There is this problem: $$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{1}{\tan\alpha -\tan\beta}$$ I started as $$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{\sin2\alpha \cos\beta-\sin2\beta \cos\alpha}{2(\cos\beta -\cos\alpha)}$$ but I'm stuck here, because I don't see how could I use $\sin(x-y)$ but also don't see how could I use any other identity without complicating this even more.

  • $\begingroup$ I think you need to divide your result that you worked out by 2 (as $\sin2x=2\sin x \cos x$) $\endgroup$ Jan 10 at 18:12
  • $\begingroup$ Write everything in terms of sin and cos. Then multiply through by all denominators to get a polynomial equation in sin and cos $\endgroup$ Jan 10 at 18:15
  • $\begingroup$ @A-LevelStudent Corrected. $\endgroup$
    – thunder
    Jan 10 at 18:22
  • 1
    $\begingroup$ What is the question ? "How to simplify..."... is the problem to prove the identity? Or to simplify the equality and solve for all possible pairs $ (\alpha, \beta)$ ? $\endgroup$ Jan 10 at 18:28
  • $\begingroup$ @AdamRubinson good point, I assumed we were proving an identity, as the question made no mention of solving an equation. If we are indeed solving an equation, then the question makes more sense, as the 'identity' is wrong. $\endgroup$ Jan 10 at 18:30

The proposed identity is not true (even though after the correction).

Indeed, take $\alpha = \pi/4$ and $\beta = 0$. Then we obtain: \begin{align*} \left(\frac{\sin(\pi/4) - \sin(0)}{\cos(0) - \cos(\pi/4)}\right)\times\cos(\pi/4)\cos(0) = \frac{\sqrt{2}}{2 - \sqrt{2}}\times\frac{\sqrt{2}}{2} = \frac{1}{2-\sqrt{2}} \end{align*}

On the other hand, one has that \begin{align*} \frac{1}{2(\tan(\pi/4) - \tan(0))} = \frac{1}{2} \end{align*}


The proposed equation is equivalent to \begin{align*} \frac{(\sin(a) - \sin(b))\sin(a-b)}{\cos(b) - \cos(a)} = 1 \end{align*}

  • $\begingroup$ My bad. I happened to try out $2$ values that made the supposed answer only out by a factor of 2, hence my suggestions. $\endgroup$ Jan 10 at 18:28
  • $\begingroup$ @A-LevelStudent No worries, it happens. $\endgroup$
    – user0102
    Jan 10 at 18:28
  • $\begingroup$ As noted by AdamRubinson the OP may in fact be attempting to solve an equation, not prove an identity. $\endgroup$ Jan 10 at 18:30
  • $\begingroup$ I will edit my answer then. Thanks for pointing out. $\endgroup$
    – user0102
    Jan 10 at 18:31
  • $\begingroup$ No problem, I also made that mistake initially. $\endgroup$ Jan 10 at 18:31

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