## sabika13 3 years ago How do I solve this trig identity: cos2x/(1+sinx^2)=tan(pie/4 -x)

1. matricked

i think ur question is cos2x/(1+sin2x) now cos2x =cos^2x -sin^2x=(cosx+sinx)(cosx-sinx) and 1+sin2x = cos^2x + sin^2x +2sinxcosx =(cosx+sinx)^2 thus cos2x/(1+sinx)^2= (cosx+sinx)(cosx-sinx)/(cosx+sinx)^2 =(cosx- sinx)/((cosx+sinx) divide both num and denom by cosx and then we have =(1-sinx/cosx)/(1+sinx/cosx) =(1-tanx)/(1+tanx) = (tan(pi/4) -tanx)/(1+tan(pi/4)tanx)=tan(pie/4 -x)

2. sabika13

for the second last step.. where did the square roots go? isnt it suppose to be: (1-sinx/cosx)(1-sinx/cos)/(1+sinx/cos)(1+sinx/cosx)

3. matricked

actually the factors (cosx+sinx) are there both in num and denom hence they cancel each other

4. sabika13

but its (cosx-sinx) in the num :s

5. matricked

(cosx+sinx)(cosx-sinx) / (cosx+sinx)^2 =(cosx-sinx) / (cosx+sinx)

6. matricked

i hope its clear now

7. matricked

why 13 u hv included in ur id ? if u don't mind

8. sabika13

if its (cosx+sinx)^2, it has to equal (cosx+sinx)(cosx+sinx)

9. sabika13

idk random.

10. matricked

(cosx+sinx)^2= (cosx+sinx)(cosx+sinx) (correct ) that is why one of the (cosx+sinx)

11. matricked

cancel each other

12. sabika13

how does cos^2x -sin^2x=(cosx+sinx)(cosx-sinx)

13. matricked

USE ..A*A -B*B =(A)^2 -(B)^2 = (A+B)(A-B)...

14. sabika13

okay, then why isnt :(cosx+sinx)^2 --> (cosx-sinx)(cosx+sinx)

15. matricked

bcoz a^2 =a*a

16. matricked

so (a+b)^2 =(a+b)(a+b)

17. matricked

but if u multiply (a+b)(a-b) u get (a^2 -b^2)

18. sabika13

so (a^2-b^2) does not equal (a-b)^2 ?

19. matricked

sahi hai..

20. sabika13

oh i get it..

21. sabika13

=(1-tanx)/(1+tanx) = (tan(pi/4) -tanx)/(1+tan(pi/4)tanx)=tan(pie/4 -x) can u explain that? .. where did pi/4 come from ?

22. Dido525

Tan (pi/4) = 1.

23. sabika13

is that a trig identity?

24. Dido525

No. If you do that on a calculator you get that.

25. Dido525

Tan(pi/4) is always 1. Look on a unit circle.

26. sabika13

Okay, thanks!!

27. Dido525

Welcome :) .