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alexeis_nicole
Question on Advanced function grade 12 Determine the value of "a" in the following equation : 2 tanx - tan2x + 2a = 1 - tan2x ･ tan^2x Hint: Double Angle Formula for Tangent is : tan2x = 2tanx/1-tan^2x
Hint: apply the double angle formula
would i matter which side i start first ? :S
i am confused why don't you just add \(\tan(2x)\) to both sides and get rid of it
oh because it is times doh
so wait... 2 tanx + 2a - 1 = tan2x - tan2x tan^2x does that work?
if i bring tan2x to the right then bring 1 to the left side?
i think it is a bunch of algebra i can try it with pencil and paper if you like
yes please? i want to understand how to do it.. i have a test coming up
im trying to do it myself as well
ok i screwed up already, let me try it again
from : tan2x = 2tanx/(1-tan^2 x) 2tanx = tan2x(1-tan^2 x) 2tanx = tan2x - tan2x tan^2 x now, apply it to left side
this has nothing whatsoever to do with trig so fora as i can tell
if we replace \(\tan(x)\) by \(x\) you get \[2x-\frac{2x}{1-x^2}+2a=1-\frac{2x^3}{1-x^2}\] now we add up on the left and right get \[\frac{2x(1-x^2)-2x+2a(1-x^2)}{1-x^2}=\frac{1-x^2-2x^3}{1-x^2}\]
ignoring the denominators, you ge t \[2x-2x^3-2x+2a-2ax^2=1-x^2-2x^3\] \[2a-2ax^2=1-x^2\]
and therefore \(a=\frac{1}{2}\)
wth! LOL that's the right answer... omg i was so off.... replacing tan(x) with "x" helped. thank you!
yeah that is a good trick to remember when you have a ton of annoying algebra to do it is good for all kinds of trig identities because almost all of the work is algebra, and you get too bogged down writing \(\sin(x)\) and \(\tan(x)\) when \(a\) or \(b\) would do