0
$\begingroup$

Let $$\left(\frac{\cos\alpha}{\sin\alpha}\right) = \left(\frac{b}{a}\right)^n \left(\frac{x_0}{y_0}\right)^\frac{1}{n - 1}$$

Now suppose if I multiply by power $(n - 1)$ to both sides then the equation will look something like this :

$$\left(\frac{\cos\alpha}{\sin\alpha}\right)^\left(n - 1\right) = \left(\frac{b}{a}\right)^n \left(\frac{x_0}{y_0}\right)$$

Have I committed some mistake here? I'm asking this because I'm confused. Also I'm not getting correct answer of my question. I think I have did all the steps correctly. I just have doubt in this one.

The doubt is whether multiplying by power $(n - 1)$ to both sides then l.h.s is easy but how will the $\left(\frac{b}{a}\right)^n$ term on r.h.s will be effected. Kindly point out my mistake.

$\endgroup$
  • $\begingroup$ you forgot to raise $(b/a)^n$ to the power as well $\endgroup$ – Zach Effman Mar 17 '16 at 16:38
  • $\begingroup$ @ZachEffman where? $\endgroup$ – Saksham Mar 17 '16 at 16:41
  • $\begingroup$ In the only step you show here $\endgroup$ – Zach Effman Mar 17 '16 at 16:42
  • $\begingroup$ Also $(\frac{x_0}{y_0})^{(\frac{1}{n-1})^{n-1}} = \frac{x_0}{y_0}$, not what you have $\endgroup$ – Zach Effman Mar 17 '16 at 16:43
  • $\begingroup$ @ZachEffman I don't think so. Kindly see the steps again. $\endgroup$ – Saksham Mar 17 '16 at 16:44
2
$\begingroup$

In general $(ab)^n$ = $a^n b^n$ and $(a^n)^m = a^{nm}$. Therefore raising both sides to the power $n-1$ yields

$$\left(\frac{\cos\alpha}{\sin\alpha}\right)^{n-1} = \left(\left(\frac{b}{a}\right)^{n} \left(\frac{x_0}{y_0}\right)^\frac{1}{n-1}\right)^{n-1} =\left(\frac{b}{a}\right)^{n(n-1)} \left(\frac{x_0}{y_0}\right).$$

$\endgroup$
  • $\begingroup$ <a href="math.stackexchange.com/a/1700351/317580">so I think this one's wrong</a> $\endgroup$ – Saksham Mar 17 '16 at 16:50
  • $\begingroup$ @user109256 I don't understand what you're saying. $\endgroup$ – Alex Provost Mar 17 '16 at 16:53
  • $\begingroup$ I think this solution has a mistake math.stackexchange.com/a/1700351/317580 see the similar steps. He manipulated them wrong. $\endgroup$ – Saksham Mar 17 '16 at 16:55
  • $\begingroup$ @user109256 I don't see a similar manipulation in his answer. $\endgroup$ – Alex Provost Mar 17 '16 at 16:57
  • $\begingroup$ the step where he equated $m_1 = m_2$ $\endgroup$ – Saksham Mar 17 '16 at 16:59

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.