If sin4 Aa+cos4 Ab=1a+b, then the value of sin8 Aa3+cos8 Ab3 is equal to
It is given that sin4 Aa+cos4 Ab=1a+b
⇒(1−cos 2 A)24a+(1+cos 2 A)24b=1a+b⇒b(a+b)(1−2 cos 2A+cos2 2A)+a(a+b)(1+2 cos 2A+cos2 2A)=4ab⇒{b(a+b)+a(a+b)}cos2 2A+2(a+b)(a−b)cos 2A +a(a+b)+b(a+b)−4ab=0⇒(a+b)2 cos2 2A+2(a+b)(a−b)cos 2A+(a−b)2=0⇒{(a+b)cos 2A+(a−b)}2=0 or cos 2A=b−ab+a
sin4 Aa+cos4 Ab=1a+b⇒1a=1a+b⇒ b=0
Hence, sin8 Aa3+cos8 Ab3=(1−cos 2A)416 a3+(1+cos 2A)416 b3=116a3[1−b−ab+a]4+116b3[1+b−ab+a]4
=16a416a3(b+a)4+16b416b3(b+a)4=1(b+a)4(a+b)=1(a+b)3
Trick:Put A=90∘, then
∴ sin8 Aa3+cos8 Ab3=1a3 which is given by option (a) Note : Students can check this question for other values of A also.