3

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..

Siminore
  • 35,136
Vic.
  • 423
  • 4
    Fortunately, we know how to rewrite a cosine as a sine of a different angle.... –  Jun 24 '12 at 10:06

3 Answers3

5

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 $$\sin \alpha = \cos \beta.$$

Siminore
  • 35,136
5

Hint: Use sum to product! $$\sin 7\phi+\sin \left(\frac{\pi}{2}-3\phi \right)=0$$

Siminore
  • 35,136
Erik
  • 166
0

$\cos 7\phi+\sin 3\phi=0$

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 $\cos 2\phi - \sin 2\phi =0$

$\sin 2\phi=\cos 2\phi$

$\tan 2\phi=1$

$2\phi = \frac \pi 4 + k\pi$

$\phi = -\frac \pi {8} + \frac{k\pi}{2}$

tomi
  • 9,594