Trigonometry help - proving identites!!!!

[Me345]

Hey, I really need some help solving this equation, or rather proving it. I have no clue how to do it. Please help! Thxs! oh, and 0 = theta, since I don't know how to really put theta.

sin0-cos0+1
sin0+cos0-1 =

sin0+1
cos0

 

[wjm11]

sin0-cos0+1
sin0+cos0-1 =

sin0+1
cos0

Working just with the left side of the eqn, multiply:

(sin0-cos0+1)*(sin0 + cos0 + 1)
--------------------------------------
(sin0+cos0-1)*(sin0 + cos0 + 1)

Then use the trig identity: sin^2 + cos^2 = 1. Can you take it from there?

 

[Me345]

I've tried using that formula, but it doesn't seem to come out to

sin(0) + 1
-----------
cos(0)

Maybe I'm cancelling out wrong? I don't know

 

[wjm11]

I've tried using that formula, but it doesn't seem to come out

Can you show us some work so we can see how far you've gotten?

 

[Me345]

well, my work is really messy cuz it's scribbled all over the page. :lol: from trying to do it so many different ways.. but I'll try to show some work.

0= theta
2= squared

sin2(0)+2sin(0)(cos-1)+cos2(0)-2cos0-2cos(0)+1
-----------------------------------------------------------
sin2(0)-(cos2(0)-2cos(0)+1)

1-cos2(0)-2sin(0)(cos0-1)+(1-sin2(0))-2cos(0)=1)
----------------------------------------------------------
1-cos2(0)-(1-sin2(0)-2cos(0)=1)

1-cos2(0)-2sin(0)(cos(0)-1)
--------------------------------
1-cos2(0)

1-cos2(0)-2sin(0)
--------------------
1-cos(0)

1-cos(0)-2sin(0) or sin2(0)-2sin(0)


That's all I got. Did I multiply wrong? cancel? I have no clue. it's too complicated for me. lol

 

[soroban]

Hello, Me345!

Prove:

. . sinθ - cosθ + 1 . . . . 1 + sinθ
. . ------------------ . = . ----------
. . sinθ + cosθ - 1 . . . . . .cosθ
.

. . . . . . . . . sinθ + (1 - cosθ)
We have: . --------------------
. . . . . . . . . sinθ - (1 - cosθ)


Multiply top and bottom by: sinθ + (1 - cosθ)

. . sinθ + (1 - cosθ) . sinθ + (1 - cosθ)
. . -------------------- . ---------------------
. . sinθ - (2 - cosθ) . .sinθ + (1 - cosθ)


The numerator is: .[sinθ + (1 - cosθ)]²

. . = . sin²θ + 2sinθ(1 - cosθ) + (1 - cosθ)²

. . = . sin²θ + 2sinθ -2sinθcosθ + 1 - 2cosθ + cos²θ

. . = . 2 + 2sinθ - 2sinθcosθ - 2cosθ

. . = . 2(1 + sinθ - sinθcosθ - cosθ)

. . = . 2[(1 + sinθ) - cosθ(1 + sinθ)]

. . = . 2(1 + sinθ)(1 - cosθ)


The denominator is: .(sinθ)² - (1 - cosθ)²

. . = . sin²θ - 1 + 2cosθ - cos²θ

. . = . (1 - cos2θ) - 1 + 2cosθ - cos²θ

. . = . 2cosθ - 2cos²θ

. . = . 2cosθ(1 - cosθ)

. . . . . . . . . . . . . . . . . . . 2(1 + sinθ)(1 - cosθ) . . . . 1 + sinθ
The fraction becomes: . -------------------------- . = . -----------
. . . . . . . . . . . . . . . . . . . . .2cosθ(1 - cosθ) . . . . . . . . cosθ

 

[Guest]

for your expansion ......

0= theta
2= squared

sin2(0)-2sin(0)(cos0)+cos2(0)+cos2(0)+2sin(0)-2cos(0)+1
-----------------------------------------------------------
sin2(0)-cos2(0)-1

then replace the top line sin2(0) + cos2(0) =1
and then take 2(.....................) for the top line.

on the bottom line rewrite -cos2(0) = sin2(0) -1
(from sin2(0) + cos2(0)=1)

over to you

 

[Guest]

sin2(0)-2sin(0)(cos0)+cos2(0)+cos2(0)+2sin(0)-2cos(0)+1
-----------------------------------------------------------
sin2(0)-cos2(0)-1 +2cos(0)


...correction for you

 

[Me345]

Hey, thank you all for your help! But I still have one question... How did you get

2(1+sin(0)-(sin0)cos0 - cos0)
to equal
2[(1+sin(0)-cos0(1+sin0)]

I get the rest of what you did, just not how that becomes.. that.

 

[Guest]

2(1+sin(0)-(sin0)cos0 - cos0)
to equal
2[(1+sin(0)-cos0(1+sin0)]

using x in place of (0)

-(sinx)cosx - cosx) take the common term of cos x out

-1 (cos x (sin x - 1) )

-1 cos x (sin x - 1)

-cos x (sin x - 1)

there you go

 

[soroban]

Hello, Me345!

How did you get: 2(1 + sin(0) - (sin0)cos0 - cos0)
to equal: 2[(1 + sin(0) - cos0(1 + sin0)]

Ignore the "2" out front.

We have: .1 + sinθ - sinθcosθ - cosθ

. . Factor -cosθ out of the last two terms: .1 + sinθ - cosθ(sinθ + 1)

. . And we have: .(1 + sinθ) - cosθ(1 + sinθ)

Then we can factor out the common (1 + sintθ).