Asked by klkl
Find the exact value of tan (11pi/12)
Answers
Answered by
mathhelper
11π/12 is the same as 165°
tan 11π/12 = tan 165°
= - tan 15°
Now tan15° = tan(45-30)°
= (tan45 - tan30) / (1 + tan45tan30)
= (1 - 1/√3) / (1 + (1)(1/√3) )
multiply top and bottom by √3
= (√3 - 1)/(√3 + 1)
so tan 11π/12 = -tan15 = <b>(1 - √3)/(√3 + 1)</b> , I changed the sign on the top
tan 11π/12 = tan 165°
= - tan 15°
Now tan15° = tan(45-30)°
= (tan45 - tan30) / (1 + tan45tan30)
= (1 - 1/√3) / (1 + (1)(1/√3) )
multiply top and bottom by √3
= (√3 - 1)/(√3 + 1)
so tan 11π/12 = -tan15 = <b>(1 - √3)/(√3 + 1)</b> , I changed the sign on the top
Answered by
klkl
can you show me by using the half-angle formula
Answered by
mathhelper
recall cos (2A) = 2cos^2 A - 1 or 1 - 2sin^2 A
cos 30 = 2cos^2 15 - 1
√3/2 + 1 = 2cos^2 15
(√3 + 2)/4 = cos^2 15
cos 15° = √(√3 + 2)/2
cos 30 = 1 - 2sin^2 15
2sin^2 15 = 1 - √3/2 = (2 - √3)/2
sin^2 15 = (2 - √3)/4
sin 15 = √(2 - √3)/2
tan 15 = sin 15/cos 15 = √(2 - √3)/2 / √(2 + √3)/2
= √(2 - √3) / √(2 + √3)
so tan 165 = - tan 15° = - √(2 - √3) / √(2 + √3)
seems messier.
cos 30 = 2cos^2 15 - 1
√3/2 + 1 = 2cos^2 15
(√3 + 2)/4 = cos^2 15
cos 15° = √(√3 + 2)/2
cos 30 = 1 - 2sin^2 15
2sin^2 15 = 1 - √3/2 = (2 - √3)/2
sin^2 15 = (2 - √3)/4
sin 15 = √(2 - √3)/2
tan 15 = sin 15/cos 15 = √(2 - √3)/2 / √(2 + √3)/2
= √(2 - √3) / √(2 + √3)
so tan 165 = - tan 15° = - √(2 - √3) / √(2 + √3)
seems messier.
Answered by
oobleck
tan(x/2) = (1-cosx)/sinx = (1 - √3/2) / (-1/2) = (√3-2)
Note that
(1-√3)/(1+√3) = (1-√3)^2 / (1-3) = (1+3-2√3)/-2 = √3-2
-√(2-√3)/(2+√3) = -√((2-√3)^2)/(4-3)) = -(2-√3) = √3-2
Note that
(1-√3)/(1+√3) = (1-√3)^2 / (1-3) = (1+3-2√3)/-2 = √3-2
-√(2-√3)/(2+√3) = -√((2-√3)^2)/(4-3)) = -(2-√3) = √3-2
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