Angle addition identities

The trigonometric angle addition identities state the following identities:

$\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)$

$\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$

$\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}$

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Proofs

Proof 1

$[asy] unitsize(216); real d = 1/cos(radians(35)); real d1 = d * cos(radians(55)); real d2 = d * sin(radians(55)); pair O = (0,0); pair A = (cos(radians(20)),0); pair B = (cos(radians(20)),sin(radians(20))); pair C = (cos(radians(20)),d2); pair D = (d1,d2); draw(O--A--B--O--D--B--O--D--C--B); dot(O); dot(B); dot(A,red); dot(C,green); dot(D,blue); label("O",O,SW); label("$\alpha$",shift(dir(10)/5)*O); label("$\beta$",shift(dir(37.5)/5)*O); label("A",A,SE,red); label("B",B,E); label("C",C,NE,green); label("D",D,dir(122.5),blue); label("$\cos \alpha$",O--A,S); label("$\sin \alpha$",A--B,E); label("1",O--B,dir(302.5)); label("$\frac{\cos \alpha \sin \beta}{\cos \beta}$",B--C,E); label("$\frac{\sin \alpha \sin \beta}{\cos \beta}$",C--D,N); label("$\frac{\sin \beta}{\cos \beta}$",B--D,dir(200)); label("$\frac{1}{\cos \beta}$",D--O,dir(325)); [/asy]$

$\fontsize{18}{27}\selectfont \sin (\alpha + \beta ) = \frac{\left( \sin \alpha + \frac{\cos \alpha \sin \beta}{\cos \beta} \right)}{\frac{1}{\cos \beta}} = \cos \beta \times \left( \sin \alpha + \frac{\cos \alpha \sin \beta}{\cos \beta} \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\fontsize{18}{27}\selectfont \cos (\alpha + \beta ) = \frac{\left( \cos \alpha - \frac{\sin \alpha \sin \beta}{\cos \beta} \right)}{\frac{1}{\cos \beta}} = \cos \beta \times \left( \cos \alpha - \frac{\sin \alpha \sin \beta}{\cos \beta} \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\fontsize{18}{27}\selectfont \tan (\alpha + \beta ) = \frac{\sin (\alpha + \beta )}{\cos (\alpha + \beta )} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} = \frac{\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta}}{1 - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

Proof 2

Let point $A(\cos\alpha, \sin\alpha)$ and point $B(\cos\beta, \sin\beta)$ be two points on the unit circle such that $\alpha > \beta$ .

By the law of cosine, we know that:

$\cos (\alpha - \beta) = \cos \angle AOB = \frac{1^2+1^2-\vert AB \vert^2}{2 \cdot 1 \cdot 1}$

Apply the distance formula to obtain the length of $AB$ :

$\begin{aligned} | A B | & = \sqrt{(\cos α - \cos β)^{2} + (\sin α - \sin β)^{2}} \\ = \sqrt{(\cos^{2} α + \sin^{2} α) + (\cos^{2} β + \sin^{2} β) - 2 \cos α \cos β - 2 \sin α \sin β} \\ = \sqrt{2 - 2 \cos α \cos β - 2 \sin α \sin β} \end{aligned}$

Substituting and rearranging to get:

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

See that the identity holds true (and makes sense geometrically) when $\alpha < \beta$ due to the fact that $\cos{(\beta - \alpha)} = \cos{(\alpha - \beta)}$ .