3

I need to prove that $$\cos^2(\beta -\gamma)+ \cos^2( \gamma - \alpha) +\cos^2(\alpha -\beta) = 1+2 \cos(\beta- \gamma) \cos( \gamma - \alpha)\cos(\alpha -\beta) $$

To do this I have used the formula $2 cos^2 \theta = 1+ \cos2\theta$

But after using the formula it seems that, the whole equation is getting complex. Can you please help me with this?

Numerical Person
  • 977
  • You should use somehow that $\beta-\gamma+\gamma-\alpha+\alpha-\beta=0$ – marwalix Apr 26 '15 at 22:36
  • 2
    A slight simplification: let $\gamma - \alpha = x, \alpha - \beta = y$. Then the identity to prove is $$\cos^2(x+y) + \cos^2 x + \cos^2 y = 1 + 2\cos(x+y)\cos x\cos y$$ – GPerez Apr 26 '15 at 22:36
  • Can you please elaborate, My head is totally cranked. I wasted too many pages. – Numerical Person Apr 26 '15 at 22:39
  • It's easier then that. See my answer. – marty cohen Apr 26 '15 at 22:44
  • 1
    Make sure to ping people with the @ symbol so they know you're talking to them! I assume that you were addressing @marwalix, but his comment is also embodied by my previous one; you can simplify the expression to only have two "free variables", instead of three. Anyway, the procedure will be the same, just expand one side until you get the other (like in the answer given). The only thing I meant to accomplish was to tidy up the expressions a bit by getting rid of a dependent variable. – GPerez Apr 26 '15 at 22:46

2 Answers2

3

I'll rewrite it as $\cos^2(b -c)+ \cos^2( c - a) +\cos^2(a -b) = 1+2 \cos(b- c) \cos( c - a)\cos(a -b) $ because I am lazy.

Since $\cos(2u) =2\cos^2(u)-1 $ and $\cos(u)\cos(v) =\frac12(\cos(u-v)+\cos(u+v)) $, applying these whenever possible (without much thinking),

$\begin{array}\\ \cos(b- c) \cos( c - a)\cos(a -b) &=\frac12(\cos(b-2c+a)+\cos(b-a))\cos(a-b)\\ &=\frac12(\cos(b-2c+a)\cos(a-b)+\cos(b-a)\cos(a-b))\\ &=\frac14((\cos(2b-2c)+\cos(2a-2c))+2\cos^2(b-a))\\ &=\frac14((2\cos^2(b-c)-1)+(2\cos^2(a-c)-1))+2\cos^2(b-a))\\ &=\frac12(\cos^2(b-c)+\cos^2(a-c)-1+\cos^2(b-a))\\ \end{array} $

Q.E.D. (to my surprise).

marty cohen
  • 107,799
1

$$F=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+1+\cos^2C$$

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $$F=\cos(A+B)\cos(A-B)+1+\cos^2C$$

If $\cos(A+B)=\cos C\iff A+B=2m\pi\pm C,$ (where $m$ is any integer)

$\displaystyle F=\cos C\cos(A-B)+\cos(A+B)\cos C+1$ $\displaystyle=1+2\cos C[\cos(A+B)+\cos(A-B)]$ $\displaystyle=1+2\cos A\cos B\cos C$

Set $\beta-\gamma=A,\gamma-\alpha=B,\alpha-\beta=C$

Community
  • 1
lab bhattacharjee
  • 274,582