Trigonometric Equation: Solving for x with Multiple Angles

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In summary, the conversation involves solving the equation (1-sin2x/cos2x = 1-tanx/1+tanx) and discussing different approaches for solving it. The correct equation is (1-sin2x)/(cos2x) = (1-tanx)/(1+tanx) and one approach is to convert the multiple angles to single angles. The conversation also includes clarifying the equation and showing work for solving it. There is also a question about determining cos4x-cos2x in terms of cosx.
  • #1
01suite
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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 alot:smile:
 
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  • #2
That equation is wrong -- I suspect you forgot to use parentheses.

What have you tried on this problem?
 
  • #3
Could you clarify the problem by using parenthesis?
Note that a+b/c+d is not the same as (a+b)/(c+d).
 
  • #4
ohh my badd ..

the equation should be...

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

sorry//./
 
  • #5
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 :smile:
 
  • #6
i tried it but i got different answer for both sides...can someone teach me the steps on solving it. ><
 
  • #7
I repeat, show your work.
 
  • #8
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...
 
  • #9
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.

[tex]
\frac{A}{B + \frac{1}{C}} \neq A * (B + C)
[/tex]

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:
  • #10
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
how do you put the entire denominator on 1 single fraction...like you mean...

1/1+(cosx/sinx)?
 
  • #12
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
ohh thankss I am going to go try it again thanks a lot =D
 
  • #14
01suite said:
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!
 
Last edited by a moderator:
  • #15
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
 
Last edited:
  • #16
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.
 

1. What is a trigonometric equation?

A trigonometric equation is an equation that involves trigonometric functions (such as sine, cosine, tangent) and an unknown variable. The goal is to solve for the unknown variable and find all possible values that satisfy the equation.

2. What are the basic trigonometric identities?

The basic trigonometric identities include the Pythagorean identities (sin^2x + cos^2x = 1 and tan^2x + 1 = sec^2x), the reciprocal identities (cscx = 1/sinx, secx = 1/cosx, cotx = 1/tanx), and the quotient identities (tanx = sinx/cosx, cotx = cosx/sinx).

3. How do I solve a trigonometric equation?

To solve a trigonometric equation, you can use algebraic techniques such as factoring, expanding, and simplifying. You can also use trigonometric identities to manipulate the equation into a simpler form. Once you have isolated the variable, you can use inverse trigonometric functions to find its values.

4. What are the common mistakes when solving trigonometric equations?

Some common mistakes when solving trigonometric equations include forgetting to check for extraneous solutions (solutions that do not satisfy the original equation), missing solutions due to using restricted domains, and making calculation errors with trigonometric functions. It is important to carefully check your work and verify that all solutions satisfy the original equation.

5. How can I check my solution to a trigonometric equation?

You can check your solution by plugging it back into the original equation and seeing if it satisfies the equation. You can also use a graphing calculator to graph both sides of the equation and see if they intersect at your solution. It is also helpful to use trigonometric identities to simplify and verify your solution.

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