# Verifying Identities

• Nov 19th 2009, 10:14 AM
Skizye
Verifying Identities
So I've been looking at this homework that's due tommorrow for a grade for the past hour and I can't seem to figure these out.

1. $sin^3x - cos^3x = (1 + sinx cosx)(sinx - cosx)$

2. $(1/(siny - 1)) - (1/(siny + 1)) = -2sec^2y$

3. $1 - 2sin^2r + sin^4r = cos^4r$

4. $tanu + (cosu/(1 + sinu)) = secu$

5. $(tanx + secx)/(secx - cosx + tanx) = cscx$

If I could get a hint or something on how to do these I would appreciate it, I don't want to ask anyone to do my homework for me (although I won't complain if you show me the whole problem) but I don't even know where to start on these problems. Thanks!
• Nov 19th 2009, 10:39 AM
I-Think
For number one, use the identity: a^3-b^3=(a-b)(a^2+ab+b^2)

Number 2
Place the L.H.S. under a common denominator and solve from there

Number 3
Use (1-sin^2r)=cos^2r on the R.H.S.

Number 4
Do the same thing as number 2

Number 5
A bit more tricky

$\frac{tanx+secx}{secx-cosx+tanx}$
becomes
$(\frac{sinx+1}{cosx})/(\frac{1-cos^2x+sinx}{cosx})$
becomes
$\frac{sinx+1}{sin^2x+sinx}$

Solve from there
• Nov 19th 2009, 10:44 AM
masters
Quote:

Originally Posted by Skizye
So I've been looking at this homework that's due tommorrow for a grade for the past hour and I can't seem to figure these out.

1. $sin^3x - cos^3x = (1 + sinx cosx)(sinx - cosx)$

2. $(1/(siny - 1)) - (1/(siny + 1)) = -2sec^2y$

3. $1 - 2sin^2r + sin^4r = cos^4r$

4. $tanu + (cosu/(1 + sinu)) = secu$

5. $(tanx + secx)/(secx - cosx + tanx) = cscx$

If I could get a hint or something on how to do these I would appreciate it, I don't want to ask anyone to do my homework for me (although I won't complain if you show me the whole problem) but I don't even know where to start on these problems. Thanks!

Hi Skizye,

Here's a start for #2

$\frac{1}{\sin y-1}-\frac{1}{\sin y+1}=-2\sec^2 y$

Work with the left side.

The common denominator will be $\sin^2-1$

$\frac{\sin y+1-\sin y+1}{\sin^2 y-1}$

Simplify the numerator.

$\frac{2}{\sin^2 y-1}$

$\frac{2}{-(1-\sin^2 y)}$

Can you take it from here?
• Nov 19th 2009, 10:52 AM
masters
Quote:

Originally Posted by Skizye
So I've been looking at this homework that's due tommorrow for a grade for the past hour and I can't seem to figure these out.

1. $sin^3x - cos^3x = (1 + sinx cosx)(sinx - cosx)$

2. $(1/(siny - 1)) - (1/(siny + 1)) = -2sec^2y$

3. $1 - 2sin^2r + sin^4r = cos^4r$

4. $tanu + (cosu/(1 + sinu)) = secu$

5. $(tanx + secx)/(secx - cosx + tanx) = cscx$

If I could get a hint or something on how to do these I would appreciate it, I don't want to ask anyone to do my homework for me (although I won't complain if you show me the whole problem) but I don't even know where to start on these problems. Thanks!

Hello again,

And for #5, work with the left side.

$\dfrac{\tan x+ \sec x}{\sec x-\cos x+\tan x}=\csc x$

$\dfrac{\dfrac{\sin x}{\cos x}+\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}-\cos x+ \dfrac{\sin x}{\cos x}}=$

Do you think you can handle it from here?
• Nov 19th 2009, 11:04 AM
Skizye
Thanks so much for the help!

On number 2 I can't figure out I would get the secant to become a sine or the other way around so they will be equal, how can I do this (or should I even be trying to do this)? Thanks!
• Nov 19th 2009, 11:16 AM
masters
Quote:

Originally Posted by masters
Hi Skizye,

Here's a start for #2

$\frac{1}{\sin y-1}-\frac{1}{\sin y+1}=-2\sec^2 y$

Work with the left side.

The common denominator will be $\sin^2-1$

$\frac{\sin y+1-\sin y+1}{\sin^2 y-1}=$

Simplify the numerator.

$\frac{2}{\sin^2 y-1}=$

$\frac{2}{-(1-\sin^2 y)}=$

Can you take it from here?

Quote:

Originally Posted by Skizye
Thanks so much for the help!

On number 2 I can't figure out I would get the secant to become a sine or the other way around so they will be equal, how can I do this (or should I even be trying to do this)? Thanks!

I had a typo in my first post. I've fixed it now.

Notice the next two steps...

$\dfrac{2}{-\cos^2 y}=$

$-2 \sec^2 y$
• Nov 19th 2009, 01:09 PM
Skizye
Alright. I finally got all of them except for 4! I found a common denominator but I'm stuck at this step

$\frac{(1 + sinu)(tanu) + cosu}{1 + sinu} = secu$

Have I done it right so far? What can I do from here?

Sorry for all the questions and thanks again for the help!
• Nov 19th 2009, 01:38 PM
masters
Quote:

Originally Posted by Skizye
Alright. I finally got all of them except for 4! I found a common denominator but I'm stuck at this step

$\frac{(1 + sinu)(tanu) + cosu}{1 + sinu} = secu$

Have I done it right so far? What can I do from here?

Sorry for all the questions and thanks again for the help!

This is how I proceeded.

$\tan u+\dfrac{\cos u}{1+ \sin u}=\sec u$

$\frac{\sin u}{\cos u}+\frac{\cos u}{1+\sin u}=$

$\frac{\sin u(1+\sin u)+\cos^2 u}{\cos u(1+\sin u)}=$

$\frac{\sin u+\sin^2u+\cos^2 u}{\cos u(1+\sin u)}=$

$\frac{\sin u+1}{\cos u(1+\sin u)}=$

$\frac{1}{\cos u}=$

$\sec u$