## alexeis_nicole 2 years ago 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

1. bahrom7893

Hint: apply the double angle formula

2. alexeis_nicole

would i matter which side i start first ? :S

3. satellite73

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

4. bahrom7893

because it's times

5. satellite73

oh because it is times doh

6. alexeis_nicole

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

7. alexeis_nicole

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

8. satellite73

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

9. alexeis_nicole

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

10. alexeis_nicole

im trying to do it myself as well

11. satellite73

ok i screwed up already, let me try it again

12. alexeis_nicole

lol i did too

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

14. satellite73

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

15. satellite73

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

16. satellite73

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

17. satellite73

and therefore $$a=\frac{1}{2}$$

18. alexeis_nicole

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

19. satellite73

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