1. ## inverse trig

I have a question on a test, but have no clue how to work the problem. My instructions are to evaluate the expression exactly:

sin (arctan 2)

A=5sq rt 2/2
b=2 sq rt5/5
c=2sq rt 5
d 5 sq rt 2

2. Originally Posted by scottsw1
...evaluate the expression exactly: sin (arctan 2)
Draw a right triangle; the particular dimensions don't matter.

Label the non-right base angle as $\displaystyle \theta$.

By definition, $\displaystyle \arctan(2)\, =\, \theta$ means that $\displaystyle \tan(\theta)\, =\, 2\, =\, \frac{2}{1}$

So label the "opposite" and "adjacent" sides of the triangle to reflect this.

Use the Pythagorean Theorem to find the value of the hypotenuse.

Since $\displaystyle \arctan(2)\, =\, \theta$, then $\displaystyle \sin(\arctan(2))$ means $\displaystyle \sin(\theta)$.

3. It looks like to me hypontenous is sq rt 5. Please help me with the equation, I didn't understand what you were trying to describe to me. Thanks for taking the time to help!!

4. Originally Posted by scottsw1
I have a question on a test, but have no clue how to work the problem. My instructions are to evaluate the expression exactly:

sin (arctan 2)

A=5sq rt 2/2
b=2 sq rt5/5
c=2sq rt 5
d 5 sq rt 2
Let $\displaystyle \alpha = \arctan 2 \Rightarrow \tan \alpha = 2$.

You need to find the value of $\displaystyle \sin \alpha$. Do you know how to use a right-angle triangle to do this (stapel has described how)?

5. I came up with the answer that sin=sq rt 5. Is this correct? If it is, it does not equal the multiple choice answers on the paper.

6. I solved the problem!! It was 2 sq rt 5.

Now here is another:

6 arcsin x= pi

Any suggestions on how to solve?

7. Originally Posted by scottsw1
I solved the problem!! It was 2 sq rt 5. Mr F says: No doubt you meant to include a vinculum ....

Now here is another:

6 arcsin x= pi

Any suggestions on how to solve?
$\displaystyle \Rightarrow \arcsin x = \frac{\pi}{6} \Rightarrow x = \sin \frac{\pi}{6} \, ....$