# Another way to express the golden ratio mathematically

In this post I’m going to be proving that…

$\varphi =\frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

So, here I go…

$x=\frac { 1+\sqrt { 5 } }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 1+2\sqrt { 5 } +5 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 6+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 6+2\sqrt { 5 } }{ 4 } -\frac { 4 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2 }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 }$

Wait for it…

$\Rightarrow \quad { x }^{ 2 }-1=1\cdot x\\ \\ \Rightarrow \quad { x }^{ 2 }-1=x\\ \\ \Rightarrow \quad { x }^{ 2 }=x+1\\ \\ \Rightarrow \quad \frac { { x }^{ 2 } }{ x } =\frac { x }{ x } +\frac { 1 }{ x } \\ \\ \Rightarrow \quad x=1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ x } } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } } \\ \\ \therefore \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

This expression for the golden ratio is quite common, however, before I produced this post – I think it would’ve been very hard to figure out how to derive it from scratch. There aren’t many quirky proofs like this one on the internet – I am quite certain. I hope you liked reading this post! 😀

# tan(A+B)=(tanA+tanB)/(1-tanAtanB)

Prove that:

$tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB }$

$LHS=tan(A+B)=\frac { sin(A+B) }{ cos(A+B) } \\ \\ =\frac { sinAcosB+cosAsinB }{ cosAcosB-sinAsinB } =\frac { \frac { sinA }{ 1 } \cdot \frac { sinB }{ tanB } +\frac { sinA }{ tanA } \cdot \frac { sinB }{ 1 } }{ \frac { sinA }{ tanA } \cdot \frac { sinB }{ tanB } -\frac { sinAsinB }{ 1 } } \\ \\ =\frac { \frac { sinAsinB }{ tanB } +\frac { sinAsinB }{ tanA } }{ \frac { sinAsinB }{ tanAtanB } -\frac { sinAsinB }{ 1 } } =\frac { sinAsinB\left( \frac { 1 }{ tanB } +\frac { 1 }{ tanA } \right) }{ sinAsinB\left( \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } \right) } \\ \\ =\frac { \frac { 1 }{ tanB } +\frac { 1 }{ tanA } }{ \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } } =\frac { \frac { tanA+tanB }{ tanAtanB } }{ \frac { 1-tanAtanB }{ tanAtanB } } =\frac { tanA+tanB }{ tanAtanB } \cdot \frac { tanAtanB }{ 1-tanAtanB } \\ \\ =\frac { tanA+tanB }{ 1-tanAtanB } =RHS$

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