1. ## [SOLVED] trig differentiating

d/dx (cosx+sinx/cosx-sinx)

I'm having a bit of trouble with trying to prove LHS = RHS (or the other way around for that matter)

I have simplified the LHS to be 2/1-2cosxsinx but how do i get the RHS?

2. Originally Posted by ipokeyou
d/dx (cosx+sinx/cosx-sinx)

I'm having a bit of trouble with trying to prove LHS = RHS (or the other way around for that matter)

I have simplified the LHS to be 2/1-2cosxsinx but how do i get the RHS?
1. You don't have a right hand side, so how can you prove that LHS=RHS.

Is it $\frac{d}{dx}\left(\frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}\right)$?

3. ah yes sorry, i left out the RHS which is meant to be sec^2(pi/4 + x)

and yes the LHS is as you quoted. sorry for the confusion

4. Originally Posted by ipokeyou
ah yes sorry, i left out the RHS which is meant to be sec^2(pi/4 + x)

and yes the LHS is as you quoted. sorry for the confusion
Dear ipokeyou,

$\frac{d}{dx}\left(\frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}\right)=sec^2\left(\frac{\pi}{4} + x\right)$

Am I correct??

5. Originally Posted by Sudharaka
Dear ipokeyou,

$\frac{d}{dx}\left(\frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}\right)=sec^2\left(\frac{\pi}{4} + x\right)$

Am I correct??
yes you are correct

6. Dear

Yep. I must be correct, since using the above expression I managed to show that the RHS=LHS.

First consider,

$\frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}=\frac{sin\left(\frac{\pi}{2}-x\right)+sinx}{sin\left(\frac{\pi}{2}-x\right)-sinx}=\frac{2sin\frac{\pi}{4}cos\left(\frac{\pi}{4 }-x\right)}{2cos\frac{\pi}{4}sin\left(\frac{\pi}{4}-x\right)}=\frac{cos\left(\frac{\pi}{4}-x\right)}{sin\left(\frac{\pi}{4}-x\right)}$

$=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)$

I hope you could continue from here!!

7. Originally Posted by Sudharaka
Dear

Yep. I must be correct, since using the above expression I managed to show that the RHS=LHS.

First consider,

$\frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}=\frac{sin\left(\frac{\pi}{2}-x\right)+sinx}{sin\left(\frac{\pi}{2}-x\right)-sinx}=\frac{2sin\frac{\pi}{4}cos\left(\frac{\pi}{4 }-x\right)}{2cos\frac{\pi}{4}sin\left(\frac{\pi}{4}-x\right)}=\frac{cos\left(\frac{\pi}{4}-x\right)}{sin\left(\frac{\pi}{4}-x\right)}$

$=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)$

I hope you could continue from here!!
Could i ask, how did you get:

$\frac{2sin\frac{\pi}{4}cos\left(\frac{\pi}{4}-x\right)}{2cos\frac{\pi}{4}sin\left(\frac{\pi}{4}-x\right)}$

and also i think you meant:

$=\cot\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)$

8. Originally Posted by ipokeyou
Could i ask, how did you get:

$\frac{2sin\frac{\pi}{4}cos\left(\frac{\pi}{4}-x\right)}{2cos\frac{\pi}{4}sin\left(\frac{\pi}{4}-x\right)}$

and also i think you meant:

$=\cot\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)$

Dear ipokeyou,

1) Do you know the sum-to-product rule?? If not better refer: Sum-to-Product formulas (Proof of these formulas could be found in many basic trignometric books.)

2) $\frac{cos\left(\frac{\pi}{4}-x\right)}{sin\left(\frac{\pi}{4}-x\right)}=\cot\left(\frac{\pi}{4}-x\right)=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)$

9. Originally Posted by Sudharaka
Dear ipokeyou,

1) Do you know the sum-to-product rule?? If not better refer: Sum-to-Product formulas (Proof of these formulas could be found in many basic trignometric books.)
ah yes, its just i havent learnt the sum-to-product rules yet but i get it. would there be a way to simplify it without it? just wondering.

and

i still dont get how you get from $\cot\left(\frac{\pi}{4}-x\right)=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)$

sorry for these silly questions

10. Originally Posted by ipokeyou
ah yes, its just i havent learnt the sum-to-product rules yet but i get it. would there be a way to simplify it without it? just wondering.

and

i still dont get how you get from $\cot\left(\frac{\pi}{4}-x\right)=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)$

sorry for these silly questions
See Complementary Angle Identities: Proofs of trigonometric identities - Wikipedia, the free encyclopedia

11. Originally Posted by mr fantastic
ah thanks for the reminder!! i got it now thanks