Hello everyone I have a question about trig.
How would I solve the following.
$\tan\left(2\arcsin(4/5)+\arccos(12/13)\right)=\frac{253}{204}$
Please help.
Hello everyone I have a question about trig.
How would I solve the following.
$\tan\left(2\arcsin(4/5)+\arccos(12/13)\right)=\frac{253}{204}$
Please help.
$ \tan(x+y+z) = \frac{\tan x+\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan x\tan z-\tan y\tan z}\tag{1} $
You want $ \tan\left(\arcsin\frac45 + \arcsin\frac45 + \arccos \frac{12}{13}\right). $ So there you have $x$, $y$, and $z$.
Draw the right triangle in which the "opposite" side is $4$ and the hypotenuse is $5$. By the Pythagorean theorem, the "adjacent" side is $3$. So $\arcsin\frac45=\arctan\frac43$, so $\tan x=\frac43$. Then draw on where the "adjacent" side is $12$ and the hypotenuse is $13$, and then find that the "opposite" side is $5$, so the tangent is $5/12$. Then you have $\tan z=\frac{5}{12}$. Plug those tangents into the right side of $(1)$ and simplify.
And recall that if you know $\tan(x+y) = \dfrac{\tan x+\tan y}{1-\tan x\tan y}$, then you can derive $(1)$ by saying $\tan(x+y+z) = \tan(\Big(x\Big)+\Big(y+z\Big))$, etc.