Question

(cos2A−cos2B)2+(sin2A+sin2B)2=4sin2(A+B)​

Answer 1

Answer:

${( \cos2A - \cos2B) }^{2} + {( \sin2A + \sin2 B }^{2}\\ = ( - 2\cos2A \cos2B + { \cos {}^{2} 2A } + \cos {}^{2} 2B) + (2 \sin2A \sin2B + \sin {}^{2} 2A + \sin {}^{2} 2B)\\ \\ { = - 2( \sin2A + \sin 2 B - \cos2 A - \cos2B) }\\ \\ { = - 2( - 4 \sin( A + B) \cos(A + B) } \\ \\ = {\tt{double \: angle \: of \: sine}}\\ \\ { = 2 \times 2(2 \sin (A + B) \cos(A + B)) }\\ \\ { = 4 \sin2(A + B)}$