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) = tan75°= sqrt(3) + 2 # 3.732 !!
tan(17π/12) = tan(π + 5π/12) = tan(5π/12)
and :
sin(5π/12) = sin(6π/12 – π/12)
= sin(π/2 – π/12)
= cos(π/12)
———————————-> sin(5π/12) = (1/2)√(2+√3)
and the same way,
cos(5π/12) = sin(π/12) and
sin(π/12) = sin15° = sin(45° – 30°) in degrees it is easier to convert…
= sin 45°· cos 30° – cos 45°· sin 30°
= = (1/√2)·(√3 /2 ) – (1/√2)·(1/2)
—————————-> cos(5π/12) = ( √3 – 1 ) / (2√2)
therefore :
tan(17π/12) = tan(5π/12) = (1/2)√(2+√3) / [( √3 – 1 ) / (2√2)]
= (1/2)√(2+√3) (2√2) / (√3 – 1 )
= √2 √(2+√3) / (√3 – 1 )
= √2 √(2+√3) / (√3 – 1 )
= √2 √(2+√3)(√3 + 1 ) / [ (√3 – 1 )(√3 + 1 ) ]
———————–> tan(17π/12) = √2 √(2+√3) (√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(17π/12)
= tan(5π/12)
= tan(3π/12 + 2π/12)
= tan(π/4 + π/6)
= tan(π/4) + tan(π/6) / [1 – tan(π/4)tan(π/6)]
= (1 + 1/√[3]) / (1 – 1/√[3])
= (√3 + 1) / (√3 – 1)

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