# Thread: Need help with trigonometri

1. ## Need help with trigonometri

solve equation: 2arcsinx = arccos(2x+x^2)

determine on easyest way posible the number:
2arctan7 + arctan(31/17).

2. Hi

$\displaystyle 2 \arcsin x = \arccos(2x+x^2)$

$\displaystyle \cos(2 \arcsin x) = \cos(\arccos(2x+x^2))$

$\displaystyle 1-2\sin^2(\arcsin x) = \cos(\arccos(2x+x^2))$

$\displaystyle 1-2x^2 = 2x+x^2$

$\displaystyle x=-1$ or $\displaystyle x=\frac{1}{3}$

You need to verify if those solutions are real solutions

3. Hello, johnnyboiy!

Hmmm, there must be an easier way . . .

Determine in the easiest way possible the number:
. . $\displaystyle \theta\;=\;2\arctan7 + \arctan\left(\tfrac{31}{17}\right)$
$\displaystyle \text{We are dealing with two angles: }\;\theta\;=\;2\underbrace{\arctan7}_{\alpha} + \underbrace{\arctan\left(\tfrac{31}{17}\right)}_{\ beta}$

And we will determine: .$\displaystyle \tan\theta \;=\;\tan(2\alpha + \beta)\;=\;\frac{\tan(2\alpha) + \tan(\beta)}{1 - \tan(2\alpha)\tan(\beta)}$ .[1]

We have: .$\displaystyle \alpha \,=\,\arctan7 \quad\Rightarrow\quad \tan\alpha \,=\,7$

Then: .$\displaystyle \tan(2\alpha) \:=\:\frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \:=\:\frac{2(7)}{1 - 7^2} \quad\Rightarrow\quad\tan(2\alpha) \:=\:-\frac{7}{24}$ .[2]

Since $\displaystyle \beta \,=\,\arctan\left(\frac{31}{17}\right)$, we have: .$\displaystyle \tan(\beta) \,=\,\frac{31}{17}$ .[3]

Substitute [2] and [3] into [1]: .$\displaystyle \tan\theta \;=\;\frac{-\dfrac{7}{24} + \dfrac{31}{17}}{1 - \left(-\dfrac{7}{24}\right)\left(\dfrac{31}{17}\right)} \;=\;\frac{\left(\dfrac{625}{408}\right)}{\left(\d frac{625}{408}\right)} \;=\;1$

Therefore:. $\displaystyle \tan\theta \:=\:1 \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{4}$