Asked by Ryu
How do you find the exact value of arctan(2-√3)
Answers
Answered by
mathhelper
so we need an angle θ , so that
tan θ = 2 - √3
( we could be sneaky and use our calculator, sure enough it says 15°)
so we will make an "intelligent guess", then prove my guess is correct
we know tan 30° = 1/√3
and I also know that tan 30° = 2 tan15°/(1 - tan^2 15°)
let tan15° = x
so tan30° = 2 tan15/(1 - tan^2 15°) becomes
1/√3 = 2x/(1-x^2)
2√3 x = 1 - x^2
x^2 + 2√3 x - 1 = 0
x = ( -2√3 ± √(12+4) )/2
= -√3 ± 2 <----- well, who would have guessed!
but we know x = tan 15
so 2 - √3 = tan 15° , and
arctan(2-√3) = 15° or π/12 radians
tan θ = 2 - √3
( we could be sneaky and use our calculator, sure enough it says 15°)
so we will make an "intelligent guess", then prove my guess is correct
we know tan 30° = 1/√3
and I also know that tan 30° = 2 tan15°/(1 - tan^2 15°)
let tan15° = x
so tan30° = 2 tan15/(1 - tan^2 15°) becomes
1/√3 = 2x/(1-x^2)
2√3 x = 1 - x^2
x^2 + 2√3 x - 1 = 0
x = ( -2√3 ± √(12+4) )/2
= -√3 ± 2 <----- well, who would have guessed!
but we know x = tan 15
so 2 - √3 = tan 15° , and
arctan(2-√3) = 15° or π/12 radians
Answered by
Bosnian
OR
Multiply and divide 2 - √3 by 1/ 2
2 - √3 = [ ( 2 - √3 ) / 2 ] / ( 1 / 2 ) =
( 1 - √3 / 2 ) / ( 1 / 2 )
Since:
sin 30°= 1 / 2
cos 30° =√3 / 2
and half-angle formula for tangent is:
tan ( θ / 2 ) = ( 1 - cos θ ) / sin ( θ )
tan ( 30° / 2 ) =
( 1 - cos 30° ) / sin 30 °=
( 1 - √3 / 2) / ( 1 / 2 ) = tan 15°
arctan ( 2 - √3 ) =
arctan ( 1 - √3 / 2) / ( 1 / 2 ) =
15° = π / 12 rad
Multiply and divide 2 - √3 by 1/ 2
2 - √3 = [ ( 2 - √3 ) / 2 ] / ( 1 / 2 ) =
( 1 - √3 / 2 ) / ( 1 / 2 )
Since:
sin 30°= 1 / 2
cos 30° =√3 / 2
and half-angle formula for tangent is:
tan ( θ / 2 ) = ( 1 - cos θ ) / sin ( θ )
tan ( 30° / 2 ) =
( 1 - cos 30° ) / sin 30 °=
( 1 - √3 / 2) / ( 1 / 2 ) = tan 15°
arctan ( 2 - √3 ) =
arctan ( 1 - √3 / 2) / ( 1 / 2 ) =
15° = π / 12 rad
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