Question

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

Answer 1

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)

Answer 2

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)

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

i hope its clear now

Answer 7

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

Answer 8

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

Answer 9

idk random.

Answer 10

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

Answer 11

cancel each other

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

bcoz a^2 =a*a

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

sahi hai..

Answer 20

oh i get it..

Answer 21

=(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 ?

Answer 22

Tan (pi/4) = 1.

Answer 23

is that a trig identity?

Answer 24

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

Answer 25

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

Answer 26

Okay, thanks!!

Answer 27

Welcome :) .