If $$\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1,$$;where abc are the angles of the triangle.! then find the value of $\sin(c)$. By trial and error put this triangle as right angled isosceles and got the answer..!! But i want a complete proof!

  • $\begingroup$ Hint: you can solve $\square \sin(c)+\square=\square$ with algebra. $\endgroup$ – anon Aug 17 '13 at 8:51
  • $\begingroup$ but how to simplify after that..?? $\endgroup$ – maths lover Aug 17 '13 at 9:01
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    $\begingroup$ (1) Are you being asked to find a numerical value for $\,\sin c\,$ ? (2) are you given that $\,a,b,c\,$ are the angles of a triangle (isosceles, even)? This may be hugely important... $\endgroup$ – DonAntonio Aug 17 '13 at 9:04

Consider the two unit vectors on $S^2$ with spherical polar coordinates $(a, 0)$ and $(b,\frac{\pi}{2}-c)$, its components in $\mathbb{R}^3$ are:

$$\begin{align}&(\sin a, 0, \cos a)\\ \text{ and }\quad&(\sin b \cos(\frac{\pi}{2}-c), \sin b\sin(\frac{\pi}{2}-c), \cos b) =(\sin b\sin c,\sin b\cos c,\cos b) \end{align}$$ The angle $\theta$ between them satisfies:

$$\cos\theta = \cos a\cos b + \sin a\sin b\sin c$$

Geometrically, $\cos\theta = 1$ if and only if these two unit vector coincides. This in turn implies $a = b$ and $c = \frac{\pi}{2}$. By sine law, the two sides opposites to angle $a$, $b$ have equal length. So the triangle is an right angled isosceles triangle and $\sin c = 1$.

Alternatively, one can use the identity

$$\begin{align} & (\sin a-\sin b \sin c)^2+(\sin b\cos c)^2 + (\cos a-\cos b)^2\\ = & 2 \left( 1 - ( \cos a\cos b + \sin a \sin b\sin c)\right) \end{align}$$

to conclude $\cos c = 0$ and hence $\sin c = 1$.


Conventionally angles are written in uppercase, the sides are denoted in lower case

As $0<A,B,C< \pi,0 <\sin A,\sin B,\sin C\le1$

$$\implies \sin A\sin B\sin C+\cos A\cos B\le \sin A\sin B+\cos A\cos B=\cos(A-B) $$

$$\implies 1\le \cos(A-B)$$ which is possible iff $\cos(A-B)=1\iff A-B=2n\pi$ where $n$ is any integer

But as $0<A,B< \pi, -\pi<A-B<\pi\implies A-B=0\iff A=B$

Then from the given relation, we have $\displaystyle \sin^2A\sin C+\cos^2A=1\iff \sin^2A\sin C=1-\cos^2A=\sin^2A$

$\displaystyle\implies \sin C=1$ as $\sin A\ne0$ as $0<A<\pi$

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    $\begingroup$ +1 Very nice! What motivated you to use an inequality to solve an equality? I would've never thought of that $\endgroup$ – Ovi Aug 17 '13 at 16:54
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    $\begingroup$ @Ovi, first of all, the right hand side $=1$ which is the upper limit of cosine/sine ratio-s. Secondly, the similarity of the Left Hand side with $\cos(A-B)$ formula $\endgroup$ – lab bhattacharjee Aug 17 '13 at 16:56
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    $\begingroup$ @mathslover, how about this method? $\endgroup$ – lab bhattacharjee Aug 17 '13 at 18:20

$$\begin{align} \sin(c)&=\frac{1-\cos(a)\cos(b)}{\sin(a)\sin(b)}\\ \sin(c)&=\frac{1}{\sin(a)\sin(b)}-\frac{\cos(a)\cos(b)}{\sin(a)\sin(b)} \end{align}$$

Can you simplify from here?

  • $\begingroup$ i want the numerical value if abc are the angles of a triangle $\endgroup$ – maths lover Aug 17 '13 at 9:14
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    $\begingroup$ @mathslover This will still work, since all you have to do is plug in the given values of $a$ and $b$. Unless there was something else you were looking for? $\endgroup$ – Ataraxia Aug 17 '13 at 9:22

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