# Angle Sum Identities of Trigonometric Functions $\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ $\sin (\alpha + \beta)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$ $\tan (\alpha + \beta) = \frac {\tan \alpha + \tan \beta} {1- \tan \alpha \tan \beta}$ ^64yfg7 ## Proof for Sine and Cosine ![[Sine and Cosine Angle Addition Identities.svg|350]] [Sine and Cosine Angle Sum Identity on desmos](https://www.desmos.com/geometry/3yoqysjwh5) - Start by constructing a triangle $ABC$ with angle $a$ and hypothenuse of length 1 - $|AC| = \cos a$ and $|BC| = \sin a$ - Then construct a second triangle $ACD$ with angle $b$ and hypothenuse of length $AC$ - $|AD| = \cos b \cdot |AC| = \cos b \cos a$ and $|CD| = \sin b \cdot |AC| = \sin b \cos a$ - Now extend $|CD|$ to form a rectangle $ADEF$ - $\angle ACD = 90\degree - b$, and $\angle BCE = 90\degree - \angle ACD = b$ - $|CE| = \cos b \cdot |BC| = \cos b \sin a$ and $|BE| = \sin b \cdot |BC| = \sin b \sin a$ - $\angle ABF = a + b$ because $AD$ and $FE$ are parallel - $|BF| = \cos (a+b) = \cos b \cos a - \sin b \sin a$ - $|AF| = \sin (a+b) = \sin b \cos a + \cos b \sin a$ ## Proof for Tangent ![[Tangent Angle Sum Identity.svg|350]] [Tangent Angle Sum Identity on desmos](https://www.desmos.com/geometry/vnuamgsvzf) - Start by constructing a triangle $ABC$ with angle $a$ and adjacent side 1 - By definition $\tan$ is opposite divided by adjacent. Our adjacent side is 1 so $|BC| = \tan a$ - $|AC| = \sqrt {1+\tan^2 a} = \sqrt {\frac {\cos^2 a + \sin^2 a} {\cos ^2 a}} = \frac 1 {\cos a}$ - $\tan b = \frac {|CD|} {|AC|} = \cos a \cdot |CD|$ and therefore $|CD| = \frac {\tan b} {\cos a}$ - $\angle ACB = 90\degree - a$, and $\angle DCE = 90\degree - \angle ACB = a$ - $|CE| = \cos a \cdot |CD| = \cos a \frac {\tan b} {\cos a} = \tan b$ - $|DE| = \sin a \cdot |CD| = \sin a \frac {\tan b} {\cos a} = \tan a \tan b$ - $|DF| = 1 - |DE| = 1 - \tan a \tan b$ - $\angle ADF = a + b$ because $AB$ and $FE$ are parallel - $\tan(a+b) = \frac {|AF|} {|DF|} = \frac {\tan a + \tan b} {1 - \tan a \tan b}$