# find the exact value of tan(17pi/12) ?

find the exact value of tan(17pi/12)

tan(17π/12) = tan(5π/12 + π)
given tan(θ +π) = tan(θ)
=> tan(17π/12) = tan(5π/12)
=> tan(17π/12) = tan(2π/12 + 3π/12)
=> tan(17π/12) = tan(π/6 + π/4)
given tan(α + β) = (tan(α) + tan(β))/(1 − tan(α)tan(β)))
=> tan(17π/12) = (tan(π/6) + tan(π/4))/(1 − tan(π/6)tan(π/4)))
=> tan(17π/12) = ((1/√3) + 1)/(1 − (1/√3)(1)))
=> tan(17π/12) = ((1/√3) + 1)/((√3/√3) − (1/√3)))
=> tan(17π/12) = √3((1/√3) + 1)/(√3 − 1)
=> tan(17π/12) = √3((1/√3) + (√3/√3))/(√3 − 1)
=> tan(17π/12) = √3(1 + √3)/[√3(√3 − 1)]
=> tan(17π/12) = (1 + √3)/(√3 − 1)
=> tan(17π/12) = (√3 + 1)/(√3 − 1) * (√3 + 1)/(√3 + 1)
using the difference of squares
=> tan(17π/12) = (√3 + 1)²/(3 − 1)
using the perfect square
=> tan(17π/12) = (3 + 2√3 + 1)/2
=> tan(17π/12) = (4 + 2√3)/2
=> tan(17π/12) = (2 + √3)

Well,
edit : see my PS : Rogue has the best answer : easy and correct !
just verify : tan(17pi/12) = tan75Â°= sqrt(3) + 2 # 3.732 !!
tan(17Ï/12) = tan(Ï + 5Ï/12) = tan(5Ï/12)
and :
sin(5Ï/12) = sin(6Ï/12 – Ï/12)
= sin(Ï/2 – Ï/12)
= cos(Ï/12)
———————————-> sin(5Ï/12) = (1/2)â(2+â3)
and the same way,
cos(5Ï/12) = sin(Ï/12) and
sin(Ï/12) = sin15Â° = sin(45Â° – 30Â°) in degrees it is easier to convert…
= sin 45Â°Â· cos 30Â° – cos 45Â°Â· sin 30Â°
= = (1/â2)Â·(â3 /2 ) – (1/â2)Â·(1/2)
—————————-> cos(5Ï/12) = ( â3 – 1 ) / (2â2)
therefore :
tan(17Ï/12) = tan(5Ï/12) = (1/2)â(2+â3) / [( â3 – 1 ) / (2â2)]
= (1/2)â(2+â3) (2â2) / (â3 – 1 )
= â2 â(2+â3) / (â3 – 1 )
= â2 â(2+â3) / (â3 – 1 )
= â2 â(2+â3)(â3 + 1 ) / [ (â3 – 1 )(â3 + 1 ) ]
———————–> tan(17Ï/12) = â2 â(2+â3) (â3 + 1 ) /2
hope it’ ll help !!
PS: pls don’t forget to give Best Answers, because according to “new rules”
only the asker may give Best Answers.
To me or to anybody else !! —> it’s about keeping people motivated to answer!
michael

As we know, multiples of pi do not change the tan function, so we can rewrite tan(17pi/12) as tan(5pi/12):
tan(5pi/12)
Rewrite tan as sin/cos:
sin(5pi/12) / cos(5pi/12)
Rewrite 5pi/12 as 8pi/12 – 3pi/12:
sin(8pi/12 – 3pi/12) / cos(8pi/12 – 3pi/12)
sin(2pi/3 – pi/4) / cos(2pi/3 – pi/4)
Expand the brackets using the compound angle formula:
[sin(2pi/3)cos(pi/4) – cos(2pi/3)sin(pi/4)] / [cos(2pi/3)cos(pi/4) + sin(2pi/3)sin(pi/4)]
Recall that cos(pi/4) and sin(pi/4) = 1 / sqrt(2):
[sin(2pi/3) * 1/sqrt(2) – cos(2pi/3) * 1/sqrt(2)] / [cos(2pi/3) * 1/sqrt(2) + sin(2pi/3) * 1 / sqrt(2)]
= [sin(2pi/3) – cos(2pi/3)] / [cos(2pi/3) + sin(2pi/3)]
Rewrite the trig functions as half angles:
= [2sin(pi/3)cos(pi/3) – cos^2(pi/3) + sin^2(pi/3)] / [cos^2(pi/3) – sin^2(pi/3) + 2sin(pi/3)cos(pi/3)]
= [sqrt(3) * 1/2 – 1/4 + 3/4] / [-1/4 + 3/4 + sqrt(3) * 1/2]
= [sqrt(3) + 1] / [sqrt(3) – 1]
And finally, rationalise the denominator:
= [sqrt(3) + 1]^2 / 2
= [3 + 2sqrt(3) + 1] / 2
= sqrt(3) + 2

tan(17Ï/12)
= tan(5Ï/12)
= tan(3Ï/12 + 2Ï/12)
= tan(Ï/4 + Ï/6)
= tan(Ï/4) + tan(Ï/6) / [1 – tan(Ï/4)tan(Ï/6)]
= (1 + 1/â) / (1 – 1/â)
= (â3 + 1) / (â3 – 1)