Determine the smallest positive value

[Baron]

Determine the smallest positive value for 'k' such that 5sin pi(x-k) = -5cos pi(x)

The answer is 1/2.

My work:
5sin pi(x-k) = -5cos pi(x)
-sin pi(x-k) = cos pi(x)
sin[-pi(x-k)] = cos pi(x) --> because sin is an odd function
sin[pi(k-x)] = cos pi(x)

Now I'm stuck.

Please show work and any necessary explanations

 

[tkhunny]

\(\displaystyle \cos\left[\pi\cdot x\right]\;=\;\sin\left[\pi\cdot\left(\frac{1}{2}-x\right)\right]\)

 

[Baron]

\(\displaystyle \cos\left[\pi\cdot x\right]\;=\;\sin\left[\pi\cdot\left(\frac{1}{2}-x\right)\right]\)

I understand how to get the answer but how does cos[pi(x)] = sin [pi*(1/2-x)] ?

 

[Deleted member 4993]

\(\displaystyle \cos\left[\pi\cdot x\right]\;=\;\sin\left[\pi\cdot\left(\frac{1}{2}-x\right)\right]\)

I understand how to get the answer but how does cos[pi(x)] = sin [pi*(1/2-x)] ?

Expand sin [pi*/2-pi*x)] using sin(A+B) = sinA*cosB+cosA*sinB and you'll see....