Solving $\sin(4k-22) = \cos(6k-13)$ My niece asked for help with an SAT prep question. We are given that 
$$\sin a = \cos b$$ 
where the angles are both acute and $a=4k-22$ and $b=6k-13$.
The only way we could think to solve it is by plotting and using fzero. But since it's an SAT problem, I assume there should be an approach that doesn't require a calculator.
Is there some trig identity I'm overlooking?
 A: You are overlooking some identities.
$$
\bbox[yellow,5pt, border:2px solid red]{
\sin A = \cos B \iff A+B = 90^\circ 
}\quad \textrm{(for A and B between 0 and 90 degrees)} \\
\bbox[yellow,5pt, border:2px solid red]{
\sin A = \cos B \iff A-B = 90^\circ 
}\quad \textrm{(for A between 0 and 90 degrees,B between 0,-90)} \\
\bbox[yellow,5pt, border:2px solid red]{
\sin A = \cos B \iff B-A = 90^\circ 
}\quad \textrm{(for B between 0 and 90 degrees,A between 0,-90)} 
$$
From here, of course, you get $4k-22 = 90-(6k-13)$
for the first case, $(6k-13) - (4k-22) = 90$ for the second case, and $(4k-22) - (6k-13) = 90$ for the third case.
A: Rewrite $\sin(4k-22) = \cos(6k-13)$ as
$$\cos(90-4k+22) - \cos(6k-13)=2\sin\frac{99-2k}2 \sin\frac{115-10k}2 =0$$
which leads to $\frac{99+2k}2 =n\pi,\>\>\>\>\> \frac{134-10k}2 =n\pi$
and the solutions $k= n\pi-\frac{99}2,\> \frac{n\pi}5-\frac{67}5$.
A: Draw a right triangle with acute angle $a$. The soh-cah-toa rule implies the other angle is $b$. Triangle angles add up to $180^\circ$.
