Trig problems

[Fullmetal_Hye]

I have to show that arctan(x) and arcsin(x) are odd functions. what do I do?

Also I need to find the exct values of these:

1.sin(arcsin(8/17)+arccos(15/17))

(I know sin(arcsin(8/17)=8/17 but I don't know how to change arccos into something else.)

2.tan[2arcsec(negative rad5)]
[/code][/list]

 

[pka]

We know that $\displaystyle y = \arctan (x)\quad \Rightarrow \quad x = \tan (y)$.
Thus to show arctangent is an odd function:
\(\displaystyle \begin{array}{rcl}
y= \arctan ( - x) & \Rightarrow & \quad - x = \tan (y) \\
& \Rightarrow & \quad x = - \tan (y) \\
& \Rightarrow & \quad x = \tan ( - y) \\
& \Rightarrow & \quad - y = \arctan (x) \\
& \Rightarrow & \quad y = - \arctan (x) \\
\end{array}\)

 

[soroban]

Hello, Fullmetal_Hye!

Find the exact values.

$\displaystyle 1)\;\sin\left(\arcsin\frac{8}{17}\,+\,\arccos\frac{15}{17}\right)$

We have: $\displaystyle \,\sin\left(\underbrace{\arcsin\frac{8}{17}}\,+\,\underbrace{\arccos\frac{15}{17}}\right)\;\;\Rightarrow\;\;\sin(\alpha\,+\,\beta)$
. . . . . . . . . . . . . $\displaystyle \alpha$ . . . . . . .$\displaystyle \beta$

We know that: $\displaystyle \,\sin(\alpha\,+\,\beta)\:=\:\sin\alpha\cos\beta\,+\,\sin\beta\cos\alpha\;$ (1)

Since $\displaystyle \alpha = \arcsin\frac{8}{17}\;\;\Rightarrow\;\;\sin\alpha = \frac{8}{17}\;\;\Rightarrow\;\;\cos\alpha = \frac{15}{17}$

Since $\displaystyle \beta = \arccos\frac{15}{17}\;\;\Rightarrow\;\;\cos\beta = \frac{15}{17}\;\;\sin\beta = \frac{8}{17}$

Substitute into (1): \(\displaystyle \,\sin(\alpha\,+\,\beta) \;= \;\left(\frac{8}{17}\right)\left(\frac{15}{17}\right)\,+\,\left(\frac{8}{17}\right)\left(\frac{15}{17}\right) \;= \;\frac{120}{289}\,+\,\frac{120}{289}\;=\;\L\frac{240}{289}\)

$\displaystyle 2)\;\tan\left[2\text{arcsec}(-\sqrt{5})\right]$

We have: $\displaystyle \,\tan\left[2\cdot\underbrace{\text{arcsec}(-\sqrt{5})}\right]\;\;\Rightarrow\;\;\tan(2\theta)$
. . . . . . . . . . . . . . . .$\displaystyle \theta$

Since $\displaystyle \theta \,= \,\text{arcsec}(-\sqrt{5})\;\;\Rightarrow\;\;\sec\theta\,=\,-\sqrt{5}$


Since $\displaystyle \tan^2\theta\:=\:\sec^2\theta\,-\,1$, we have: \(\displaystyle \,\tan^2\theta\:=\-\sqrt{5})^2\,-\,1\:=\:4\)
$\displaystyle \;\;$Hence: $\displaystyle \,\tan\theta\,=\,\pm2\;\;\Rightarrow\;\;\theta\:=\:\arctan(\pm 2)$

The problem becomes:\(\displaystyle \L\,\tan(2\theta) \:=\:\frac{2\tan\theta}{1 - \tan^2\theta} \:=\:\frac{2(\pm2)}{1\,-\,(\pm2)^2} \;=\;\frac{\pm4}{-3}\;=\;\pm\frac{4}{3}\)