# Lets push trigonometry to limits

Calculus Level 5

$\displaystyle{{ f^{ k }\left( \theta \right) =\sum _{ r=1 }^{ n }{ { \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ { 2 }^{ r } } \right) }^{ k } } }+\cfrac { 1 }{ 3 } \sum _{ r=1 }^{ n }{ \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ 8^{ r } } \right) } }$

Consider a function $$f$$ defined as above. And we again define a function $$L$$ below.

$\displaystyle{L(\theta )=\lim _{ n\rightarrow \infty }{ \left( \sum _{ k=1 }^{ 3 }{ f^{ k }\left( \theta \right) } \right) } }$

If $\displaystyle{{ { L\left( \cfrac { \pi }{ 6 } \right) =\cfrac { a }{ { \pi }^{ 3 } } -\cfrac { b }{ { { \pi } }^{ 2 } } +\cfrac { c }{ { \pi } } -\cfrac { 1 }{ d } +p(1-\sqrt { q } ) }-\sqrt { q } }}$

Find $$a+b+c+d+p+q$$.

Details and assumptions:

• Here $$a,b,c,d,p,q$$ are positive integers .

• $$\text{Hint: Do you know how to convert product into sum?}$$

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