1. ## arccos(1/3)+2*arccos(-2sqrt(2)/3)

Without calculator, find value of:
$\arccos\frac{1}{3}+2\arccos(\frac{-2\sqrt{2}}{3})$

2. ## Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

I presume you know that $\left(\frac{1}{3}\right)^2= \frac{1}{9}$ but you might not have noticed that $\left(\frac{2\sqrt{2}}{3}\right)^2= \frac{8}{9}$ and so $\left(\frac{1}{3}\right)^2+ \left(\frac{2\sqrt{2}}{3}\right)^2= 1$.

3. ## Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

I have try, but I think this is not correct. Look !

4. ## Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

It is that "2" in " $2 arccos\left(\frac{-2\sqrt{2}}{3}\right)$ that is the problem. I am tempted to suggest that it is an error!
Obviously $arccos\left(\frac{1}{3}\right)+ arccos\left(\frac{-2\sqrt{2}}{3}\right)$ would be simple.

5. ## Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

Yes, it is 2. It is not an error

Darn!

7. ## Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

When in situations likes this, I love drawing pictures.

For the second term, we have a triangle with a base of 1, a hypotenuse of 3, so the other side using pythagoras is $\pm 2\sqrt{2}$ given, that we are talking distance, we take the positive.

Hence, we can say.
$arccos(\frac{1}{3} = arcsin(\frac{2\sqrt(2)}{3})$

Now, we set the original equation to theta, as there is some angle that is works out to be,

$arccos(\frac{-2\sqrt(2)}{3}) = \theta - arcsin(\frac{2\sqrt(2)}{3})$

Take the inverse of the inverse.
$\frac{-2\sqrt(2)}{3} = cos(\theta - arcsin(\frac{2\sqrt(2)}{3}) )$

Recall cos(a-b) = cos(a)cos(b) - sin(a)sin(b)
$\frac{-2\sqrt(2)}{3} = \frac{cos(\theta)}{3} - sin(\theta)\frac{2\sqrt(2)}{3}$

Multiply through by 3

$-2\sqrt(2) = cos(\theta) - 2\sqrt(2)sin(\theta)$

$-2\sqrt(2) = \frac{cos(\theta) }{1-sin(\theta)}$

$8 = \frac{cos^2(\theta) }{(1-sin(\theta))^2}$

This is about all I can whittle it down to..

Brutal question for non-calc.

8. ## Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

It is really tough question for me and i am not able to solve this type of question.