Evaluate this trigonometric identity

[SammyS]

Homework Statement:: If tanx = 4/3 and π < x < 3π/2 , evaluate the following
(Sinx-2cosx) / (cotx-sinx)
Relevant Equations:: Tan x=sinx/cosx
Cotx=1/tan

One of the difficulties you have stems from incorrect algebra.

(Sinx-2cosx)/ (cotx - sinx)
Substitute tan instead of cot

You did this fine in your "blackboard" image. You get:

$\displaystyle \frac{\sin x - 2\cos x}{\dfrac{1}{\tan x}-\sin x}$

(Tanx(sinx-2cosx)/(1-sinx)

That result is incorrect. Presumably you are multiplying the numerator and denominator by tan x , but you didn't distribute tan x through the denominator. Distributing correctly gives:

$\displaystyle \frac{\tan x (\sin x - 2\cos x)}{1 - \tan x \cdot \sin x}$ , which you might as well write as:

$\displaystyle \tan x \ \ \frac{\sin x - 2\cos x}{1 - \tan x \cdot \sin x}$

This gets you close to what @PeroK has been suggesting. Just multiply numerator and denominator by sec(x), also known as 1/cos(x) .

 

[lioric]

One of the difficulties you have stems from incorrect algebra.
You did this fine in your "blackboard" image. You get:

$\displaystyle \frac{\sin x - 2\cos x}{\dfrac{1}{\tan x}-\sin x}$
That result is incorrect. Presumably you are multiplying the numerator and denominator by tan x , but you didn't distribute tan x through the denominator. Distributing correctly gives:

$\displaystyle \frac{\tan x (\sin x - 2\cos x)}{1 - \tan x \cdot \sin x}$ , which you might as well write as:

$\displaystyle \tan x \ \ \frac{\sin x - 2\cos x}{1 - \tan x \cdot \sin x}$

This gets you close to what @PeroK has been suggesting. Just multiply numerator and denominator by sec(x), also known as 1/cos(x) .

That made a lot of sense. I was trying to simply do the reciprocal since it was 1/tan. But the -sinx is the issue here.
Thank you very much

 

[SammyS]

That made a lot of sense. I was trying to simply do the reciprocal since it was 1/tan. But the -sinx is the issue here.
Thank you very much

I'm not sure what you mean by:
−sin x is the issue here.

In post #33 you had posted two images. Here's a snip from the first:

The last line there should be $\dfrac{\sin^2 x - 2\sin{x}\cos x}{\cos{x} - \sin^2{x}}$, which is equivalent to what @etotheipi posted in Post #27.

Presumably, you intended to multiply the numerator and denominator of original expression by sin x , but you failed to distribute that in the denominator.

Algebra !

 

[neilparker62]

Homework Statement:: If tanx = 4/3 and π < x < 3π/2 , evaluate the following
(Sinx-2cosx) / (cotx-sinx)
Relevant Equations:: Tan x=sinx/cosx
Cotx=1/tan

(Sinx-2cosx)/ (cotx - sinx)
Substitute tan instead of cot
(Tanx(sinx-2cosx)/(1-sinx)
What do I do from here
I don't think what I did there is correct
That's why I didn't expand the tan to sin/cos

You could divide through by cos(x) and then most of the terms would be known from the given(s). Just need one additional ratio which you can extract via Pythagoras thm (and quadrant data) as others have advised.

 

[kuruman]

Fully aware that too many cooks spoil the broth, I will summarize a straightforward solution based on statements that have been by others, but is independent of the ratio of the sides of the implied right triangle. My intent is to help wrap up this thread.
1. Let $u=\tan x$.
2. Because angle $x$ lies in the third quadrant, both $\sin x$ and $\cos x$ are negative. We write$$\sin x=-\frac{u}{\sqrt{1+u^2}}~;~~~\cos x=-\frac{1}{\sqrt{1+u^2}}.$$We can deduce these expressions by noting that the sum of their squares must be $1$ and that their ratio must be $u.$
3. Use the given value $u=\frac{4}{3}$ to find numerical values for the sine and cosine and substitute them in the expression remembering that two negative signs make a positive sign.
4. Simplify the fraction to a ratio of integers.

 

[neilparker62]

From "chef WA"

and @SammyS I must admit the suggestion I made came from WA in the first place.