0
$\begingroup$

The question was

Solve. $\sqrt{2}\cos x + \cot x = 0,\ x\in [-\pi, \pi]$

So she did

$\sqrt{2}\cos x + \frac{\cos x}{\sin x} = 0$ which is fine

But the next step confuses me. I understand where she went from this next step, but not how she got to it.

$\cos x(\sqrt{2} + \frac{1}{\sin x}) = 0$.

What I'm not understanding is, is how the $\sqrt{2}$ left the '$\cos x$' and joined the $\frac{1}{\sin x}$, and how the $\frac{\cos x}{\sin x}$ even became $\frac{1}{\sin x}$?

Any help would be great.

$\endgroup$
  • $\begingroup$ Your teaching factored out the $\cos(x)$. Namely this step is from the distributive rule. $\endgroup$ – MathNewbie Jun 15 '15 at 5:18
4
$\begingroup$

This is a use of the distributive law which states that, for any real numbers $a, b, c$ we have $a(b+c) = ab+ac $.

In your case, you have $0 = \sqrt{2}\cos x + \dfrac{\cos x}{\sin x} $.

Your teacher then used the distributive law (and other laws) to write

$\begin{array}\\ 0 &=\sqrt{2}\cos x + \dfrac{\cos x}{\sin x}\\ &=\cos x \sqrt{2} + \cos x\dfrac1{\sin x}\\ &=\cos x \left(\sqrt{2} + \dfrac1{\sin x}\right)\\ \end{array} $

If you still do not understand this, please get some help, because you have to understand algebra at this level to have any chance of understanding the other material you will be taught.

$\endgroup$
  • $\begingroup$ Honestly, I've never heard of 'distributive law', but after this explanation, I make myself look dumb. I guess just adding in the sinx and cosx makes it look harder than it really is. Next time, I'll just use a let statement. Thanks for the explanation. $\endgroup$ – DoubleHelix Jun 15 '15 at 5:23
  • $\begingroup$ Do you understand my explanation? $\endgroup$ – marty cohen Jun 15 '15 at 5:54
  • $\begingroup$ Yes, I understand it. Basically like, a = sqrt(2) b=cosx c=sinx ..... ba + b x 1/c..... Divide b out..... b(a + 1/c) $\endgroup$ – DoubleHelix Jun 15 '15 at 6:16
  • $\begingroup$ Good. Trig functions can turn into each other in a number of ways, so you just have to get experience manipulating them. $\endgroup$ – marty cohen Jun 15 '15 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.