Potentially incorrect answer key to trig difference identity practice I am trying to answer this question:
Given cos(x) = 5/13 and sin(x) is negative, find:
sin(x)
sin(x-π)
cos(x-π)
sin(x-π/2)
cos(x-π/2)
The answers that I got were:
sin(x)=-sqrt(13^2-5^2)=-12
sin(x-π)=sin(x)cos(π)-cos(x)sin(π)=(-1)(-12)-(5/13)(0)=12
cos(x-π)=cos(x)cos(π)+sin(x)sin(π)=(-1)(5/13)+(-12)(0)=-5/13
sin(x-π/2)=sin(x)cos(π/2)-cos(x)sin(π/2)=(-12)(0)-(5/13)(1)=-5/13
cos(x-π/2)=cos(x)cos(π/2)+sin(x)sin(π/2)=(5/13)(0)+(-12)(1)=-12
I have triple checked my work, but I only get credit for parts 3 and 4. Have I in fact made an error, or is it the answer key that is erroneous?
 A: You forgot (1st, 2nd and5th lines) to divide by $13$. You also should use  some fundamental trigonometric identities to shorten computations:
\begin{align*}
\sin(\pi-x)&=\sin x&\sin(-x)&=-\sin x & \sin\Bigl(\frac\pi2-x\Bigr)&=\cos x\\
\cos(\pi-x)&=-\cos x&\cos(-x)&=\cos x &\cos\Bigl(\frac\pi2-x\Bigr)&=\sin x\\
\tan(\pi-x)&=-\tan x&\tan(-x)&=-\tan x&\tan\Bigl(\frac\pi2-x\Bigr)&=\cot x
\end{align*}
A: Notice that $\sin(x)$ and $\cos(x)$ can never go outside the rang of $[-1,1]$, so you can recognize that you have made mistakes in your answers. I think the problem occurred because you forgot a denominator of $13^2$ in the calculation of $\sin(x)$. The process for everything else looks correct. 
For the first part (since the $\sin$ is negative) we have
$$\sin(x)=-\sqrt{1-\cos^2x}=-\sqrt{1-\frac{5^2}{13^2}}=-\sqrt{\frac{13^2-5^2}{13^2}}=-\sqrt{\frac{144}{13^2}}=-\frac{12}{13}$$
With this new value of $\sin(x)$, your other problems should work out (note that they worked out in parts 3 and 4 because here we multiplied $\sin(x)$ by $0$).
