# Inverse trigonometry question

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1. May 22, 2016

### takando12

1. The problem statement, all variables and given/known data
If 1/2<=x<=1
, then cos-1[ x/2 + (√(3-3x2))/2] + cos-1 x is equal to?

2. Relevant equations

3. 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)

Last edited: May 22, 2016
2. May 22, 2016

### Staff: Mentor

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.

3. May 22, 2016

### takando12

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. May 22, 2016

### Staff: Mentor

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.

5. May 22, 2016

### takando12

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. May 22, 2016

### Staff: Mentor

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. May 22, 2016

### takando12

it's $\frac{\cos(\theta)}{2}$ .
I'm sorry, I've made more corrections to the question now. Is it clear?

8. May 22, 2016

### Staff: Mentor

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?

9. May 22, 2016

### takando12

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.

Last edited: May 22, 2016
10. May 22, 2016

### Staff: Mentor

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. May 22, 2016

### takando12

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. May 22, 2016

### Staff: Mentor

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. May 22, 2016

### takando12

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. May 23, 2016

### Staff: Mentor

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. May 23, 2016

### takando12

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.?

Last edited: May 23, 2016
16. May 23, 2016

### Staff: Mentor

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.

17. May 23, 2016

### takando12

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,
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. May 23, 2016

### Staff: Mentor

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

19. May 23, 2016

### ehild

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. May 24, 2016

### takando12

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?