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Question

If sin4Aa+cos4Ab=1a+b, then sin8Aa3+cos8Ab3=


Solution

The correct option is A
sin4+Aa+cos4Ab=1a+b1a(1cos2A2)2+1b(1+cos2A2)2=1a+b

b(a+b)(1cos2A)2+a(a+b)(1+cos2A)2=4ab
[b(a+b)+a(a+b)]cos2 2A+[a(a+b)b(a+b)]2cos2A+b(a+b)+a(a+b)4ab=0
(a+b)2cos22A+2(a2b2)cos2A+(ab)2=0[(a+b)cos2A+(ab)]2=0

(a+b)cos2A+(ab)=0cos2A=bab+a

sin8+Aa3+cos8Ab3=1a3(1cos2A2)4+1b3(1+cos2A2)4

=116a3(1bab+a)4+116b3(1+bab+a)4=116a3(2ab+a)4+116b3(2bb+a)4

=16a416a3(a+b)4+16b416b3(a+b)4=a(a+b)4+b(a+b)4=1(a+b)3
 

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