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

tan-1 is inverse tan

cos-1 is inverse cos

Printable View

- Jul 8th 2008, 10:52 AMdavis16this 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, 10:59 AMTKHunny
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, 10:12 AMdavis16
still need help. atleast tell me how to start

- Jul 9th 2008, 11:02 AMflyingsquirrel
Hi

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

By the way, $\displaystyle \arctan (-1)=-\arctan 1$ and you should know the value of $\displaystyle \arctan 1$. - Jul 9th 2008, 01:25 PMTKHunny
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 9th 2008, 11:31 PMMoo
- Jul 9th 2008, 11:45 PMflyingsquirrel
- Jul 9th 2008, 11:48 PMMoo
Yes, you're right :D