i have two problems that i can't figure out how to do...
1. Give the exact real number value for Tan inverse [tan 7pi/12]
and 2. Evaluate: (show your work. do not give a decimal approximation)
Sin[cos inverse(1/3) + sin inverse(-2/5)]
On a restriced interval $\displaystyle \arctan(x)$ is an inverse function of $\displaystyle \tan(x)$ so $\displaystyle \tan(\arctan(x))=x$ and for the second one use the fact that $\displaystyle \sin(\arccos(x))=\sqrt{1-x^2}$ and $\displaystyle \sin(\arcsin(x))=x$ after using the identity $\displaystyle \sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\sin\left(B\right)\cos\left(A\right)$
$\displaystyle \begin{aligned}\sin\left(\arccos\left(\frac{1}{3}\ right)\right)+\arcsin\left(\frac{-2}{5}\right)&=\sin\left(\arccos\left(\frac{1}{3}\r ight)\right)\cos\left(\arcsin\left(\frac{-2}{5}\right)\right)\\
&+\sin\left(\arcsin\left(\frac{-2}{5}\right)\right)\cos\left(\arccos\left(\frac{1} {3}\right)\right)\\
&=\sqrt{1-\left(\frac{1}{3}\right)^2}\sqrt{1-\left(\frac{-2}{5}\right)^2}+\frac{1}{2}\cdot\frac{-2}{5}\\
&=\frac{2\sqrt{42}-2}{15}
\end{aligned}$
Let $\displaystyle \cos^{-1} \left( \frac{1}{3} \right)= \alpha \Rightarrow \cos \alpha = \frac{1}{3}$.
Let $\displaystyle \sin^{-1} \left(-\frac{2}{5}\right) = \beta \Rightarrow \sin \beta = - \frac{2}{5}$.
Then $\displaystyle \sin (\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \, .... $
Note that arctangent and tangent are inverses of each other.
Thus, if we have $\displaystyle \tan^{-1}\left(\tan(\varphi)\right)$, it just reduces to $\displaystyle \varphi$. The same thing applies to $\displaystyle \tan\left(\tan^{-1}(\varphi)\right)$. It also equals $\displaystyle \varphi$.
Knowing this, what do you think the answer is?