# Inverse trigonometry question

• takando12
In summary, if 1/2<=x<=1, then cos-1[ x/2 + (√(3-3x2))/2] + cos-1 x is equal to θ + cos-1 [cos (π/3) cos θ + sin (π/3) sinθ].

## Homework Statement

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

## 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)

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takando12 said:

## 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:

## 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

Mark44 said:
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.

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

Mark44 said:
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.

takando12 said:
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!

Mark44 said:
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?

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?

Mark44 said:
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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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.

Mark44 said:
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?

takando12 said:
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.

Mark44 said:
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.

takando12 said:
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.

Mark44 said:
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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takando12 said:
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.

takando12 said:
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?
Mark44 said:
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,
takando12 said:
= 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?

cos(A)cos(B) + sin(A)sin(B) = cos(A - B)

Mark44 said:
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)})##.

takando12
ehild said:
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.
takando12 said:
= 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.
ehild said:
means 0.5≤cos(θ)≤1, that is 0≤θ≤π/3
Why are the inequality signs being reversed?

takando12 said:
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.

takando12
ehild said:
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.

I am glad that I could help.

takando12

## What is inverse trigonometry?

Inverse trigonometry is a branch of mathematics that deals with finding the angle measure of a triangle when the lengths of its sides are known. It is also known as arc trigonometry.

## What is the difference between regular trigonometry and inverse trigonometry?

Regular or direct trigonometry deals with finding the side lengths or angle measures of a triangle using the given side lengths and angle measures. Inverse trigonometry, on the other hand, deals with finding the angle measure when the side lengths are known.

## How do I solve an inverse trigonometry question?

To solve an inverse trigonometry question, you first need to identify the given side lengths and angle measures. Then, use the inverse trigonometric function (such as arcsine, arccosine, or arctangent) to find the angle measure. Finally, use the inverse trigonometric ratio to solve for the missing side length.

## What are the common inverse trigonometric functions?

The most common inverse trigonometric functions are arcsine (sin-1), arccosine (cos-1), and arctangent (tan-1). These functions are used to find the angle measure of a triangle when the side lengths are known.

## What are some real-life applications of inverse trigonometry?

Inverse trigonometry is used in various fields such as engineering, physics, and navigation. It is used to determine the angles of elevation and depression in construction, to calculate the height of buildings or mountains, and to navigate ships and airplanes using trigonometric functions.