Without calculator, find value of:
$\displaystyle $\arccos\frac{1}{3}+2\arccos(\frac{-2\sqrt{2}}{3})$$
Please help.
I presume you know that $\displaystyle \left(\frac{1}{3}\right)^2= \frac{1}{9}$ but you might not have noticed that $\displaystyle \left(\frac{2\sqrt{2}}{3}\right)^2= \frac{8}{9}$ and so $\displaystyle \left(\frac{1}{3}\right)^2+ \left(\frac{2\sqrt{2}}{3}\right)^2= 1$.
I have try, but I think this is not correct. Look !
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It is that "2" in "$\displaystyle 2 arccos\left(\frac{-2\sqrt{2}}{3}\right)$ that is the problem. I am tempted to suggest that it is an error!
Obviously $\displaystyle arccos\left(\frac{1}{3}\right)+ arccos\left(\frac{-2\sqrt{2}}{3}\right)$ would be simple.
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 $\displaystyle \pm 2\sqrt{2} $ given, that we are talking distance, we take the positive.
Hence, we can say.
$\displaystyle 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,
$\displaystyle arccos(\frac{-2\sqrt(2)}{3}) = \theta - arcsin(\frac{2\sqrt(2)}{3}) $
Take the inverse of the inverse.
$\displaystyle \frac{-2\sqrt(2)}{3} = cos(\theta - arcsin(\frac{2\sqrt(2)}{3}) )$
Recall cos(a-b) = cos(a)cos(b) - sin(a)sin(b)
$\displaystyle \frac{-2\sqrt(2)}{3} = \frac{cos(\theta)}{3} - sin(\theta)\frac{2\sqrt(2)}{3}$
Multiply through by 3
$\displaystyle -2\sqrt(2) = cos(\theta) - 2\sqrt(2)sin(\theta)$
$\displaystyle -2\sqrt(2) = \frac{cos(\theta) }{1-sin(\theta)}$
$\displaystyle 8 = \frac{cos^2(\theta) }{(1-sin(\theta))^2}$
This is about all I can whittle it down to..
Brutal question for non-calc.