# Is this derivation I did for the expansion of $\cos (\alpha+\beta)$ correct and how can I get rid of the $\pm$ symbol in it?

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

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 :

[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?

• Set $\beta=-\alpha$ to obtain: $1=\cos^2\alpha\mp\sin^2\alpha$. What sign do you choose? :)
• I changed it because $\sin(-\alpha)=-\sin(\alpha)$ (see above).