Solving $\sin 7\phi+\cos 3\phi=0$ The question is find the general solution of this equation:$$\sin(7\phi)+\cos(3\phi)=0$$
I tried to use the "Sum-to-Product" formula, but found it only suitable for $\sin(a)\pm   \sin(b)$ or $\cos(a)\pm  \cos(b)$. So I tried to expand $\sin 7\phi$ and $\cos 3\phi$, but the equation became much more complicated..
I'm self studying BUT There's nothing about how to solve this type of equations on my textbook..
reeeaaaaally confused now..
 A: A very basic fact that most students seem to consider like a magician's trick is that $$\sin \alpha = \cos \left( \frac{\pi}{2} - \alpha \right).$$ By this elementary identity, we can easily solve equations like $$\sin \alpha = \cos \beta.$$
A: Hint: Use sum to product!
$$\sin 7\phi+\sin \left(\frac{\pi}{2}-3\phi \right)=0$$
A: $\cos 7\phi+\sin 3\phi=0$
Note that $7\phi = 5\phi+2\phi$ and that $3\phi=5\phi-2\phi$
Rewrite original statement as: $\cos (5\phi+2\phi)+\sin (5\phi-2\phi)=0$
$\cos 5\phi \cos 2\phi - \sin 5\phi \sin 2\phi + \sin 5\phi \cos 2 \phi - \cos 5\phi \sin 2\phi=0$
$\cos 5\phi \cos 2\phi - \cos 5\phi \sin 2\phi + \sin 5\phi \cos 2 \phi- \sin 5\phi \sin 2\phi  =0$
$\cos 5\phi \left(\cos 2\phi - \sin 2\phi \right)+\sin 5\phi \left(\cos 2\phi-\sin 2\phi \right)=0$
$\left(\cos 5\phi +\sin 5\phi \right) \left(\cos 2\phi - \sin 2\phi \right)=0$
Either $\cos 5\phi +\sin 5\phi =0$
$\sin 5\phi=-\cos 5\phi$
$\tan 5\phi=-1$
$5\phi = -\frac \pi 4 + k\pi$
$\phi = -\frac \pi {20} + \frac{k\pi}{5}$
or $\cos 2\phi - \sin 2\phi =0$
$\sin 2\phi=\cos 2\phi$
$\tan 2\phi=1$
$2\phi = \frac \pi 4 + k\pi$
$\phi = -\frac \pi {8} + \frac{k\pi}{2}$
