Arrange in ascending order $\tan45^\circ,\tan80^\circ$ and $\tan100^\circ$
Mia Lopez
Arrange in ascending order $\tan45^\circ,\tan80^\circ$ and $\tan100^\circ.$
We know that $\tan45^\circ=1$ because a right triangle with angle equal to $45^\circ$ is isosceles.
I don't know the exact values of $\tan80^\circ$ and $\tan 100^\circ.$ Can you give me a hint?
$\endgroup$3 Answers
$\begingroup$Hint:
$\tan x \lt 0$ for $90^\circ\lt x \lt 180^\circ$ and $\tan x$ is an increasing function on $[0,90^\circ)$.
$\tan \theta$ is defined to be the orange line in the diagram. What happens to this line as $\theta$ goes from $0^\circ$ to $90^\circ$?
$\endgroup$ 7 $\begingroup$Write $\tan=\dfrac{\sin}{\cos}$
$$\tan80^\circ-\tan45^\circ=\cdots=\dfrac{\sin(80-45)^\circ }{\cos80^\circ\cos45^\circ}>0$$
We know $\tan45^\circ=?$
and $\tan100^\circ=\tan(180^\circ-80^\circ)=-\tan80^\circ<0$
$\endgroup$ 3 $\begingroup$$\tan(x)$ is defined on $\bigcup_{n\in \Bbb Z} (90(2n-1), 90(2n+1))$ (when working in degrees) and increasing inside each such set.
Moreover, recall that $\tan(x)\equiv\tan(x+180k)$ for each $k\in \Bbb Z$
So, we may say that $\tan(x)$ is increasing on the range $(-90, 90)$, as seen when $n=0$, and $\tan(100)=\tan(-80)<\tan(45)<\tan(80)$
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