# Evaluate this trigonometric identity

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

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lioric
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

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

lioric
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

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

lioric
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
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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
lioric
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

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?

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

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})##.

lioric
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

Could you emphasis on that last point again

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
lioric

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 into the mix and finding the value at the end

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 into 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?

lioric
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

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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##.

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
lioric
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

lioric
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 into this situation

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

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

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

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Here is what I got so far. Now where do I go from here?View attachment 265531

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

lioric
I know the triangle method. As posted above etotheipi has shown that very well and I understand that.
id also want to know the identity method of doing this because I’m weak at that

In any case I can't see how the second line follows from the first. If anything, it should be ##\frac{\sin^2 x - \sin{2x}}{\cos{x} - \sin^2{x}}##, no?

Delta2
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I know the triangle method. As posted above etotheipi has shown that very well and I understand that.
id also want to know the identity method of doing this because I’m weak at that

Is it really so difficult to express ##\cos x## in terms of ##\tan x##? I gave you the formula in a previous post:

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

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I know the triangle method. As posted above etotheipi has shown that very well and I understand that.
id also want to know the identity method of doing this because I’m weak at that
I'd like to see you get the answer by that method. At least that means you know when you get the right answer by any other method.

lioric
In any case I can't see how the second line follows from the first. If anything, it should be ##\frac{\sin^2 x - \sin{2x}}{\cos{x} - \sin^2{x}}##, no?
How did that cos become a sin?
The cos was given in the question

You changed $$\frac{\sin{x} - 2\cos{x}}{\cot{x}-\sin{x}}$$to$$\frac{\sin^2{x} - 2\cos{x}}{\cos{x} - \sin{x}}$$

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How did that cos become a sin?
No, 2 cos x got multiplied by sin x to give sin 2x.

lioric
there I corrected that part

and this is cos in terms of cos

What do I do frome here

Last edited:
lioric
I’m sort of stuck here.

Your expression is still not right, since you didn't distribute the ##\sin{x}## to both terms in the denominator.

If you still want to do it the long way, you need to find ##\sin{x}## and ##\cos{x}## in terms of ##\tan{x}##. To get you started,$$\sec^2{x} = 1+\tan^2{x}$$ $$\cos^2{x} = \frac{1}{1+\tan^2{x}}$$ $$\sin^2{x} = 1- \frac{1}{1+\tan^2{x}}$$Simplify those, N.B. you need to use the quadrants to figure out the signs after you take the square root.

Delta2 and PeroK