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Did I approach this question correctly.

I answered a trigonometric question in C3, and i wanted to know if i went about answering it in a way that would be acceptable by the examiners. The question was.

"Given that tan(x+π3)=12 tan(x + \frac{\pi}{3} ) =\frac{1}{2} show than tan(x)=853tan(x)=8-5\sqrt{3}."

Ok then so at first i tried to solve for tanx and it became very messy, i don't no if it was because i was going about it the wrong way or because i just wasn't meant to solve for tanx at all, at one point i tried rationalising the denominator as it became a surd, and that made it even more cumbersome.

So eventualy I subsititued 853 8-5\sqrt{3} in place of tanx tanx into the addition formula for tan(x+π3) tan(x + \frac{\pi}{3} ) , getting ofc 12 \frac{1}{2} as my answer, Stating that "therefore tanx=853tanx = 8-5\sqrt{3}" , will i be penalised for using this method in an exam?
(edited 13 years ago)
Reply 1
What you've shown is "if tan(x) = 8 - 5 \sqrt 3" then "tan(x+pi/3) = 1/2", which is the other way round from what you were asked. I think you'd lose some marks as a consequence.

Easiest thing here is to write y = x+pi/3 so the question becomes "given tan y = 1/2, show tan(yπ/3)=853\tan(y - \pi/3) = 8 - 5 \sqrt 3". Which you can then do using the same tan(A+B) formula, only this time going in the right direction. The arithmetic should be easier too.
Reply 2
Original post by DFranklin
What you've shown is "if tan(x) = 8 - 5 \sqrt 3" then "tan(x+pi/3) = 1/2", which is the other way round from what you were asked. I think you'd lose some marks as a consequence.

Easiest thing here is to write y = x+pi/3 so the question becomes "given tan y = 1/2, show tan(yπ/3)=853\tan(y - \pi/3) = 8 - 5 \sqrt 3". Which you can then do using the same tan(A+B) formula, only this time going in the right direction. The arithmetic should be easier too.


Thanks il attempt it in this way, btw how did you come to the conclusion that this would make the arithmetic easier, so i can break questions down this way in the future.
Reply 3
wow i solved this without even having to touch my pencil, was it correct of me to substitue 12 \frac{1}{2} for tany and 3 \sqrt{3} for {tex] tan\frac{\pi}{3} ? Th calculator just gave me 853 8 -5\sqrt3
Reply 4
I can't tell from what you're written what you did, but you basically need to calculate tan(y-pi/3) which is (tan(y) - tan(pi/3)/(1+tan(y)tan(pi/3)). (And then substitute values as you say).
Reply 5
Original post by DFranklin
I can't tell from what you're written what you did, but you basically need to calculate tan(y-pi/3) which is (tan(y) - tan(pi/3)/(1+tan(y)tan(pi/3)). (And then substitute values as you say).


Yes this is what i did, and it made short work of the question thanks for your help. Also would you mind explaining the proof of the addition formula for cos(A-B) please, it took me so long to try and understand i skipped it and just used the formula in relation to what ever the question asked. My main issue with it is that even though
(edited 13 years ago)
Reply 6
My main issue with it is that even though OP makes angle and OQ makes angle b and B is a larger ange than A angle POQ =(A-B) even though angle B is clearly larger than A.
Reply 7
Use the double angle formula for tan(a+b) x=a ,b=pi/3 , expand it, once you have that equate it to 1/2, then simplify get tanx on one side, rationalise the other side by mulitplying by the conjugate and it comes out tanx=8-5root3

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