Prove that (cos 2A - cos 2B)^2 + (sin 2A - sin 2B)^2=4sin^2(A-B)....please help me to solve this trigo question...thank u

Hint: within the parentheses, use sum to product identities, and you will have it in a few steps.

$\displaystyle (\cos 2A - \cos 2B)^2 + (\sin 2A - \sin 2B)^2 = 4 \sin^2 (A-B)$

Use
$\displaystyle \cos(2A) - \cos(2B) = -2 \sin \left(\frac{2A+2B}{2}\right)\sin\left(\frac{2A-2B}{2}\right)$
$\displaystyle = - 2 \sin(A+B)\sin(A-B)$
$\displaystyle \sin(2A) - \sin(2B) = 2 \cos \left(\frac{2A+2B}{2}\right)\sin\left(\frac{2A-2B}{2}\right)$
$\displaystyle = 2 \cos(A+B)\sin(A-B)$

Can you proceed now?

im still having problem with this question....

Originally Posted by sharmala
im still having problem with this question....

Hello, sharmala!

We will need these two identities:

. . $\displaystyle \cos x\cos y + \sin x\sin y \:=\:\cos(x-y)$

. . $\displaystyle \frac{1-\cos2\theta}{2} \:=\:\sin^2\theta$

$\displaystyle \text{Prove: }\:(\cos2A - \cos2B)^2 + (\sin2A - \sin2B)^2\;=\;4\sin^2(A\!-\!B)$

Expand the left side:

$\displaystyle \cos^22A - 2\cos2A\cos2B + \cos^22B + \sin^22A - 2\sin2A\sin2B + \sin^22B$

. . . $\displaystyle =\;\underbrace{\sin^22A + \cos^22A}_1 \:+\: \underbrace{\sin^22B + \cos^22B}_1 \:-\: 2\big(\cos2A\cos2B + \sin2A\sin2B\big)$

. . . $\displaystyle =\;2 - 2\cos(2A\!-\!2B)$

. . . $\displaystyle =\;2\big[1 - \cos(2[A\!-\!B])\big]$

. . . $\displaystyle =\;4\left[\frac{1-\cos(2[A\!-\!B])}{2}\right]$

. . . $\displaystyle =\;4\sin^2(A\!-\!B)$

$\displaystyle [- 2 \sin(A+B)\sin(A-B)]^2+[2 \cos(A+B)\sin(A-B)]^2$
$\displaystyle = 4\sin^2(A+B)\sin^2(A-B)+4\cos^2(A+B)\sin^2(A-B)$
$\displaystyle = 4 \sin^2(A-B)[\sin^2(A+B)+\cos^2(A+B)]=4\sin^2(A-B)$