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Sin2A + cos4A = cos2A + sin4A​

User Napuzba
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1 Answer

5 votes

Answer:

Hello,

Explanation:

I suppose the question is to solve the equation.


Remember:\\\\sin(a)-sin(b)=2cos((a+b)/(2) )*sin((a-b)/(2) )\\cos(a)-cos(b)=2sin((a+b)/(2) )*cos((b-a)/(2) )\\\\\\sin(2a)+cos(4a)=cos(2a)+sin(4a)\\\\\Longleftrightarrow sin(4a)-sin(2a)=cos(4a)-cos(2a)\\\\\Longleftrightarrow 2cos((4a+2a)/(2))*sin((4a-2a)/(2))=2cos((4a+2a)/(2))*sin((2a-4a)/(2))\\\\\Longleftrightarrow cos(3a)*sin(a)=cos(3a)*(-sin(a))\\\\\Longleftrightarrow 2cos(3a)*sin(a)=0\\\\


sin (a)=0\ or\ cos(3a)=0\\\\\boxed{k\in\ \mathbb{Z}\ : a=k*\pi\ or\ a=((2k+1)*\pi)/(2) }\\

User Kayshawn
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