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greendayluvr13 · Nov 18, 2011, 11:32 AM

cos(a)-cos(b)^(2) sin(a)-sin(b)^(2) = (4sin^(2)((a-b)/2))

ebaines · Nov 18, 2011, 11:59 AM

I interpret what you wrote as this:

$ \cos(a)-\cos^2(b)\sin(a) - \sin^2(b) = 4 \sin^2 (\frac {a-b} 2) $

But these are not equal. Please re-type the equation and be careful with parentheses.

Aurora2000 · Nov 18, 2011, 02:02 PM

Even if interpreted as
$(\cos(a)-\cos(b))^2(\sin(a)-\sin(b))^2=4\sin^2(\frac{a-b}{2})$
is false too, just try some particular value for $a,\ b$ .
(Hint: RHS can reach 4 in module.)

ebaines · Nov 18, 2011, 02:39 PM

Got it!

What the OP meant to write is this:

$ (\cos a - \cos b)^2+ (\sin a - \sin b)^2 = 4 \sin (\frac {a-b} 2) $

(hint - quite often when OP's cut and paste a formula from another source the '+' signs don't copy in properly).

OK - if you multiply this out and apply the identy cos^2x + sin^2x = 1 you get to a fairly simple result. Then the trick is to use the fact that

$ \cos (a-b) = \cos (2 (\frac {a-b} 2 )) = 1 - sin^2 (\frac {a-b} 2) $

From there it falls right out.

Aurora2000 · Nov 18, 2011, 02:55 PM

Just a minor note, $\cos(2x)=\cos^2(x)-\sin^2(x)=1-2\sin(x)$ :-)