Solving for x using 1 - tan(theta) in a Trigonometric Equation

[James Brady]

This problem is:

\frac{sin(45)cos(\theta) - cos(45)sin(\theta)}{x} = 1 - tan(\theta)

I've been trying to find x mostly just by multiplying things by 1-tan(theta), but so far nothing I've tried yields the numerator.

Also, sin(45)cos(\theta) - cos(45)sin(\theta) = sin(45 - \theta), it that helps...

 

[rock.freak667]

If you are to solve for 'x' why can't you just cross-multiply by 'x' and then divide by '1-tanθ '?

Do you need to simplify or something like that?

 

[Mark44]

This problem is:

\frac{sin(45)cos(\theta) - cos(45)sin(\theta)}{x} = 1 - tan(\theta)

I've been trying to find x mostly just by multiplying things by 1-tan(theta), but so far nothing I've tried yields the numerator.

Also, sin(45)cos(\theta) - cos(45)sin(\theta) = sin(45 - \theta), it that helps...

Is your goal to solve for x? To reduce confusion, that's why we ask posters to use the template and enter the complete problem description.

If you're trying to solve for x, start by multiplying both sides of the equation by x.

 

[Curious3141]

This problem is:

\frac{sin(45)cos(\theta) - cos(45)sin(\theta)}{x} = 1 - tan(\theta)

I've been trying to find x mostly just by multiplying things by 1-tan(theta), but so far nothing I've tried yields the numerator.

Also, sin(45)cos(\theta) - cos(45)sin(\theta) = sin(45 - \theta), it that helps...

Note that $\sin(45)\cos(\theta) - \cos(45)\sin(\theta) = \sin(45 - \theta)$ and that $\tan(45 - \theta) = \frac{1 - \tan\theta}{1 + \tan\theta}$.

 

[James Brady]

If you are to solve for 'x' why can't you just cross-multiply by 'x' and then divide by '1-tanθ '?

Do you need to simplify or something like that?

Wow, that seems really obvious to me now that you brought that up, haha. I'm having a pretty bad time doing the long division though... Check it out ->

x = \frac{sin(45)cos(\theta) - cos(45)sin(\theta)}{-tan(\theta) + 1}

I'm doing the leading terms in polynomial long division and I end up with...

\frac{-sin(45)cos(\theta)}{tan(\theta}

Ignoring the sin(45) term, I'm getting \frac{cos(\theta)}{tan(\theta)} = \frac{adjacent}{hypotenuse} * \frac{adjacent}{opposite} = \frac{adjacent^2}{hypotenuse*opposite}

The \frac{adjacent^2}{hypotenuse*opposite} term means nothing to me and doesn't really help solve. It's pretty difficult to figure this one out.

 

[Mentallic]

The \frac{adjacent^2}{hypotenuse*opposite} term means nothing to me and doesn't really help solve. It's pretty difficult to figure this one out.

Which is exactly why that approach won't help you in this problem.

Take a look at Curious3141's post and see what you can do with that.

 

[James Brady]

Take a look at Curious3141's post and see what you can do with that.

Oh I gotcha... because \frac{sine(x)}{cosine(x)} = tan(x)... So I can just divide by cos(45 - \theta) to end up with tan(45 - \theta) which is equal to \frac{1 - tan(\theta)}{1 + tan(\theta)}

Yeah that makes a sense, thanks everyone.

 

[chwala]

I am in agreement with post$4$unless i am missing something here, ...what is the ultimate aim of the problem?express $x$ as subject? ...here we have two unknown variables i.e $θ$ and $x$, from my way of looking at it, we can also have,
$\dfrac {\cos(\theta)-\sin(\theta)}{1-\tan(\theta)}$=$\sqrt{2}$$x$

 

[robphy]

I am in agreement with post$4$unless i am missing something here, ...what is the ultimate aim of the problem?express $x$ as subject? ...here we have two unknown variables i.e $θ$ and $x$, from my way of looking at it, we can also have,
$\dfrac {\cos(\theta)-\sin(\theta)}{1-\tan(\theta)}$=$\sqrt{2}$$x$

The left-hand-side can simplify.

 

[Hall]

Wow, that seems really obvious to me now that you brought that up, haha. I'm having a pretty bad time doing the long division though... Check it out ->

x = \frac{sin(45)cos(\theta) - cos(45)sin(\theta)}{-tan(\theta) + 1}

I'm doing the leading terms in polynomial long division and I end up with...

\frac{-sin(45)cos(\theta)}{tan(\theta}

Ignoring the sin(45) term, I'm getting \frac{cos(\theta)}{tan(\theta)} = \frac{adjacent}{hypotenuse} * \frac{adjacent}{opposite} = \frac{adjacent^2}{hypotenuse*opposite}

The \frac{adjacent^2}{hypotenuse*opposite} term means nothing to me and doesn't really help solve. It's pretty difficult to figure this one out.

If you're are working strictly in a right angled triangle and have assumed one of the angles to be 45 degrees, your $theta$ will get equal to 45 degrees, making you adjacent = opposite.

So, $\frac{\cos \theta }{\tan \theta} = \frac{adjacent^2 }{hypotenuse \times opposite}$
And that would simplify to:
$$
\frac{\cos \theta }{\tan \theta}= \frac{adjacent}{hypotenuse} = \cos \theta = \frac{opposite}{hypotenuse} = \sin \theta$$

[Oh my Lord! What I have written for so long.]

But believe me, it's not something we should really try to make sense.

 

[robphy]

If you're are working strictly in a right angled triangle and have assumed one of the angles to be 45 degrees, your $theta$ will get equal to 45 degrees, making you adjacent = opposite.

There is nothing that forces one to make this assumption about \theta.

It seems to me that the problem in the OP could be the result of a calculation, possibly arising from a rotation of axes by 45 degrees.

If one solves for x in the OP and plots it on desmos, the answer will be revealed... but one should get it by hand.

I think the result by @chwala is the most efficient approach.
There's just one more step.

 

[chwala]

There is nothing that forces one to make this assumption about \theta.

It seems to me that the problem in the OP could be the result of a calculation, possibly arising from a rotation of axes by 45 degrees.

If one solves for x in the OP and plots it on desmos, the answer will be revealed... but one should get it by hand.

I think the result by @chwala is the most efficient approach.
There's just one more step.

Which step is that? I do not seem to see it...

also i would desist from going to old posts unless necessary...this should be my last post on any old posts...i have been told to avoid that...my apologies...

 

[robphy]

Which step is that? I do not seem to see it...

From (\cos(\theta)-\sin(\theta)), factor out \cos(\theta).

 

[chwala]

ok from,
$\dfrac {\cos(\theta)-\sin(\theta)}{1-\tan(\theta)}$=$\sqrt{2}$$x$
$\cos(\theta)$×$\dfrac {\cos(\theta)-\sin(\theta)}{\cos(\theta)-\sin(\theta)}$=$\sqrt{2}$$x$
⇒$\cos(\theta)$=$\sqrt{2}$$x$ therefore
$x$=$\dfrac {\cos(\theta)}{\sqrt{2}}$...
now solving this gives us; When $\theta=0, x=0.707$ this satisfies our original problem in post $1$