Posted by Juan on Friday, June 3, 2011 at 1:12am.
345°=360°-15°
tan(A-B)=(tan A - tan B)/(1 - (tan A)*(tan B))
tan(360°)=0
tan(360°-15°)= tan(360°)-tan(15°)/(1-(tan(360°)*(tan(15°))
tan(360°-15°)= 0-tan(15°)/(1-0*(tan(15°))
tan(360°-15°)=tan(345°)= -tan(15°)/(1-0)
tan(345°)= -tan(15°)/1
tan(345°)= -tan(15°)
tan(x/2) = [1 - cos(x)]/sin(x)
tan(30°/2)=tan(15°)
sin(30°)= 1/2
cos(30°)= sqroot(3)/2
tan(30°/2)=tan(15°)= [1 - cos(30°)]/sin(30°)
= (1 - sqroot(3)/2)/(1/2)
= 2 - sqroot(3)
tan(15°) = 2 - sqroot(3)
tan(345°)= -tan(15°)
tan(345°)= -(2 - sqroot(3))
tan(345°)= sqroot(3)-2
or ...
345° is coterminal with -15°
so tan 345° = - tan15
= - tan(45-30)
= - (tan45 - tan30)/(1 + tan45tan30)
= -(1-1/√3)/(1+1/√3) , multiply top and bottom by √3
= (1 - √3)(√3 + 1)
You can rationalize if necessary to get Bosnian's answer
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