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Evaluate this trigonometric identity

  • #26
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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
 
  • #27
etotheipi
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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?
 
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  • #28
PeroK
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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##.
 
  • #29
PeroK
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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.
 
  • #30
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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
 
  • #31
etotheipi
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You changed $$\frac{\sin{x} - 2\cos{x}}{\cot{x}-\sin{x}}$$to$$\frac{\sin^2{x} - 2\cos{x}}{\cos{x} - \sin{x}}$$
 
  • #32
haruspex
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How did that cos become a sin?
No, 2 cos x got multiplied by sin x to give sin 2x.
 
  • #33
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there I corrected that part

89D8BD67-489C-43AE-8C85-0AA2FE682F9B.png


and this is cos in terms of cos
F95D0FA8-BBAF-4428-8AFA-AD6E0D39A2E5.png


What do I do frome here
 
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  • #34
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I’m sort of stuck here.
 
  • #35
etotheipi
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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.
 
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  • #36
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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.
Thank you very much
 
  • #37
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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.
I know you might say
” I already told this guy how to do this. Why wouldn’t he just go and finish the question the way we taught him”
its just that I see how lot of you guys simply with such ease, and I also know that there are several methods to doing one question, and I just feel I’d understand the concept better when I do it from different methods to converge on the answer. So please don’t be frustrated with me.
i thank you for your patience with me
 
  • #38
etotheipi
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I know you might say
” I already told this guy how to do this. Why wouldn’t he just go and finish the question the way we taught him”
its just that I see how lot of you guys simply with such ease, and I also know that there are several methods to doing one question, and I just feel I’d understand the concept better when I do it from different methods to converge on the answer. So please don’t be frustrated with me.
i thank you for your patience with me
Well sure, but the idea is that given enough time to google the formulae and do the algebra I'm pretty sure you would be able to get the answer the long way. It's more helpful for you to learn more efficient ways to solve things like this.
 
  • #39
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Well sure, but the idea is that given enough time to google the formulae and do the algebra I'm pretty sure you would be able to get the answer the long way. It's more helpful for you to learn more efficient ways to solve things like this.
I know
i just have a hard time figuring out which goes where.
 
  • #40
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Just to clarify, am I on the right track so far?
9B058EE2-EDCE-46DE-BF08-F981BAEF2E40.png
 
  • #41
etotheipi
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You still haven't distributed the ##\sin{x}## into the denominator
 
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  • #42
PeroK
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Just to clarify, am I on the right track so far?
View attachment 265540
You were better off with the function you started with. Generating ##\sin(2x)## is a step in the wrong direction altogether.
 
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  • #43
etotheipi
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Yes I'm not even sure why you would want to multiply by ##\frac{\sin{x}}{\sin{x}}## in the first place
 
  • #44
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ok ppl
let me start off again
 
  • #45
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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.
Wait so are you saying that I have to root these things, and use the quadrants, find the value and signs, and substitute it in to the main question?
It's pretty much the same as the triangle version right?
 
  • #46
etotheipi
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It's pretty much the same as the triangle version right?
You bet
1593524544041.png
 
  • #47
PeroK
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Wait so are you saying that I have to root these things, and use the quadrants, find the value and signs, and substitute it in to the main question?
It's pretty much the same as the triangle version right?
You can plug those things in and try to simplify. Maybe you get a nice simple answer and maybe you don't. I suspect that in this case you won't get a nice simple formula in terms of ##\tan x##.
 
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  • #48
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So there is no way to do this without plugging values right?
I was thinking of simplify it without any values
But as you said we need the values to proceed. So just like the triangle. Thank you very much.
 
  • #49
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Everytime I come here, my mind flourishes. I'd like to thank all the people here who help people like us to get our brains thinking like that
 
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  • #50
FactChecker
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Even if you can find trig identities to solve the problem, that is not what the problem was expecting you to do. The solution where you use the 3,4,5 right triangle (or determine an angle from trig tables) is the fundamental thing to learn and what the problem expected. Do not go down the "rabbit hole" of trig identities when that is not necessary.
 
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