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spayn [35]
2 years ago
13

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

Mathematics
1 answer:
lbvjy [14]2 years ago
8 0

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)}

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