## AonZ 2 years ago Pls Pls Pls help... I have been stuck for a while!!! Prove tan3x = (3tanx - tan^3 x) / ( 1 - 3tan^2 x) using sin3x/ cos3x = tan3x sin3x = 3sinx - 4sin^3 x cos3x = 4cos^3 x - 3cosx

1. .Sam.

Have you tried using$\tan(3x)=\tan(2x+x)$ Then $\tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$

2. AonZ

the question asked to solve it using tan3x = sin3x / cos3x :/

3. AonZ

|dw:1369139564173:dw|

4. AonZ

can you pls post it :) May help me solve it :)

5. trumpetmaster7777

factor out sin and cos

6. trumpetmaster7777

(3sinx - 4sin^3 x)/(4cos^3 x - 3cosx) (sinx)(3-4sin^2(x))/((cosx)(4cos^2(x)-3)

7. AonZ

my way was to factor out sin and cos so it was tanx times something then i divided the numerators by cos^2 x then im stuck

8. trumpetmaster7777

Pick a trig function, sine or cosine

9. AonZ

any :)

10. trumpetmaster7777

Thanks, loser66

11. trumpetmaster7777

so, let's go with sine cos^2(x)=1-sin^2(x), right?

12. AonZ

yep

13. trumpetmaster7777

plug in (1-sin^2(x)) for cos^2(x) (sinx)(3-4sin^2(x))/((cosx)(4cos^2(x)-3)

14. AonZ

yea i actually tried that :P

15. AonZ

after doing that would you divide by cos^2 x?

16. trumpetmaster7777

wait up look at the tan(3x) expression can't you clean that up a bit?

17. AonZ

possibly... But you could always change it both ways

18. AonZ

this is the last step i got to|dw:1369140948450:dw|

19. trumpetmaster7777

1/cos^2(x) is sec^2(x), right?

20. trumpetmaster7777

use the tangent-secant identity

21. trumpetmaster7777

1+tan^2(x) = sec^2(x)

22. AonZ

!!!! :D ur awsome

23. trumpetmaster7777

Thank you :D

24. AonZ

on denominator its still sec^2 - 4tan^2 ... errr im lost again

25. RolyPoly

Would that help? sec^2 = 1 + tan^2

26. AonZ

oh yes :)