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Inverse trigonometry question

  • #1
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Homework Statement


If 1/2<=x<=1
, then cos-1[ x/2 + (√(3-3x2))/2] + cos-1 x is equal to?



Homework Equations




The Attempt at a Solution


Substituting x=cosθ
= cos-1(cosθ) + cos-1 [ cos(θ)/2 + (√(3(1-cos2(θ)))/2]
= θ + cos-1 [cos (π/3) cos θ + sin (π/3) sinθ]
= θ + cos-1 [cos (θ -(π/3)]
= 2θ -(π/3)
= 2 cos-1x -(π/3)
But the correct answer according to the book is π/3. Please help me understand what I've done wrong.
 
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Answers and Replies

  • #2
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Homework Statement


1/2<=x<=1
cos-1[ x/2 + (√3-3x2)/2]] + cos-1 x
The square root is for the entire term.
Then use parentheses to indicate this, as in √(3-3x2)/2. From what you wrote I can't tell if you mean this:
$$\frac{\sqrt{3 - 3x^2}} 2$$

or this:$$\sqrt{\frac{3 - 3x^2}{2}}$$

Also, what is the question? The problem statement should include what it is you need to do.
takando12 said:

Homework Equations




The Attempt at a Solution


Substituting x=cosθ
= cos-1(cosθ) + cos-1 [ cosθ/2 + (√3(1-cos2θ))/2]
= θ + cos-1 [cos π/3 cos θ + sinπ/3 sinθ]
= θ + cos-1 [cos (θ -π/3)]
= 2θ -π/3
= 2 cos-1x -π/3
But the correct answer according to the book is π/3. Please help me understand what I've done wrong.
 
  • #3
123
5
Then use parentheses to indicate this, as in √(3-3x2)/2. From what you wrote I can't tell if you mean this:
$$\frac{\sqrt{3 - 3x^2}} 2$$

or this:$$\sqrt{\frac{3 - 3x^2}{2}}$$

Also, what is the question? The problem statement should include what it is you need to do.
It is the first one. Square root for the numerator.
We are asked to find what that expression is equal to. I've reworded it now.
 
  • #4
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4,998
takando12 said:
Substituting x=cosθ
= cos-1(cosθ) + cos-1 [ cosθ/2 + √(3(1-cos2θ))/2]
= θ + cos-1 [cos π/3 cos θ + sinπ/3 sinθ]
How did you get the term on the right, above? I don't see where cos(π/3) and sin(π/3) come in.

I haven't worked the problem yet, but I suspect that your substitution x = cos(θ) isn't the best one to use.
takando12 said:
= θ + cos-1 [cos (θ -π/3)]
= 2θ -π/3
= 2 cos-1x -π/3
But the correct answer according to the book is π/3. Please help me understand what I've done wrong.
 
  • #5
123
5
How did you get the term on the right, above? I don't see where cos(π/3) and sin(π/3) come in.

I haven't worked the problem yet, but I suspect that your substitution x = cos(θ) isn't the best one to use.
1/2= cosπ/3
√3/2 =sinπ/3
So I thought i'll plug those in and use the cos(x+y)=cosxcosy -sinxsiny.
 
  • #6
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1/2= cosπ/3
√3/2 =sinπ/3
So I thought i'll plug those in and use the cos(x+y)=cosxcosy -sinxsiny.
My question is this: How does cosθ/2 + √(3(1-cos2θ))/2 become cos π/3 cos θ + sinπ/3 sinθ?

Also, you are continuing to write things ambiguously. When you write cosθ/2 do you mean ##\cos(\frac{\theta}{2})## or ##\frac{\cos(\theta)}{2}##? And the same question for the sine term.
Use parentheses to make things clear!
 
  • #7
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My question is this: How does cosθ/2 + √(3(1-cos2θ))/2 become cos π/3 cos θ + sinπ/3 sinθ?

Also, you are continuing to write things ambiguously. When you write cosθ/2 do you mean ##\cos(\frac{\theta}{2})## or ##\frac{\cos(\theta)}{2}##? And the same question for the sine term.
Use parentheses to make things clear!
it's ##\frac{\cos(\theta)}{2}## .
I'm sorry, I've made more corrections to the question now. Is it clear?
 
  • #8
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4,998
It's clearer now, but I still don't understand something.
takando12 said:
Substituting x=cos
cos-1(cosθ) + cos-1 [ cos(θ)/2 + (√(3(1-cos2(θ)))/2]
= θ + cos-1 [cos π/3 cos θ + sinπ/3 sinθ]
How does this -- cos(θ)/2 + (√(3(1-cos2(θ)))/2 -- turn into this -- cos π/3 cos θ + sinπ/3 sinθ?
 
  • #9
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It's clearer now, but I still don't understand something.

How does this -- cos(θ)/2 + (√(3(1-cos2(θ)))/2 -- turn into this -- cos π/3 cos θ + sinπ/3 sinθ?
cos(θ) * 1/2 + sin(θ) *√3/2 ====> cos (θ) cos (π/3) + sin (θ) sin(π/3).
as √(1-cos2θ) =sin(θ)
Few more corrections added to the question.
 
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  • #10
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OK, now I see. One thing you didn't take into account when you went from ##\sqrt{1 - \cos^2(\theta)}\frac{\sqrt{3}} 2## to ##\sin(\theta) \frac{\sqrt{3}} 2## is that ##\sqrt{1 - \cos^2(\theta)}## isn't equal to ##\sin(\theta)## -- it's equal to ##|\sin(\theta)|##.
If ##0 \le \theta \le \frac{\pi} 3## then ##\sin(\theta) \ge 0## but if ##-\frac{\pi} 3 \le \theta \le 0##, then ##\sin(\theta) \le 0##.

I'm not sure if that's where the problem lies -- I still haven't worked through this problem. I've just been trying to follow what you're doing.
 
  • #11
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OK, now I see. One thing you didn't take into account when you went from ##\sqrt{1 - \cos^2(\theta)}\frac{\sqrt{3}} 2## to ##\sin(\theta) \frac{\sqrt{3}} 2## is that ##\sqrt{1 - \cos^2(\theta)}## isn't equal to ##\sin(\theta)## -- it's equal to ##|\sin(\theta)|##.
If ##0 \le \theta \le \frac{\pi} 3## then ##\sin(\theta) \ge 0## but if ##-\frac{\pi} 3 \le \theta \le 0##, then ##\sin(\theta) \le 0##.

I'm not sure if that's where the problem lies -- I still haven't worked through this problem. I've just been trying to follow what you're doing.
Well according tot he question 1/2<=x<=1
So, (π/3) ≤(cos-1x or θ) ≤ 0 ? That doesn't seem right. The inequality signs seem to be reversed. Am I missing something?
 
  • #12
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Well according tot he question 1/2<=x<=1
So, (π/3) ≤(cos-1x or θ) ≤ 0 ? That doesn't seem right. The inequality signs seem to be reversed. Am I missing something?
I wrote ##-\frac{\pi} 3 \le \theta \le 0##

In your substitution, ##x = \cos(\theta)##, and you're given that 1/2 <= x <= 1. If you graph ##x = \cos(\theta)##, the usual thing to do would be to have x on the vertical axis, and ##\theta## the horizontal axis. If ##1/2 \le x \le 1##, then one arch of the cosine graph gives the interval for ##\theta## that I wrote.
 
  • #13
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I wrote ##-\frac{\pi} 3 \le \theta \le 0##

In your substitution, ##x = \cos(\theta)##, and you're given that 1/2 <= x <= 1. If you graph ##x = \cos(\theta)##, the usual thing to do would be to have x on the vertical axis, and ##\theta## the horizontal axis. If ##1/2 \le x \le 1##, then one arch of the cosine graph gives the interval for ##\theta## that I wrote.
Looking at the cos-1 graph, the corresponding interval for 1/2≤x≤1 is 0≤θ≤π/3. How is it −π3≤θ≤0 ?The cos-1 graph's range doesn't even cover those values.
 
  • #14
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Looking at the cos-1 graph, the corresponding interval for 1/2≤x≤1 is 0≤θ≤π/3. How is it −π3≤θ≤0 ?The cos-1 graph's range doesn't even cover those values.
The graph of x = cos(θ), with θ on the horizontal axis and x on the vertical axis, isn't one-to-one, so it doesn't have an inverse that is itself a function. To make x = cos(θ) one-to-one, the domain must be restricted. The usual convention is to restrict the domain to [0, π], but that interval is somewhat arbitrary.

But your substitution was x = cos(θ), with the condition that 1/2 ≤ x ≤ 1. If you look at the graph of x = cos(θ) (which looks exactly like the graph of y = cos(x) ), when 1/2 ≤ x ≤ 1, the x values under the arch closest to the origin lie between -π/3 and +π/3. There are many other concave down arches for whiich 1/2 ≤ x ≤ 1, but I was talking about the one closest to the origin.
 
  • #15
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5
The graph of x = cos(θ), with θ on the horizontal axis and x on the vertical axis, isn't one-to-one, so it doesn't have an inverse that is itself a function. To make x = cos(θ) one-to-one, the domain must be restricted. The usual convention is to restrict the domain to [0, π], but that interval is somewhat arbitrary.

But your substitution was x = cos(θ), with the condition that 1/2 ≤ x ≤ 1. If you look at the graph of x = cos(θ) (which looks exactly like the graph of y = cos(x) ), when 1/2 ≤ x ≤ 1, the x values under the arch closest to the origin lie between -π/3 and +π/3. There are many other concave down arches for whiich 1/2 ≤ x ≤ 1, but I was talking about the one closest to the origin.
I am sorry, but I just can't get it. "x = cos(θ) (which looks exactly like the graph of y = cos(x)" , yes but with the axis reversed. So in the cos-1, graph, are you asking me to pick another section where the function is invertible? ie [0, -π]? ie. other half of the concave section where -π/3≤x≤0 also lies?
How does all this finally tie up with the question.?
 
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  • #16
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I am sorry, but I just can't get it. "x = cos(θ) (which looks exactly like the graph of y = cos(x)" , yes but with the axis reversed. So in the cos-1, graph, are you asking me to pick another section where the function is invertible? ie [0, -π]? ie. other half of the concave section where -π/3≤x≤0 also lies?
How does all this finally tie up with the question.?
This side discussion is about my response in post #10.
One thing you didn't take into account when you went from ##\sqrt{1 - \cos^2(\theta)}\frac{\sqrt{3}} 2## to ##\sin(\theta) \frac{\sqrt{3}} 2## is that ##\sqrt{1 - \cos^2(\theta)}## isn't equal to ##\sin(\theta)## -- it's equal to ##|\sin(\theta)|##.
##\sqrt{1 - \cos^2(\theta)} = \sqrt{\sin^2(\theta)}##
The last expression above isn't equal to ##\sin(\theta)##; it's equal to ##|\sin(\theta)|##. You have to take into account the possibility that ##\sin(\theta)## could be negative.
 
  • #17
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I am sorry, but I just can't get it. "x = cos(θ) (which looks exactly like the graph of y = cos(x)" , yes but with the axis reversed. So in the cos-1, graph, are you asking me to pick another section where the function is invertible? ie [0, -π]? ie. other half of the concave section where -π/3≤x≤0 also lies?
This side discussion is about my response in post #10.

##\sqrt{1 - \cos^2(\theta)} = \sqrt{\sin^2(\theta)}##
The last expression above isn't equal to ##\sin(\theta)##; it's equal to ##|\sin(\theta)|##. You have to take into account the possibility that ##\sin(\theta)## could be negative.
Yes, I get that, but even if sin θ is negative , it results in the final answer to be 2θ + (π/3). Which is 2cos-1(x) + (π/3).
So considering both cases of sin(θ) being positive or negative doesn't give me the final answer as π/3.
It seems as though the only way is to write this bit as,
= θ + cos-1 [cos (π/3) cos θ + sin (π/3) sinθ]
= θ + cos-1 [cos (θ -(π/3)]
cos ( (π/3) - θ).
Coming to think of it, how do we decide whether to choose cos( (π/3) -θ) or cos ( (π/3) +θ) from that previous expression?
 
  • #18
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cos(A)cos(B) + sin(A)sin(B) = cos(A - B)
 
  • #19
ehild
Homework Helper
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cos(A)cos(B) + sin(A)sin(B) = cos(A - B)
which is the same as cos(B-A). The range of the arccos function is [0, pi], so you have to choose the non-negative one from A-B and B-A.
As the range of the arccos function is [0, pi], 0.5≤x≤1 means 0.5≤cos(θ)≤1, that is 0≤θ≤π/3. In that range of θ, sin(θ)≥0, so
##\sqrt{1-\cos^2(\theta)}=\sin(\theta)##. There is only one possible value for
## \cos^{-1} (\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-\cos^2(x)})##.
 
  • #20
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which is the same as cos(B-A). The range of the arccos function is [0, pi], so you have to choose the non-negative one from A-B and B-A.
As the range of the arccos function is [0, pi], 0.5≤x≤1 means 0.5≤cos(θ)≤1, that is 0≤θ≤π/3. In that range of θ, sin(θ)≥0, so
##\sqrt{1-\cos^2(\theta)}=\sin(\theta)##. There is only one possible value for
## \cos^{-1} (\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-\cos^2(x)})##.
Finally......
I shall summarize it again, please check if it's right.
√(1-cos2θ) = +sin(θ) , as (π/3)≤θ≤0.
= θ + cos-1 [cos (θ -(π/3)]
This part is wrong, as the range of arccos is [0,π] and since (π/3)≤θ≤0, θ-(π/3) is -ve and I can write only the +ve (π/3)-θ.
Just one more question.
means 0.5≤cos(θ)≤1, that is 0≤θ≤π/3
Why are the inequality signs being reversed?
 
  • #21
ehild
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Finally......
I shall summarize it again, please check if it's right.
√(1-cos2θ) = +sin(θ) , as (π/3)≤θ≤0.

This part is wrong, as the range of arccos is [0,π] and since (π/3)≤θ≤0, θ-(π/3) is -ve and I can write only the +ve (π/3)-θ.
Just one more question.

Why are the inequality signs being reversed?
I don't understand what disturbs you. Cosine is monotonously decreasing function in the interval [0, pi] It decreases from 1 to 0 when theta increases from 0 to pi. So cos-1(x) decreases when x increases.
 
  • #22
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I don't understand what disturbs you. Cosine is monotonously decreasing function in the interval [0, pi] It decreases from 1 to 0 when theta increases from 0 to pi. So cos-1(x) decreases when x increases.
Right.Got it. Nothing disturbs me anymore.
Thank you for the help.
Peace.
 
  • #23
ehild
Homework Helper
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I am glad that I could help.
 

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