Good God! Just how am I supposed to prove this identity?! Help please!?

Question

Hi! I'm supposed to prove a trigonometric identity with this pile of crap, and I'm not sure exactly what to do. I thought that I had a pretty solid understanding of how to do this, but then I came to this problem! I'm now starting to think that maybe I just lucked out with the past 500+ problems that I did because I can't figure this one out for the life of me! It seems so easy, but everything that I tried just falls short of what I expected. Maybe someone can help!

( 1 - 2cos²(y) ) ' ' ' ' ' ' ' ' ' ' ( sin(y) + cos(y) )
--------------------------- = --------------------------
( 1 - 2cos(y)sin(y) ) ' ' ' ' ' ' ( sin(y) - cos(y) )


Thanks so much! I'll pick a best answer TODAY!

Answer 1

OK... i'll give it a go.

sin²y - cos²y'''''''''''''sin²y - cos²y
--------------------- = ----------------------------
(1 - sin2y)''''''''''''''''''(siny - cosy)^2

(multiplied numerator and denominator on the right side by (siny-cosy), and converted the ( 1 - 2cos²(y)) to sin²y - cos²y, because if sin²y + cos²y = 1, therefore sin²y - cos²y = 1-2cos²(y)), as 2cos²(y) is the difference between sin²y + cos²y and sin²y - cos²y. Also used double angle identity on the denominator on the left.

sin²y - cos²y''''''''' sin²y - cos²y
------------------- = -------------------------
(1-2sin2y)''''''''''''' sin²y - 2sinycosy + cos²y

(multiplied denominator out)

sin²y - cos²y''''''''' sin²y - cos²y
------------------- = -------------------------
(1-2sin2y)''''''''''''' 1 - (2sin2y)

added the sin²y and cos²y to get 1, and used double angle identity again.

whew! there ya go.

Answer 2

To go from the right hand side to the left hand side:
(siny + cosy) / (siny - cosy)
multiply numerator and denominator by (siny - cosy)
= (siny + cosy)(siny - cosy) / (siny - cosy)^2
= (sin²y - cos²y) / (sin²y - 2sinycosy + cos²y)
substitute sin²y + cos²y = 1 and sin²y = 1-cos²y
= (1 - cos²y - cos²y) / (1 - 2sinycosy)
= (1 - 2cos²y) / (1 - 2sinycosy)

Answer 3

start from the right hand side and multiply numerator and denominator by sin(y) - cos(y). I think once you simplify, you will easily be able to get to the left hand side.

Math Rules!

Answer 4

Let cos(y) = c and sin(y) = s. We have:

(1 - 2c²)(s - c) = (s + c)(1 - 2cs)

s - c - 2c²s + 2c^3 = s + c - 2cs² - 2c²s

- 2c + 2cs² + 2c^3 = 0

- 1 + s² + c² = 0

c² = 1 - s²

QED

Answer 5

you remember
cos2y = 2cos^2(y) -1 by trig identity
multiply both sides by -1
-cos2y = -2cos^2(y) +1
arrange it
[-cos2y = 1-2cos^2(y),] and remember sin^2(x) +cos^2(x) =1
and cos2y = cos^2(y)-sin^2(y) trig identity
start he left hand side to prove that equal the right hand side
1-2cos^2(y)
----------------
1-2cos(y)siny

-cos2y
---------------------------------------------- cuz -cos2y= 1-2cos^2(y)
sin^2(y)+cos^2(y) -2cosysiny

-(cos^2(y)-sin^2(y))
----------------------------------------- cuz cos2y= cos^2(y)- sin^2(y)
sin^2(y) -2sinycosy+cos^2(y)

now remember different square cos^2(y)-sin^2(y)
will equal to (cosy-siny)(cosy+siny)
and perfect square sin^2(y)-2sinycosy+cos^2(y) will be equal
(siny-cosy)^2

-(cosy-siny)(cosy+siny)
-------------------------------
(siny-cosy)^2

distribute -(cosy-siny) will be equal siny-cosy
(siny-cosy)(cosy+siny)
------------------------------
(siny-cosy)(siny-cosy)

cosy+siny
------------- cuz siny-cosy will be cancelled out
siny-cosy
rerange
siny+cosy
------------ ............... same on right hand side prove
siny-cosy