# Mathematics - Really Simple Trig Identities that I can't get!

• Nov 10th 2007, 03:35 AM
Macleef
Mathematics - Really Simple Trig Identities that I can't get!
Prove:
(tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx)

What I have so far:

L.S.
= (sinx / cosx) sinx / (sinx / cosx) + sinx
= (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx)
= (sin^2x / cosx) / (cosx / sinx + sinxcosx)

What do I do now and if I did something wrong, where did I go wrong and how should I go about to fix it?
• Nov 10th 2007, 03:58 AM
kalagota
Quote:

Originally Posted by Macleef
Prove:
sin^2x - sin^4x = cos^2x - cos^4x

What I have:

L.S.
= sin^2x - sin^4x
= sin^2x (-sin^2x)
= 1 - cos^2x (-1 + cos^2x)
= -1 +cos^2x + cos^2x - cos^4x
= cos^2x - 1 +cos^2x - cos^4x
= sin^2x + cos^2x - cos^4x

Where did I go wrong?

$\displaystyle sin^2x - sin^4x = sin^2x (1 - sin^2 x) = (1 - cos^2 x) cos^2 x = RHS$
---

Quote:

Originally Posted by Macleef
Prove: sinx - 1 / sinx + 1 = - cos^2x / (sinx + 1)^2

What I have so far:

R.S.
= - (1 - sin^2x) / (sinx + 1) (sinx + 1)
= -1 + sin^2x / (sinx + 1) (sinx + 1)

What do I do now?

$\displaystyle \frac{- cos^2x }{(sinx + 1)^2} = \frac{-(1 - sin^2 x)}{(sinx + 1)^2}= \frac{-(1 - sinx)(1 + sinx)}{(sinx + 1)^2}$
can you continue now?
---
Quote:

Originally Posted by Macleef
Prove:
(tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx)

What I have so far:

L.S.
= (sinx / cosx) sinx / (sinx / cosx) + sinx
= (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx)
= (sin^2x / cosx) / (cosx / sinx + sinxcosx)

What do I do now and if I did something wrong, where did I go wrong and how should I go about to fix it?

$\displaystyle LHS = \frac{\frac{sin^2x}{cosx}}{\frac{sinx}{cosx}+sinx} = \frac{\frac{sin^2x}{cosx}}{\frac{sinx + sinx cosx}{cosx}}$

$\displaystyle = \frac{sin^2x}{sinx + sinx cosx} = \frac{1 - cos^2x}{sinx(1 + cosx)} = \frac{(1 - cosx)(1 + cosx)}{sinx(1 + cosx)}$

$\displaystyle \frac{1 - cosx}{sinx}$, multiplying $\displaystyle \frac{tanx}{tanx}$ yields RHS..
• Nov 10th 2007, 04:12 AM
Macleef
Prove:
sin^2x - sin^4x = cos^2x - cos^4x

$\displaystyle sin^2x - sin^4x = sin^2x (1 - sin^2 x) = (1 - cos^2 x) cos^2 x = RHS$

^^^

How did you get "sin^2x(1 - sin^2 x)" in the second line? When I did this, I got "sin^2x (-sin^2x)"
Can you tell me what I did wrong ? I don't know how you got the "1" from ...

---

$\displaystyle \frac{- cos^2x }{(sinx + 1)^2} = \frac{-(1 - sin^2 x)}{(sinx + 1)^2}= \frac{-(1 - sinx)(1 + sinx)}{(sinx + 1)^2}$
can you continue now?

When I did this, I get:
= (-1 + sinx) (1 + sinx) / (sinx + 1) (sinx +1)
= (-1 + sinx) / (sinx + 1)

And now I'm stumped? I don't know what to do ...
• Nov 10th 2007, 04:24 AM
kalagota
Quote:

Originally Posted by Macleef
How did you get "sin^2x(1 - sin^2 x)" in the second line? When I did this, I got "sin^2x (-sin^2x)"
Can you tell me what I did wrong ? I don't know how you got the "1" from ...

---
remember factoring.. (a+ab)=a(1+b), so when you factored out sin^2 x from sin^2 x - sin^4 x, it must be sin^2 x (1 - sin^2 x).. you should remain 1 from the parenthesis..

Quote:

Originally Posted by Macleef
When I did this, I get:
= (-1 + sinx) (1 + sinx) / (sinx + 1) (sinx +1)
= (-1 + sinx) / (sinx + 1)

And now I'm stumped? I don't know what to do ...

you are done.. that's what you want to show right?
• Nov 10th 2007, 04:46 AM
Macleef
..
Quote:

Originally Posted by kalagota
---
remember factoring.. (a+ab)=a(1+b), so when you factored out sin^2 x from sin^2 x - sin^4 x, it must be sin^2 x (1 - sin^2 x).. you should remain 1 from the parenthesis..

you are done.. that's what you want to show right?

The first one, I remember it now, thanks

And as for the second one,

I'm proving

sinx - 1 / six + 1 = -cos^2x / (sinx + 1)^2

But ... when I did RS = RS, I get this:

= (-1 + sinx) / (sinx + 1)
= - (1 - sinx) / (sinx + 1)

Why do I have a negative in the numerator next to the bracket? How do I remove this to prove the identity?
• Nov 10th 2007, 04:50 AM
kalagota
Quote:

Originally Posted by Macleef
The first one, I remember it now, thanks

And as for the second one,

I'm proving

sinx - 1 / six + 1 = -cos^2x / (sinx + 1)^2

But ... when I did RS = RS, I get this:

= (-1 + sinx) / (sinx + 1)
= - (1 - sinx) / (sinx + 1)

Why do I have a negative in the numerator next to the bracket? How do I remove this to prove the identity?

wait, did u mean like this?

$\displaystyle sinx - \frac{1}{sinx} + 1$ or $\displaystyle sinx - \frac{1}{sinx + 1}$ or $\displaystyle \frac{sinx - 1}{sinx + 1}$

first, second or third?
• Nov 10th 2007, 05:06 AM
Macleef
Quote:

Originally Posted by kalagota
wait, did u mean like this?

$\displaystyle sinx - \frac{1}{sinx} + 1$ or $\displaystyle sinx - \frac{1}{sinx + 1}$ or $\displaystyle \frac{sinx - 1}{sinx + 1}$

first, secodn or third?

I mean like

= - (1 - sinx) over (sinx + 1)

or

- (1 - sinx)
_____________
sinx + 1
• Nov 10th 2007, 05:09 AM
kalagota
i am asking what is your left hand side.. first second or third?

coz if its the third one, then you are done with

$\displaystyle \frac{-1 + sinx}{sinx + 1}$..
remember that addition of real number is commutative so
$\displaystyle \frac{-1 + sinx}{sinx + 1} = \frac{sinx - 1}{sinx + 1}$ which is the third one..
• Nov 10th 2007, 05:12 AM
Macleef
wow I see it now, thanks, I feel so stupid right now -_-