# this problem is hard see if you can help me out

• Jul 8th 2008, 11:52 AM
davis16
this problem is hard see if you can help me out
find exact value of cos[tan-1(-1)+cos-1(-3/5)]

tan-1 is inverse tan
cos-1 is inverse cos
• Jul 8th 2008, 11:59 AM
TKHunny
Here we are again. Have you done a, b, and d on the last one, yet?

For this one, you also need this formula, which you should know.

cos(a+b) = cos(a)cos(b)-sin(a)sin(b)

After that, you'll need those Right Triangles again.
• Jul 9th 2008, 11:12 AM
davis16
still need help. atleast tell me how to start
• Jul 9th 2008, 12:02 PM
flyingsquirrel
Hi
Quote:

Originally Posted by davis16
atleast tell me how to start

You've already been told how to start :D : expand $\cos\left[\arctan(-1)+\arccos\left(-\frac35\right)\right]$ using $\cos(a+b)=\cos a \cos b-\sin a \sin b$ with $a=\arctan(-1)$ and $b=\arccos\left(-\frac35\right)$.

By the way, $\arctan (-1)=-\arctan 1$ and you should know the value of $\arctan 1$.
• Jul 9th 2008, 02:25 PM
TKHunny
Quote:

Originally Posted by davis16
still need help. atleast tell me how to start

You have posted seven (7) problems in four (4) posts. You have not shown ANY work at all. You have provided NO feedback to indicate whether you are learning anything. So far, you have managed to get others to do your homework for you. This will NOT serve you well on an examination. What's it going to be? Learning or just laziness? I can be harsh if you would like. So far, I'm just trying to get your attention. You WILL fail your class if you keep on this way. Pick a different path. Sooner is better.
• Jul 10th 2008, 12:31 AM
Moo
Quote:

Originally Posted by flyingsquirrel
Hi

You've already been told how to start :D : expand $\cos\left[\arctan(-1)+\arccos\left(-\frac35\right)\right]$ using $\cos(a+b)=\cos a \cos b-\sin a \sin b$ with $a=\arctan(-1)$ and $b=\arccos\left(-\frac35\right)$.

By the way, $\arctan (-1)=-\arctan 1$ and you should know the values of $\arctan 1$.

(Giggle)
• Jul 10th 2008, 12:45 AM
flyingsquirrel
Quote:

Originally Posted by Moo
(Giggle)

What is $\arctan 1$ then ? (we're not talking about the solutions of $\tan x=1$ for $x\in\mathbb{R}$...)
• Jul 10th 2008, 12:48 AM
Moo
Yes, you're right :D