# Trig Identities and Double Angle Formulas

1. Nov 5, 2007

### rum2563

[SOLVED] Trig Identities and Double Angle Formulas

1. The problem statement, all variables and given/known data
If tan(a) = 1/5 and tan(b) = 1/239, find tan(4a - b)

2. Relevant equations
tan2x = 2tanx/1-tan^2x

tan(x-y) = tanx - tany/ 1 +tanx tany

3. The attempt at a solution

tan (4a - b) = (tan4a - tanb)/(1 + tan4a tanb)
= ((2tan2a/1-tan^2 2a) - tanb)/ (1-(2tan^2 2a/1-tan^2 2a)tanb)

= ((2tan2(1/5)/1-tan^2 2(1/5)) - tan(1/239)) / (1 - (2tan^2 2(1/5))tan(1/239))
= ((2tan(2/5))/1-tan^2 (2/5)) - tan(1/239)) / (1 - 2tan^2 (2/5)tan(1/239))

After this I get stuck. Please someone help. I need this really quick.

2. Nov 5, 2007

### rock.freak667

tan(a) = 1/5 means that where ever you see tan(a) you can replace it with 1/5

tan2(a)=2tan(a)/{1-tan^2(a)}
you can find a value for tan2a...but if I were you, I would not bother to expand out tan4a. I would just find the value for tan4a and substitute it

3. Nov 5, 2007

### rum2563

ok, Here is my new try:

tan4a - tanb / 1 + tan4a tanb

tan(4/5) - tan(1/239) / 1 + tan4(1/5) tan(1/239)
tan(951/1195) / 1 + tan(951/1195)

4. Nov 5, 2007

### rock.freak667

well you still aren't getting it

$$tan4a=\frac{2tan2a}{1-tan^22a}$$ ....call this (*)

now separately you know that $$tan2a=\frac{2tana}{1-tan^2a}$$

so then substituting $tana=\frac{1}{5}$

$$tan2a= \frac{\frac{2}{5}}{1-\frac{1}{25}}$$
$tan2a=\frac{5}{12}$

now that you have a value for tan2a can you sub this value into * and find the value for tan4a ?

5. Nov 6, 2007

### rum2563

Yes, I finally got it. Thanks to rock.freak667.

tan (4a - b) = tan4a - tanb / 1 + tan4a tanb

tan4a = 2tan2a / 1-2tan^2 a

tan2a = 2tana / 1- tan^2a
= 2(1/5) / 1 - (1/25)
= (2/5) / (24/25)
= 5/12

tan4a = 2tan2a / 1-2tan^2 a
= 2(5/12) / (1 - 25/144)
= (5/6) / (119/144)
= 120/119

tan (4a - b) = tan4a - tanb / 1 + tan4a tanb
= (120/119) - (1/239) / 1 + (120/119)(1/239)
= (28561/28441) / 1 + 120/28441
= (28561/28441) / (28561/28441)
= 1

Thanks very much to rock.freak667. I really needed help on this.