Positive!! That was a good example. But, I do have a question. Is there a way to rule out the negative sign without examples? I just find drawing conclusions from examples a little ambiguous (this particular example is good compared to others, though). $\endgroup$ – Rajdeep Sindhu Jun 3 '20 at 19:03
  • $\begingroup$ Wrong! Negative (in your formula it is opposite). Here example is a good friend, as you should choose one possibility from two. It is similar to choosing positive square root if one computes a distance. $\endgroup$ – user Jun 3 '20 at 19:04
  • $\begingroup$ I don't get how it'd be negative, can you please elaborate? Thanks for the second part of your comment. $\endgroup$ – Rajdeep Sindhu Jun 3 '20 at 19:07
  • 1
    $\begingroup$ I changed it because $\sin(-\alpha)=-\sin(\alpha)$ (see above). $\endgroup$ – user Jun 3 '20 at 19:12

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