Trigonometric solution

Question

How to get exact solution to the following problems:

$$\tan(\arccos(2/7))$$ and $$\arcsin(\sin(8π/9))$$

I tried all kinds of stuff, but just can't get to the solution.

Answer 1

Hint:

Let $\arcsin\left(\sin\dfrac{8\pi}9\right)=x$

$\implies\sin x=\sin\dfrac{8\pi}9$

Using this, $-\dfrac\pi2\le x\le\dfrac\pi2\ \ \ \ (1)$

$\implies x=m\pi+(-1)^n\dfrac{8\pi}9$ where $m$ is an integer such that $(1)$ is is satisfied.

If $m$ is even $=2n$(say), $$-\dfrac\pi2\le2n\pi+\dfrac{8\pi}9\le\dfrac\pi2$$

$$\iff-9\le36n+16\le9\iff-1<-\dfrac{25}{36}\le n\le-\dfrac7{36}<0$$

But $n$ is an integer.

What if $m$ is odd $=2n+1?$

Answer 2

Hint for the first one.

Notice that

$$\tan(\arccos(x)) = \frac{\sqrt{1-x^2}}{x}$$

hence you immediately get the numerical result for this.

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For the second one, notice that $$\sin\frac{8\pi}{9} = \sin\frac{\pi}{9} \approx \frac{\pi}{9}$$

You can uses eventually Taylor Series for the arcsine:

$$\arcsin(x) \approx x+\frac{x^3}{6}+O\left(x^4\right)$$

And plug $\pi/9$ inside.