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Posted by **Juan** on Friday, June 3, 2011 at 1:12am.

- Math -
**Bosnian**, Friday, June 3, 2011 at 2:22am345°=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°)

- Math -
**Bosnian**, Friday, June 3, 2011 at 2:36amtan(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

- Math -
**Reiny**, Friday, June 3, 2011 at 7:17amor ...

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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