Re: Does this Look right?

I would use:

$\displaystyle \sin(u)=-\sqrt{1-\cos^2(u)}$

$\displaystyle \sin(u-\pi)=-\sin(\pi-u)=-\sin(u)$

$\displaystyle \cos(u-\pi)=\cos(\pi-u)=-\cos(u)$

$\displaystyle \sin\left(u-\frac{\pi}{2} \right)=-\sin\left(\frac{\pi}{2}-u \right)=-\cos(u)$

$\displaystyle \cos\left(u-\frac{\pi}{2} \right)=\cos\left(\frac{\pi}{2}-u \right)=\sin(u)$

Re: Does this Look right?

Quote:

Originally Posted by

**MarkFL2** I would use:

$\displaystyle \sin(u)=-\sqrt{1-\cos^2(u)}$

$\displaystyle \sin(u-\pi)=-\sin(\pi-u)=-\sin(u)$

$\displaystyle \cos(u-\pi)=\cos(\pi-u)=-\cos(u)$

$\displaystyle \sin\left(u-\frac{\pi}{2} \right)=-\sin\left(\frac{\pi}{2}-u \right)=-\cos(u)$

$\displaystyle \cos\left(u-\frac{\pi}{2} \right)=\cos\left(\frac{\pi}{2}-u \right)=\sin(u)$

Thank You very much, So the initial cos (u)=5/13 has nothing to do with the other problems? I thought I had to relate each one to the original somehow

Re: Does this Look right?

But the other problem is when I input your answers the program tells me variable "u" is not defined in this context?

meaning these answers it will consider wrong

Re: Does this Look right?

You are given the value of cos(u), then from this you may find sin(u) using the first formula, and since you now have sin(u) and cos(u), the rest come from that.

Re: Does this Look right?

Re: Does this Look right?

Which would mean sin(u-pi) should be Sin(u-pi)= sin-(12/13)cos(pi)-cos-(12/13)(sin(pi)

Re: Does this Look right?

If sin(u) = -12/13, then -sin(u) = 12/13.