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Given a triangle $ABC$ with angles a,b & c, prove that if $\sin^2(a) + \sin^2(b) + \sin^2(c) = 2$ then the triangle is right angled (has an angle of $90^o$).

If I assume the triangle is right angled and have AB, AC and BC as sides, with BC being the base, then I can say that statement is true, since $\sin^2(a)$ would be 1 where $a=90^o$.

Also, since it's a right angled triangle we can say that $\sin^2(b) = AC^2/BC^2$ and $\sin^2(c) = AB^2/BC^2$. The sum of $\sin^2(b) + \sin^2(c)$ would be equal to $(AC^2+AB^2)/BC^2$ which is 1 (pythagoras theorem). Hence the sum of squares of sines of all 3 angles is indeed 2.

But how do I prove this? As in, I'm only given the starting equation, how do I get from there to a right angled triangle?

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  • $\begingroup$ Note that, $\sin^2b=\cos^2a+\cos^2c$ or any other combination. $\endgroup$ – Mann Jun 15 '15 at 12:27
  • $\begingroup$ i think you mean $sin^2(c) = AB^2/BC^2$ and not $sin(c) = AB^2/BC^2$ $\endgroup$ – supinf Jun 15 '15 at 12:36
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    $\begingroup$ Yup, thanks for the catch, that's what I meant. Fixed now. $\endgroup$ – MikhaelM Jun 15 '15 at 12:38
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Notice, in $\Delta ABC$ , we have $a+b+c=180^o$

Given that $$\sin^2 a+\sin^2 b+\sin^2 c=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (180^o-(a+b))=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (a+b)=2$$ $$\implies (\sin a\cos b+\cos a\sin b)^2=2-\sin^2 a-\sin^2 b$$ $$\implies \sin^2 a\cos^2 b+\cos^2 a\sin^2 b+2\sin a\sin b\cos a\cos b=\cos^2 a+\cos^2 b$$ $$\implies -(1-\sin^2 a)\cos^2 b-(1-\sin^2 b)\cos^2 a+2\sin a\sin b\cos a\cos b=0$$ $$\implies -\cos^2 a\cos^2 b-\cos^2 a\cos^2 b+2\sin a\sin b\cos a\cos b=0$$ $$\implies -2\cos a\cos b(\cos a\cos b-\sin a\sin b)=0$$ $$\implies \cos a\cos b\cos(a+b)=0$$ $$ \color{blue}{\text{if } \quad \cos a=0 \implies a=90^o}$$ $$ \color{blue}{\text{if } \quad \cos b=0 \implies b=90^o}$$ $$ \color{blue}{\text{if } \quad \cos (a+b)=0 \implies a+b=90^o \iff c=90^o}$$ We find that above three cases show that $\color{blue}{\Delta ABC}$ is $\color{blue}{\text{right angled}}$.

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  • $\begingroup$ Just checked your response and I find it really really good! Thanks for the reply, this solution is really good and easy to follow. $\endgroup$ – MikhaelM Jun 28 '15 at 13:54
  • $\begingroup$ It's worth noting that, since $a+b+c=180^\circ$, then $\cos(a+b) = -\cos c$. $\endgroup$ – steven gregory Jan 23 '17 at 15:28
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We have: $$\sin^2 A + \sin^2 B+\sin^2 C = 3-\sum_{cyc}\cos^2 A =\frac{1}{2}\left(3-\sum_{cyc}\cos(2A)\right)$$ hence $LHS=2$ is equivalent to $\sum_{cyc}\cos(2A)=-1$ or to: $$ \cos^2 A+\cos(B+C)\cos(B-C) = 0 \tag{1}$$ but since $A+B+C=\pi$, $\cos(B+C)=\cos(\pi-A)=-\cos(A)$ and the previous line is equivalent to: $$ \cos(A)=\cos(B-C)\quad\text{or}\quad \cos(A)=0,\tag{2} $$ so $A=\frac{\pi}{2}$ or: $$ -\cos(B)\cos(C)+\sin(B)\sin(C) = \cos(B)\cos(C)+\sin(B)\sin(C)\tag{3}$$ that gives $\cos(B)\cos(C)=0$.

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  • $\begingroup$ I don't quite understand your first line. What does that sum mean? $\endgroup$ – MikhaelM Jun 15 '15 at 12:49
  • $\begingroup$ @MikhaelM: $\sum_{cyc}\cos^2 A$ is just a shorthand notation for $\cos^2 A+\cos^2 B+\cos^2 C$. $\endgroup$ – Jack D'Aurizio Jun 15 '15 at 13:09
  • $\begingroup$ Thank you! Your solution is the easiest to understand, at least for me. $\endgroup$ – MikhaelM Jun 15 '15 at 14:56
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$$F=\sin^2A+\sin^2B+\sin^2C=1-(\cos^2A-\sin^2B)+1-\cos^2C$$

Method $\#1:$

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$F=2-\cos(A+B)\cos(A-B)-\cos^2C$$

Method $\#2:$ Alternatively using Double angle & Prosthaphaeresis formula ,

$$\sin^2A+\sin^2B=1-\dfrac{\cos2A+\cos2B}2=1-\cos(A+B)\cos(A-B)$$

Now $\cos(A+B)=\cos(\pi-C)=-\cos c$

$\implies F=2-[-\cos C\cos(A-B)+\cos^2C]$

$=2-\cos C[-\cos(A-B)+\cos C]$

$=2-\cos C[-\cos(A-B)-\cos(A+B)]$ as $\cos(A+B)=-\cos c$

$$\implies\sin^2A+\sin^2B+\sin^2C=2+\cos C[2\cos A\cos B]$$

Can you reach home from here?

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$$\sin(a)=\sin(\pi-b-c)=\sin(b+c),$$ Then with obvious shorthands, $$\begin{align}0&=2-s_a^2-s_b^2-s_c^2\\ &=c_b^2+c_c^2-s_a^2\\ &=c_b^2+c_c^2-(s_a c_b+c_a s_b)^2\\ &=c_b^2+c_c^2-s_b^2c_c^2-2s_b c_b c_c s_c-c_b^2s_c^2\\ &=c_b^2c_c^2+c_b^2c_c^2-2s_b c_b c_c s_c\\ &=2c_b c_c(c_b c_c-s_b s_c)\\ &=2c_bc_cc_{b+c}.\end{align}$$

The product is null with one of $$b=\frac\pi2,$$ $$c=\frac\pi2,$$ $$b+c=\pi-a=\frac\pi2.$$

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    $\begingroup$ i think he is asking for the other direction $\endgroup$ – supinf Jun 15 '15 at 12:34
  • $\begingroup$ @supinf: you are quite right, I am trying to fix. $\endgroup$ – Yves Daoust Jun 15 '15 at 12:36
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    $\begingroup$ I think this just proves that my sum is 2 in a right angled triangle. I think I need to prove that if the sum is 2, then the triangle is right angled. $\endgroup$ – MikhaelM Jun 15 '15 at 12:39
  • $\begingroup$ @MikhaelM: I have updated. $\endgroup$ – Yves Daoust Jun 15 '15 at 13:25

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