# Homework Help: Help Trig Equation (Again )

1. Jun 11, 2006

### 01suite

I have another equation that i dont know how to solve

Prove:

1-sin2x/cos2x = 1-tanx/1+tanx

thanks in advance to whoever that's solving it, thanks alot

2. Jun 11, 2006

### Hurkyl

Staff Emeritus
That equation is wrong -- I suspect you forgot to use parentheses.

What have you tried on this problem?

3. Jun 11, 2006

### TD

Could you clarify the problem by using parenthesis?
Note that a+b/c+d is not the same as (a+b)/(c+d).

4. Jun 11, 2006

### 01suite

the equation should be...

(1-sin2x)/(cos2x) = (1-tanx)/(1+tanx)

sorry//./

5. Jun 11, 2006

### TD

Now, there are a couple approaches - for example:

- starting at the RHS: converting the tan(x)'s to sin(x)/cos(x) and simplifying.
- starting at the LHS: using the double-angle formulas to go to the single angle x.

Give it a shot

6. Jun 11, 2006

### 01suite

i tried it but i got different answer for both sides...can someone teach me the steps on solving it. ><

7. Jun 11, 2006

### Hurkyl

Staff Emeritus

8. Jun 11, 2006

### 01suite

umm...

L.S = (1- sin2x)/cos2x
= 1- [sin (x+x)]/[cos (x+x)]
= (1- 2sinxcosx)/(2cos^2x -1)
= (1- 2sinxcosx)/(1-2sin^2x)
..then i dunno how to simplify that anymore

then...

R.S = (1-tanx)/(1+tanx)
= [1-(sinx/cosx)]/[1+(sinx/cosx)]
= [1-(sinx/cosx)] x [1+(cosx/sinx)] <--(???)
= 1+ (cosx/sinx) - (sinx/cosx)
= 1 + [(cos^2x-sin^2x)/sinxcosx]
= 1 + [(2cos^2x -1)/sinxcosx]
then...i dunno how to solve it after that either...

9. Jun 11, 2006

### Hurkyl

Staff Emeritus
Okay, then set them equal, and see if you can simplify that.

If you can, then once you get to something true, you can get the answer by reversing your work!

But, you still have to correct your work on the R.H.S.

$$\frac{A}{B + \frac{1}{C}} \neq A * (B + C)$$

You can't just flip part of the denominator -- if you want to flip it to turn division it into multiplication, the denominator must be a single fraction, not a sum of things.

Last edited: Jun 11, 2006
10. Jun 11, 2006

### TD

The line with (???) is wrong indeed.
First, put the entire denominator on 1 single fraction, as well as the numerator.
Then you can apply (a/b)/(c/d) = (a/b)*(d/c).

11. Jun 11, 2006

### 01suite

how do you put the entire denominator on 1 single fraction...like you mean...

1/1+(cosx/sinx)?

12. Jun 11, 2006

### TD

Like this: 1+x/y = y/y+x/y = (y+x)/y.

You make the denominators the same, then you can add the numerators.

13. Jun 11, 2006

### 01suite

ohh thankss im gonna go try it again thanks alot =D

14. Jun 11, 2006

### HallsofIvy

I think I would have been inclined to use cos(2x)= cos^2(x)- sin^2(x)=
(cos(x)- sin(x))(cos(x)+ sin(x)).
And, of course, 1= sin^2(x)+ cos^2(x) so 1- 2sin(x)cos(x)= cos^2(x)- 2sin(x)cos(x)+ sin^2(x)= (cos(x)- sin(x))^2.
Then the left hand side is (cos(x)- sin(x))^2/((cos(x)- sin(x))(cos(x)+ sin(x))= (cos(x)- sin(x)/(cos(x)+ sin(x))

Doesn't multiplying both numerator and denominator by cos(x) make sense here? [cos(x)- sin(x)]/(cos(x)+ sin(x)) and I think you're done!

Last edited by a moderator: Jun 11, 2006
15. Jun 11, 2006

### 01suite

so yea..i got the R.S..i got -1 as the answer...it looks right....but i still dont get the LS
L.S = (1- sin2x)/cos2x
= 1- [sin (x+x)]/[cos (x+x)]
= (1- 2sinxcosx)/(2cos^2x -1)
= (1- 2sinxcosx)/(1-2sin^2x)

wut do i do after that ??

and one more question...wut do they mean when they say...

determine cos4x-cos2x in terms of cosx

Last edited: Jun 11, 2006
16. Jun 12, 2006

### TD

Just -1 for the RHS? Perhaps you should show your work there.

On the other question: it means you have to convert the multiple angles (4x and 2x) to single angles (x), which will produce higher powers.