this problem is hard see if you can help me out

**davis16** · July 8th 2008, 10:52 AM

find exact value of cos[tan-1(-1)+cos-1(-3/5)]

tan-1 is inverse tan
cos-1 is inverse cos

**TKHunny** · July 8th 2008, 10:59 AM

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.

**davis16** · July 9th 2008, 10:12 AM

still need help. atleast tell me how to start

**flyingsquirrel** · July 9th 2008, 11:02 AM

Hi

Originally Posted by davis16

atleast tell me how to start

You've already been told how to start

: 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 $\arctan 1$ .

**TKHunny** · July 9th 2008, 01:25 PM

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.

**Moo** · July 9th 2008, 11:31 PM

Originally Posted by flyingsquirrel

Hi

You've already been told how to start

: 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 $\arctan 1$ .

**flyingsquirrel** · July 9th 2008, 11:45 PM

Originally Posted by Moo

What is $\arctan 1$ then ? (we're not talking about the solutions of $\tan x=1$ for $x\in\mathbb{R}$ ...)

**Moo** · July 9th 2008, 11:48 PM

Yes, you're right

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