Question

Prove the following trigonometric identity:
(1-2sin^2x)/(cosx+sinx)=cosx-sinx
Thanks for any help.

Answer 1

do u know \(\large \sin^2x+\cos^2x=1\)
???

Answer 2

\[\frac{ 1-2\sin ^{2}x }{ \cos x+\sin x }=\cos x-\sin x\]

Answer 3

Yes I do

Answer 4

then substitute that 1 in the numerator by
sin^2 x + cos^2 x
what do u get ?

Answer 5

sin^2x+cos^2x-2sin^2x?

Answer 6

yes, what does that simplify to ?

Answer 7

-sin^2x+cos^2x

Answer 8

correct!
now thats in \(\large a^2-b^2 \)
form , right >
can you factorize ?

Answer 9

like a^2-b^2 = (a+b)(a-b)

Answer 10

So (sinx+cosx)(sinx-cosx)?

Answer 11

wouldn't the 2nd term be
cos x - sinx ?

because we have cos^2 x - sin^2 x
right ?

Answer 12

What exactly do you mean by 2nd term?

Answer 13

(sinx+cosx)(sinx-cosx)
the sin x - cos x term
should be cos x - sin x

Answer 14

-sin^2x+cos^2x
= cos^2 x- sin^2 x
= (cos x + sin x)(cos x - sin x)
got this ?

Answer 15

Yes i follow it

Answer 16

and what gets cancelled ?

Answer 17

from numerator and denominator ?

Answer 18

cosx+sinx?

Answer 19

yes!
and what remains ?

Answer 20

cosx-sinx Thank you so much for your help!

Answer 21

welcome ^_^