Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$

Question

I want to prove this identity:

$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$

Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be the right.

Answer 1

dont know of an elegant method: on LHS substitute $c = a - b$

$LHS = \sin^2 a + \sin^2 b + (\sin a\cos b - \cos a \sin b)^2 $

$= \sin^2 a + \sin^2 b + \sin^2a \cos^2b + \cos^2a \sin^2b - 2\cos a\cos b \sin a \sin b$

$ = \sin^2 a + \sin^2 b + (1 - \cos^2a)\cos^2b + \cos^2a(1-\cos^2b)- 2\cos a\cos b \sin a \sin b $

$ = 2 - 2\cos^2a\cos^2b - 2\cos a\cos b \sin a \sin b$

$= 2 - 2 \cos a \cos b(\cos a\cos b + \sin a \sin b)$

$= 2 (1 - \cos a \cos b\cos(a-b)) = 2(1 - \cos a \cos b\cos c)$

Answer 2

If $a = b+ c$ then

$\cos a = \cos b\cos c - \sin b\sin c$

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

So solving $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$

involves solving

$\sin^2 b + \sin^2 c + \sin^2 b\cos^2 c + 2\cos b\sin b\cos c\sin c + \cos^2 b \sin^2 c = 2(1 - \cos b^2 \cos^2 c + \cos b\cos c\sin b\sin 2)$

$\sin^2 b + \sin^2 c + \sin^2 b\cos^2 c +\cos^2 b \sin^2 c +\cos^2 b \cos^2 c+\cos^2 b \cos^2 c= 2$

$\sin^2 b + \sin^2 c + \cos^2 c(\sin^2 b +\cos^2 b)+\cos^2 b(\sin^2 c + \cos^2 c) = 2$

$\sin^2 b +\sin^2 c +\cos^2 c + \cos^2 b = 2$

which is always true.

So all $a = b+c$ will solve the equation.

...

So I have to wonder if the problem wasn't supposed to be not that we were told $a = b+c$ but we were suppose to solve the equation to find that $a = b+c$ is a solution.

Well, no point restarting from scratch. If we *don't know that $a = b+c$ ...

Let $d = b+c$.

So

$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$

$\sin^2 a + \sin^2 d + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) +\sin^2 d$

$\sin^2 a + 2(1 - \cos d\cos b \cos c) = 2(1-\cos a \cos b \cos c) +\sin^2 d$ (from above)

$1 - \cos^2 a - 1 + \cos^2 d = 2\cos b\cos c(\cos d - \cos a)$

$(\cos d - \cos a)(\cos d + \cos a) = 2\cos b\cos c(\cos d - \cos a)$.

So If $\cos d = \cos a$ then $a = \pm d = \pm (b + c)$ are one system of solutions.

Other wise

$\cos d + \cos a = 2\cos b\cos c$

$\cos b\cos c - \sin b\sin c + \cos a = 2\cos b\cos c$

$\cos a = \cos b\cos c + \sin b\sin c = \cos (b-c)$ so

$a = \pm (b-c)$.

So (modulo $2\pi$) all solution are $a = \pm b \pm c$.

Answer 3

Let $\, X := e^{ia}, Y := e^{ib}, Z := e^{ic}. \,$ Now expand and factor $$ W \!:=\! \sin^2 a \!+\! \sin^2 b \!+\! \sin^2 c \!-\! 2(1 \!-\! \cos a \cos b \cos c) \!=\! \frac{(X Y \!-\! Z) (Y Z \!-\! X) (Z X \!-\! Y) (X Y Z \!-\! 1)}{(2 X Y Z)^2}.\,$$ This leads to $$\, W = 4 \sin\frac{a+b-c}2 \sin\frac{a-b+c}2 \sin\frac{-a+b+c}2 \sin\frac{a+b+c}2 \,$$ which is zero iff at least one of $\, a+b-c,\, a-b+c,\, -a+b+c,\, a+b+c \,$ are multiples of $\,2\pi.\,$

Answer 4

Ignoring the condition $a=b+c$, we can see that the equation can be manipulated into an interesting form:

$$\begin{align} \sin^2 a + \sin^2 b + \sin^2 c &= 2 ( 1 - \cos a\cos b\cos c ) \\ ( 1 - \cos^2 a ) + ( 1 - \cos^2 b ) + ( 1 - \cos^2 c ) &= 2 - 2 \cos a\cos b\cos c \\ 1 - \cos^2 b - \cos^2 c &= \cos^2 a - 2 \cos a\cos b\cos c \\ \qquad 1- \cos^2 b - \cos^2 c + \cos^2 b\cos^2 c &= \cos^2 a - 2 \cos a \cos b\cos c + \cos^2 b \cos^2 c \\ (1-\cos^2b)(1-\cos^2 c) &= ( \cos a - \cos b \cos c )^2 \\ \sin^2 b\sin^2 c &= ( \cos a - \cos b \cos c )^2 \\ 0 &= ( \cos a - \cos b \cos c )^2 -\sin^2 b\sin^2 c \\ 0 &= (\cos a - \cos b \cos c + \sin b \sin c) \\ &\phantom{=}\cdot( \cos a - \cos b \cos c - \sin b\sin c) \\ 0 &= (\cos a - \cos(b+c))( \cos a - \cos(b-c)) \end{align}$$

Thus, $\cos a = \cos(b\pm c)$, so that

$$a = \pm b \pm c + 2\pi k \quad\text{(with independent $\pm$s)}$$

In particular, $a=b+c$ is one of those cases.

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It's perhaps worth noting that the original equation reduces even further using the "difference-to-product" identity: $$\cos\theta - \cos\phi = - 2 \sin\frac{\theta+\phi}{2}\cdot\sin\frac{\theta-\phi}{2}$$ to $$\sin\frac{a+b+c}{2}\cdot\sin\frac{-a+b+c}{2}\cdot\sin\frac{a-b+c}{2}\cdot\sin\frac{a+b-c}{2} = 0$$ which bears an interesting resemblance to Heron's formula for the area of a triangle. Moreover, had the sign of the $2\cos a\cos b\cos c$ term been reversed, then we'd have instead $$(\cos a + \cos(b+c))( \cos a + \cos(b-c)) = 0$$ which gives rise to another Heron-like form: $$\cos\frac{a+b+c}{2}\cdot\cos\frac{-a+b+c}{2}\cdot\cos\frac{a-b+c}{2}\cdot\cos\frac{a+b-c}{2} = 0$$ This relation is satisfied, for instance, when $a+b+c=180^\circ$; that is, when $a$, $b$, $c$ are the angles of a triangle.

I guess my point is that the equation in question isn't entirely random.