Ever wondered what is the relation of these angles?

This formula derivation was inspired by a math exercise I did in class (Which is weird cause the exercises in Math are almost all elementary).

You need to know basic trigonometry for this one.


The formula does:

Input:θi \boxed{\text{Input:} \theta_{i}}

Output:θe \boxed{\text{Output:} \theta_{e}}

This photo initializes the variables used for the following proof:

Note that θi\theta_{i} can obtain such values that [0<θi<360] [0 < \theta_{i} < 360 ] . 360360 is just a full circle and beyond it is overlap.


To find θe\theta_{e} we need to relate it between the hh and OAOA:

cos(θe)=hOA \cos (\theta_{e}) = \frac{h}{OA}

By the Pythagorean theorem\text{Pythagorean theorem}:

h2+r2=OA2h^{2} + r^{2} = OA^{2}

h=OA2r2h= \sqrt{OA^{2} - r^{2}}

Replace hh in the cos(θe)=hOA \cos (\theta_{e}) = \frac{h}{OA} we get:

cos(θe)=OA2r2OA \cos (\theta_{e}) = \frac{\sqrt{OA^{2} - r^{2}}}{OA}

We can find rr by making a relation with the surface of the 2 objects which does not change during transformation from fig.1\text{fig.1} to fig.2\text{fig.2}; call this SsideS_{side}. SsideS_{side} can be found in 2 ways:

  • Sside=π×(OA)2×θi360S_{side} = \pi \times (OA)^{2} \times \frac{ \theta_{i} }{360} , by fig.1\text{fig.1}

  • Sside=π×(OA)×rS_{side} = \pi \times (OA) \times r , by fig.2\text{fig.2}

Let's equal these 2 expressions:

π×(OA)×r=π×(OA)2×θi360\pi \times (OA) \times r = \pi \times (OA)^{2} \times \frac{ \theta_{i} }{360} We can simplify this to:

r=(OA)×θi360 r = (OA) \times \frac{ \theta_{i} }{360}

Replace rr in the cos(θe)=OA2r2OA \cos (\theta_{e}) = \frac{\sqrt{OA^{2} - r^{2}}}{OA} we get:

cos(θe)=OA2((OA)×θi360)2OA \cos (\theta_{e}) = \frac{\sqrt{OA^{2} - ((OA) \times \frac{ \theta_{i} }{360})^{2}}}{OA}

Let's simplify this expression. Using the property a2b2=(a+b)(ab) a^{2} - b^{2} = (a+b)(a-b) in OA2((OA)×θi360)2\sqrt{OA^{2} - ((OA) \times \frac{ \theta_{i} }{360})^{2}} :

cos(θe)=([OA]+[OA]×θi360)×([OA][OA]×θi360)OA\cos (\theta _{ e })=\frac { \sqrt { ([OA]+[OA]\times \frac { \theta _{ i } }{ 360 } )\times ([OA]-[OA]\times \frac { \theta _{ i } }{ 360 } ) } }{ OA }

I did it in square brackets "[OA][OA]" to not confuse too much the reader.

Factorise ([OA]+[OA]×θi360)([OA]+[OA]\times \frac { \theta _{ i } }{ 360 } ) and in ([OA][OA]×θi360)([OA]-[OA]\times \frac { \theta _{ i } }{ 360 } ) a [OA][OA]:

cos(θe)=[OA]2(1+θi360)(1θi360)OA\cos (\theta _{ e })=\frac { \sqrt { { [OA]^{ 2 }(1+\frac { \theta _{ i } }{ 360 } )(1-\frac { \theta _{ i } }{ 360 } ) } } }{ OA }

The [OA][OA] -s cancel out, so we are left with the final formula (Yes, θe\theta _{ e } is independent from [OA][OA]) :

cos(θe)=(1+θi360)(1θi360) \boxed {\cos (\theta _{ e })=\sqrt { { (1+\frac { \theta _{ i } }{ 360 } )(1-\frac { \theta _{ i } }{ 360 } ) } } }

Quite an elegance!

Note by Nikolas Кraj
2 weeks, 3 days ago

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Please report here for any mistake i have done (in reply here). Thanks

Nikolas Кraj - 2 weeks, 2 days ago

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