## anonymous 5 years ago the value of tan(2tan^-1 (1/5)-pi/4) is?

1. anonymous

Let$\theta=2\tan^{-1} (1/5)-(\pi/4)$Form a triangle with tan theta

2. anonymous

how?

3. anonymous

How what?!

4. anonymous

wrong!

5. anonymous

let x= $\tan^{-1} \frac{1}{5}$

6. anonymous

so our expression becomes $\tan(2x-\frac{\pi}{4})$

7. anonymous

now apply difference of angle formula

8. anonymous

$\tan(A-B) = \frac{\tan(A)-\tan(B) } {1+\tan(A)\tan(B) }$

9. anonymous

so we get

10. anonymous

$\frac{\tan(2x) -1}{1+\tan(2x)}$

11. anonymous

but ,$\tan(2x) = \frac{\sin(2x)}{\cos(2x)} = \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}$

12. anonymous

now, we go back to our substitution x= tan^-1(1/5) take tan of both sides , and we get tan(x) = 1/5 now we can draw up a general right angle triangle and mark an angle x , and fill in the lengths 1 and 5 for thre opposite and adjacent sides respectively remember tan = opposite/adjacent

13. anonymous

note the hypotenuse is sqrt(26)

14. anonymous

now from that triangle we can find the value of sin(x) and cos(x) , the then you just plug them into the expression we had above

15. anonymous

so $\sin(x) = \frac{1}{\sqrt{26}}$ $\cos(x) = \frac{5}{\sqrt{26}}$ just by using the definitions

16. anonymous

so , our whole expression was$\tan(2x- \frac{\pi}{4}) = \frac{ \frac{ 2\sin(x)\cos(x) }{\cos^2(x)-\sin^2(x) } -1}{ 1+ \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}}$

17. anonymous

then sub in the values for sin(x) and cos(x) above, and simplify

18. anonymous

you can do that

19. anonymous

good. it worked out well. thanx man!