Proving trigonometric equations:

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April 21st 2009, 02:26 PM
etheen

Can someone help me prove the following identities:

(1+cos^2x/1+tan^2x) tanx=cotx

e) secx-cosx=sinxtanx f)sinxcosx/sinx=1+tanx\tanx g)cosx/sinx-sinx/secx=cos^2xcotx

h)cotx-tanx/cotx+tanx=2cos^2x-1 i)3sin^2x+4cos2x=3+cos^2x j)9sec^2x-5tan^2x=5+4sec^2x

k) (2cos^2x-1)^2/sin^4x-cos^4x=2sin^2x-1 L)(sin^2x/1-cosx)-1=cosx

m)1+cos2x/sin2x=cotx

Okay, here are these ones for now... the rest are not the same so I want to have these down first:) Thank you so much for helping by the way! I wish I was a math wizz too!
April 21st 2009, 10:11 PM
Grandad

Trig Identities

Hello etheen

Quote:

Originally Posted by etheen

Can someone help me prove the following identities:

(1+cos^2x/1+tan^2x) tanx=cotx

This is not true. Try it when $x = \tfrac{\pi}{4}$ for example.

Quote:

e) secx-cosx=sinxtanx

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

Can you do it from here? (Use $\sin^2x + \cos^2x =1$ )

Quote:

f)sinxcosx/sinx=1+tanx\tanx

Again, this is not true. The LHS is $\cos x$ and the RHS is $\cot x + 1$ , and they are not identical.

Quote:

g)cosx/sinx-sinx/secx=cos^2xcotx

$\frac{\cos x}{\sin x}-\frac{\sin x}{\sec x}= \frac{\cos x}{\sin x}-\sin x\cos x$

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

Factorise the numerator and then use $\sin^2x + \cos^2x =1$ .

Hopefully, you've got the idea now, and can do some of the remaining questions yourself. Show us your attempts if you want more help.

Grandad
April 22nd 2009, 12:57 PM
etheen

okay for e) um....1-cos^2x =sin^2x and cosx=cottsint ...okay I'm stuck now:|

by the way what do you mean that f) is not true, grandad?
April 22nd 2009, 01:14 PM
masters

Quote:

Originally Posted by etheen

okay for e) um....1-cos^2x =sin^2x and cosx=cottsint ...okay I'm stuck now:|

Hi etheen,

Here's the finish to (e). Work with the LHS

$\sec x -\cos x={\color{red}\sin x \tan x}$

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

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

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

$\sin x \left(\frac{\sin x}{\cos x}\right)=$

${\color{red}\sin x \tan x}$
April 22nd 2009, 01:42 PM
etheen

Hm, I see alittle!

can youplease exaplin exactly what explains when we do? 1/cosx-cosx?

you do LCD so we had to do 1 times cosx to get the denominator to be the same and once it was the same, we did cosx times cosx because we did that to the bototm right?
April 22nd 2009, 09:58 PM
Grandad

Trig Identities

Hello etheen

Quote:

Originally Posted by etheen

by the way what do you mean that f) is not true, grandad?

You are asking us to prove that

Quote:

f)sinxcosx/sinx=1+tanx\tanx

is true for all values of $x$ - and it simply isn't!

Are you sure you have written the question down correctly?

Grandad
April 23rd 2009, 11:47 AM
masters

Quote:

Originally Posted by masters

Hi etheen,

Here's the finish to (e). Work with the LHS

$\sec x -\cos x={\color{red}\sin x \tan x}$

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

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

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

$\sin x \left(\frac{\sin x}{\cos x}\right)=$

${\color{red}\sin x \tan x}$

Quote:

Originally Posted by etheen

Hm, I see alittle!

can youplease exaplin exactly what explains when we do? 1/cosx-cosx?

you do LCD so we had to do 1 times cosx to get the denominator to be the same and once it was the same, we did cosx times cosx because we did that to the bototm right?

Does this help?

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