\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\def\arcsec{{\rm arcsec}}
\def\arccot{{\rm arccot}}

\SN{6.8}
Useful angles:
\itemitem{}
$\cos0=1$, $\sin0=0$
\itemitem{}
$\cos\frac\pi2=0$, $\sin\frac\pi2=1$
\itemitem{}
$\cos\pi=-1$, $\sin0\pi=0$
\itemitem{}
$\cos(\frac\pi6)=\frac12$, $\sin(\frac\pi6)=\frac{\sqrt3}2$
\itemitem{}
$\cos(\frac\pi4)=\frac{\sqrt2}2$, $\sin(\frac\pi4)=\frac{\sqrt2}2$
\itemitem{}
$\cos(\frac\pi3)=\frac{\sqrt3}2$, $\sin(\frac\pi3)=\frac{1}2$
\itemitem{}

\pn2.a) $\arctan(-1)=-\frac{\pi}4$ since $\tan(\frac{-\pi}4)=-1$ and
$-\frac\pi2<-\frac\pi4<\frac\pi2$.\par
b) $\arctan{\sqrt3}=\frac\pi6$ since $\tan(\frac\pi6)=\sqrt3$\par
c) $\arctan{\frac{-1}{\sqrt3}}=-\frac\pi3$
since $\tan(-\frac\pi3)=-\frac1{\sqrt3}$

\pn19. $\tan\br2{\arcsin(-\frac12)}=$?
But $\arcsin(-\frac12)=-\frac\pi3$ since $\sin(-frac\pi3)=-frac12$,
so $\tan\br2{\arcsin(-\frac12)}=\tan(-\frac\pi3)=-\tan(\frac\pi3)=
\frac{-1}{\sqrt3}$.

\pn31. $\tan\bigl(\arcsec(3y)\bigr)=?$.
Set up a right triangle with one angle $\theta$
with $\sec\theta=3y$.
Then $(3y)^2=\tan^2(3y)+1$, so
$\tan(3y)=\sqrt{9y^2-1}$.

\pn41. $\displaystyle\lim_{x\to1^-}\arcsin x=\frac\pi2$ since the 
inverse of a continuous function is continuous and\par
$\displaystyle\lim_{x\to\frac\pi2^-}\sin x=1$.

\pn43.
$\displaystyle\lim_{x\to\infty}\arctan x=\frac\pi2$ since
$\displaystyle\lim_{x\to\frac\pi2^-}\tan x=\infty$.

\pn49.
Let $\beta$ be the angle with sides $x$ and $3$ 
from the picture on page 511 in the book.
Then $\cot\beta=\frac x3$ and
$\cot(\alpha+\beta)=\frac x{3+12}=\frac x{15}$ so
$\alpha+\beta=\arccot(\frac x{15})$ and
$\beta=\arccot(\frac x3)$ so
$\alpha=\arccot(\frac x{15})-\arccot(\frac x{3})$.



\end

6.8  p. 510:   2, 19, 31, 41, 43, 49
