Question

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

Answer 1

Hint: apply the double angle formula

Answer 2

would i matter which side i start first ? :S

Answer 3

i am confused
why don't you just add \(\tan(2x)\) to both sides and get rid of it

Answer 4

because it's times

Answer 5

oh because it is times
doh

Answer 6

so wait...
2 tanx + 2a - 1 = tan2x - tan2x tan^2x
does that work?

Answer 7

if i bring tan2x to the right then bring 1 to the left side?

Answer 8

i think it is a bunch of algebra
i can try it with pencil and paper if you like

Answer 9

yes please? i want to understand how to do it.. i have a test coming up

Answer 10

im trying to do it myself as well

Answer 11

ok i screwed up already, let me try it again

Answer 12

lol i did too

Answer 13

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

Answer 14

this has nothing whatsoever to do with trig so fora as i can tell

Answer 15

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
\[\frac{2x(1-x^2)-2x+2a(1-x^2)}{1-x^2}=\frac{1-x^2-2x^3}{1-x^2}\]

Answer 16

ignoring the denominators, you ge t
\[2x-2x^3-2x+2a-2ax^2=1-x^2-2x^3\]
\[2a-2ax^2=1-x^2\]

Answer 17

and therefore \(a=\frac{1}{2}\)

Answer 18

wth! LOL that's the right answer... omg i was so off.... replacing tan(x) with "x" helped.
thank you!

Answer 19

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