There's this step question which is really starting to annoy me. I'll post the whole question:

show if (tanx)²=2tanx+1, then tan2x=-1 [easy bit]

find all solutions to tanx=2+tan3x, for 0 < x < 2p

also solve cotx=2cot3x, for 3p/2 < x < p/2

this question looks reasonable when i started it but i can't seem to to solve it and it is really bugging me. if anyone has anyone has any tips i will be grateful.

luqman

Luqman,

There are several ways to derive the expansion of tan(nx) in terms of tan(x). Are you familiar with the expansion of sin(a+b) and cos(a+b) in terms of sin(x), cos(x)?

If you are, then take the ratio of the two, and set a=b=x, divide the right hand side through by cos(a) and cos(b) and you have the expression for tan(2x) in terms of tan(x). This gives you the answer to your first problem.

For the second, set a=2x,b=x. Again take the ratio - this gives you tan(3x) in terms of sin(x), cos(x), sin(2x), cos(2x). By setting a=b=x gives us expansions for sin(2x) and cos(2x).

Multiplying everything out and using tanx = 2 + tan3x you get:

0=[1+tan(x)-3tan² (x)+tan³ (x)]/[1-3tan² (x)]

The numerator can be factored ...

Andre