Question

HELP PLEASE!
find the exact value of
`tan(-(\pi)/(12) )`

Answer 1

tan ( - pie / 12 ) =

- tan ( pie / 12 ) =

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Remainder formula :

tan^2 (x) = [ 1 - 2Cos(2x) ] ÷ [ 1 + 2Cos(2x) ]

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Thus :

tan^2 ( pie/12) =

[ 1 - Cos( 2pie/12 ) ] ÷ [ 1 + Cos( 2pie/12 ) ] =

[ 1 - Cos( pie/6 ) ] ÷ [ 1 + Cos( pie/6 ) ] =

[ 1 - √3/2 ] ÷ [ 1 + √3/2 ] =

[ 2 - √3 / 2 ] ÷ [ 2 + √3 / 2 ] =

[ 2 - √3 ] ÷ [ 2 + √3 ] =

2 - √3 / 2 + √3 =

(2 - √3)(2 - √3) / (2 + √3)(2 - √3) =

(2 - √3)^2 / 4 - 3 =

(2 - √3)^2 / 1 =

(2 - √3)^2

So :

tan^2 ( pie/12 ) = (2 - √3)^2

Take a square root from both sides

tan( pie/12 ) = 2 - √3

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Thus ;

- tan ( pie/12 ) = - ( 2 - √3 )

- tan ( pie/12 ) = - 2 + √3

Approximately :

- tan ( pie/12 ) = - 2 + 1.732

- tan ( pie/12 ) = - 0.268