# Evaluate this trigonometric identity

## 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)
(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

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PeroK
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Have you ever heard of a 3-4-5 triangle?

• FactChecker
Have you ever heard of a 3-4-5 triangle?
Ya triangles which have their lengths in that ratio which fits them into pythagorean triples

etotheipi
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So can you draw a triangle involving the angle ##x##?

So can you draw a triangle involving the angle ##x##?
I can understand what you're saying because tanx = 4/3 where 4 is opposite and 3 is adjacent
Ok I have drawn a triangle

etotheipi
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So now can you evaluate ##\sin{x}## and ##\cos{x}##?

So now can you evaluate ##\sin{x}## and ##\cos{x}##?
Sinx = 4/5
Cox = 3/5
So that question is asking me to just put in those values???

Is there a trigonometric identity method for this where we substitute tan and stuff???

• PeroK
PeroK
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Sinx = 4/5
Cox = 3/5
So that question is asking me to just put in those values???

Is there a trigonometric identity method for this where we substitute tan and stuff???
On a given range, ##\tan x## is one-to-one. If you know the value of ##\tan x## you know the value of ##x##, hence you know the value of ##\sin x## and ##\cos x##. All you need is Pythagoras.

Do you know the term "quadrant"?

• etotheipi
On a given range, ##\tan x## is one-to-one. If you know the value of ##\tan x## you know the value of ##x##, hence you know the value of ##\sin x## and ##\cos x##. All you need is Pythagoras.

Do you know the term "quadrant"?
Ya that circle thing which shows which of the trigonometric function are positive or negative

etotheipi
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Ya that circle thing which shows which of the trigonometric function are positive or negative
So what are the signs of those functions in the third quadrant?

So what are the signs of those functions in the third quadrant?
180 and 270 the third quadrant

etotheipi
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180 and 270 the third quadrant
Indeed, but if the angle to the positive ##x## axis of a line segment from the origin to a point on the unit circle is ##\theta##, then the coordinates of that point are ##(\cos{\theta}, \sin{\theta})##.

Indeed, but if the angle to the positive ##x## axis of a line segment from the origin to a point on the unit circle is ##\theta##, then the coordinates of that point are ##(\cos{\theta}, \sin{\theta})##.

Could you emphasis on that last point again

etotheipi
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Could you emphasis on that last point again  That should help you to work out the signs!  Just kidding. The question tells you that the angle is in the third quadrant, and from the unit circle you can easily determine whether each of sin/cos/tan is positive or negative in that quadrant. Some people call it a "CAST diagram": • • DifferentialGalois and archaic  That should help you to work out the signs!  Just kidding. The question tells you that the angle is in the third quadrant, and from the unit circle you can easily determine whether each of sin/cos/tan is positive or negative in that quadrant.
Ahh ok

Now I like this method
But I was wondering if there is another method of doing this by a simplification sort of way like by substituting tan in to the mix and finding the value at the end

• PeroK
PeroK
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Ahh ok

Now I like this method
But I was wondering if there is another method of doing this by a simplification sort of way like by substituting tan in to the mix and finding the value at the end
You mean a method where you don't have to think, but just look up a formula and plug in the numbers?

You mean a method where you don't have to think, but just look up a formula and plug in the numbers?
No no no. Not like that. Like you put in tan expand it to sin/cos and cancel some stuff like that

PeroK
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No no no. Not like that. Like you put in tan expand it to sin/cos and cancel some stuff like that
It should be fairly obvious how to do that. Remember that ##\sin^2 x + \cos^2 x = 1##.

And ##\sec^2 x = 1 + \tan^2 x##.

etotheipi
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2019 Award
Sometimes a few right angled triangles can save you from working with lots of identities. For instance, consider the Weierstrass substituion, ##t = \tan{\frac{\theta}{2}}##: Of course, you still need to remember some double angle formulae.

• PeroK
Sometimes a few right angled triangles can save you from working with lots of identities. For instance, consider the Weierstrass substituion, ##t = \tan{\frac{\theta}{2}}##:

View attachment 265530
Of course, you still need to remember some double angle formulae.
Yes I know this is a very nice method. But in this case this is very specific for this question because of the parameters given.
But I'm weak in the identity department. So I'd like to work on that too. Could you please show me

It should be fairly obvious how to do that. Remember that ##\sin^2 x + \cos^2 x = 1##.

And ##\sec^2 x = 1 + \tan^2 x##.
Could you show me how that fits in to this situation

PeroK
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Could you show me how that fits in to this situation
I think it's time you made your best effort to solve this problem yourself.

I think it's time you made your best effort to solve this problem.
Ok thank you I'll post my progress

• etotheipi
Ok thank you I'll post my progress
Here is what I got so far. Now where do I go from here? PeroK