More Trig/Pre-Cal Identity Proofs

**Purpledog100** · February 3rd 2009, 02:46 PM

Sorry to ask again, but I need help with these problems:

(secx+1)/tanx = sinx/(1-cosx)
I got (1+cosx)/sinx which is the opposite of what I need.

tan^4x + tan^2x = sec^4 - sec^2x

tan^2x - sin^2x = tan^2x * sin^2x

sinx/(1+cosx) + (1+cosx)/sinx = 2cscx

I could only figure out about half of my homework and I would ask the teacher, but she isn't at school in the morning for tutoring.

**Danny** · February 3rd 2009, 02:59 PM

Originally Posted by Purpledog100

Sorry to ask again, but I need help with these problems:

(secx+1)/tanx = sinx/(1-cosx)
I got (1+cosx)/sinx which is the opposite of what I need. (This is good)

tan^4x + tan^2x = sec^4 - sec^2x

tan^2x - sin^2x = tan^2x * sin^2x

sinx/(1+cosx) + (1+cosx)/sinx = 2cscx

I could only figure out about half of my homework and I would ask the teacher, but she isn't at school in the morning for tutoring.

Now multiply top and bottom by $1- \cos x$

$\frac{1 + \cos x}{\sin x} \cdot \frac{1 - \cos x}{1 - \cos x}$

expand the numerator, use some identites and simplify.

**skeeter** · February 3rd 2009, 03:02 PM

(secx+1)/tanx = sinx/(1-cosx)
I got (1+cosx)/sinx which is the opposite of what I need.

go from there ...

$\frac{1+\cos{x}}{\sin{x}} \cdot \frac{1 - \cos{x}}{1 - \cos{x}} =$

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

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

$\frac{\sin{x}}{1 - \cos{x}}$

**Amanda H** · February 3rd 2009, 03:11 PM

For the third one:

$\tan^2 x -\sin^2 x$

$= \frac{\sin^2 x}{\cos^2 x} - \tan^2 x \cos^2 x$
$= \frac{\sin^2 x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \cos^2x$
$= \frac{\sin^2 x - \sin^2 x \cos^2 x} {\cos^2 x}$
$= \frac {\sin^2 x (1- \cos^2 x)}{\cos^2 x}$
$= \frac {\sin^2 x}{\cos^2 x} (1- \cos^2 x)$
$= \tan^2 x \sin^2 x$

**Danny** · February 3rd 2009, 03:19 PM

Originally Posted by danny arrigo

Now multiply top and bottom by $1- \cos x$

$\frac{1 + \cos x}{\sin x} \cdot \frac{1 - \cos x}{1 - \cos x}$

expand the numerator, use some identites and simplify.

For the second, factor a $\tan^2 x$ from the LHS, i.e.

$\tan^2x \left( 1 + \tan^2x \right)$

then use the identity

$1 + \tan^2x = \sec^2x$ to eliminate the tan

For the forth one, multiply the first term by

$\frac{1 - \cos x}{1 - \cos x}$

then follow the idea what skeeter did above.

**Purpledog100** · February 3rd 2009, 03:43 PM

Originally Posted by Amanda H

For the third one:

$\tan^2 x -\sin^2 x$

$= \frac{\sin^2 x}{\cos^2 x} - \tan^2 x \cos^2 x$
$= \frac{\sin^2 x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \cos^2x$
$= \frac{\sin^2 x - \sin^2 x \cos^2 x} {\cos^2 x}$
$= \frac {\sin^2 x (1- \cos^2 x)}{\cos^2 x}$
$= \frac {\sin^2 x}{\cos^2 x} (1- \cos^2 x)$
$= \tan^2 x \sin^2 x$

Why did you change $-sin^2x$ to $-tan^2x*cos^2x$

For the forth one, multiply the first term by

then follow the idea what skeeter did above.

I'm sorry, but I still don't understand how to do it. I have:
${sinx(1-cosx)}/sin^2x + (1+cosx)/sinx$

Do I multiply the second fraction by sinx to have equal denominators?

**Amanda H** · February 3rd 2009, 05:08 PM

It comes from
$\frac {\sin x}{\cos x} = \tan x$
rearranged it becomes
$\sin x = \tan x \cos x$

**Jhevon** · February 3rd 2009, 05:20 PM

Originally Posted by Amanda H

It comes from
$\frac {\sin x}{\cos x} = \tan x$
rearranged it becomes
$\sin x = \tan x \cos x$

this works, but it's overkill. you could go directly from $\frac {\sin^2 x}{\cos^2 x} - \sin^2 x \to \frac {\sin^2 x - \sin^2 x \cos^2 x}{\cos^2 x}$ . just combining fractions, as we know $\frac ab \pm \frac cd = \frac {ad \pm bc}{bd}$ . from there, it's nice

we could also do

$\frac {\sin^2 x}{\cos^2 x} - \sin^2 x$

$= \sin^2 x \left( \frac 1{\cos^2 x} - 1 \right)$

$= \sin^2 x \cdot \frac {1 - \cos^2 x}{\cos^2 x}$

$= \sin^2 x \tan^2 x$

@ OP: the nice thing about trig identities is that there are usually several ways to solve them. it is good practice to use your imagination to come up with alternative methods of proving the identities than the ones posted here. just take note of the techniques used, like changing everything to sines and cosines. usually a good way to go.

**Purpledog100** · February 3rd 2009, 05:55 PM

Originally Posted by Amanda H

It comes from
$\frac {\sin x}{\cos x} = \tan x$
rearranged it becomes
$\sin x = \tan x \cos x$

Oh, right. Duh. I had that in my notes and still couldn't figure it out.

@ OP: the nice thing about trig identities is that there are usually several ways to solve them. it is good practice to use your imagination to come up with alternative methods of proving the identities than the ones posted here. just take note of the techniques used, like changing everything to sines and cosines. usually a good way to go.

Yeah, but I usually pick the wrong ways to solve them.

Thanks to everyone for all of their help!!

**Jhevon** · February 3rd 2009, 06:42 PM

Originally Posted by Purpledog100

Yeah, but I usually pick the wrong ways to solve them.

that's the point. as long as your moves are mathematically sound, and you can get from one side to the other, there is no wrong way. a problem here can be done several valid ways. the difference is, one way might be easier or quicker than another. if you do enough practice, this won't matter though, as you will either be able to move so quickly that the extra steps hardly slow you down, or you will naturally be able to spot the easier ways first

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