1. ## trig help....

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)]

2. Originally Posted by ilaura
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 $\arctan(x)$ is an inverse function of $\tan(x)$ so $\tan(\arctan(x))=x$ and for the second one use the fact that $\sin(\arccos(x))=\sqrt{1-x^2}$ and $\sin(\arcsin(x))=x$ after using the identity $\sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\sin\left(B\right)\cos\left(A\right)$

3. i'm still really confused.....

i tried doing something with the second one,

i don't think its right but i got:

6(square root of 2) + 3(square root of 21) -1
____________________
15

4. Originally Posted by ilaura
i'm still really confused.....

i tried doing something with the second one,

i don't think its right but i got:

6(square root of 2) + 3(square root of 21) -1
____________________
15
\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}

5. Originally Posted by Mathstud28
......and for the second one use the fact that $\sin(\arccos(x))=\sqrt{1-x^2}$ and $\sin(\arcsin(x))=x$

i don't understand where these come from....
are they some kind of identities?

6. Originally Posted by ilaura
i don't understand where these come from....
are they some kind of identities?
Yes List of trigonometric identities - Wikipedia, the free encyclopedia

7. Originally Posted by ilaura
[snip]
and 2. Evaluate: (show your work. do not give a decimal approximation)

Sin[cos inverse(1/3) + sin inverse(-2/5)]
Let $\cos^{-1} \left( \frac{1}{3} \right)= \alpha \Rightarrow \cos \alpha = \frac{1}{3}$.

Let $\sin^{-1} \left(-\frac{2}{5}\right) = \beta \Rightarrow \sin \beta = - \frac{2}{5}$.

Then $\sin (\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \, ....$

8. Originally Posted by Mathstud28
\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}
where did you get the 1/2? is it part of the identity?

9. Originally Posted by ilaura
where did you get the 1/2? is it part of the identity?
Clearly it's a typo and should be 1/3.

10. Originally Posted by ilaura
My homework says:

1. Give the exact real number value for arctan [tan (7pi/12)]

steps would be appreciated. thanks
Note that arctangent and tangent are inverses of each other.

Thus, if we have $\tan^{-1}\left(\tan(\varphi)\right)$, it just reduces to $\varphi$. The same thing applies to $\tan\left(\tan^{-1}(\varphi)\right)$. It also equals $\varphi$.

Knowing this, what do you think the answer is?