Trigonometric Equation: Solving for x with Multiple Angles

[01suite]

I have another equation that i don't know how to solve

Prove:

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


thanks in advance to whoever that's solving it, thanks a lot

 

[Hurkyl]

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

What have you tried on this problem?

 

[TD]

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

 

[01suite]

ohh my badd ..

the equation should be...

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

sorry//./

 

[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

 

[01suite]

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

 

[Hurkyl]

I repeat, show your work.

 

[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 don't know 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 don't know how to solve it after that either...

 

[Hurkyl]

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.

 

[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).

 

[01suite]

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

1/1+(cosx/sinx)?

 

[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.

 

[01suite]

ohh thankss I am going to go try it again thanks a lot =D

 

[HallsofIvy]

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 don't know how to simplify that anymore

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))

then...

R.S = (1-tanx)/(1+tanx)
= [1-(sinx/cosx)]/[1+(sinx/cosx)

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!

 

[01suite]

so yea..i got the R.S..i got -1 as the answer...it looks right...but i still don't 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

 

[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.