Let us consider the equation cos4⁡xa+sin4⁡xb=1a+b,x∈0,π2 ;a,b>0cos8xa3+sin8xb3 isThe value of sin2x in terms of a and b is

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1/a+b

1a+b2

1a+b3

1a+b4

ba+b

b2−a2a2+b2

a2+b2b2−a2

answer is , .

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Detailed Solution

we have, cos4⁡xa+sin4⁡xb=1a+b=cos2⁡x+sin2⁡xa+b⇒ cos2⁡xcos2⁡xa−1a+b=sin2⁡x1a+b−sin2⁡xb⇒ cos2⁡xbcos2⁡x−asin2⁡xa(a+b)=sin2⁡xbcos2⁡x−asin2⁡xb(a+b)⇒ cos2⁡xa=sin2⁡xb⇒ 1−sin2⁡xa=sin2⁡xb⇒ sin2⁡x=ba+b and cos2⁡x=aa+b∴ sin8⁡xb3+cos8⁡xa3=b4b3(a+b)4+a4a3(a+b)4=1(a+b)3we have, cos4⁡xa+sin4⁡xb=1a+b=cos2⁡x+sin2⁡xa+b⇒ cos2⁡xcos2⁡xa−1a+b=sin2⁡x1a+b−sin2⁡xb⇒ cos2⁡xbcos2⁡x−asin2⁡xa(a+b)=sin2⁡xbcos2⁡x−asin2⁡xb(a+b)⇒ cos2⁡xa=sin2⁡xb⇒ 1−sin2⁡xa=sin2⁡xb⇒ sin2⁡x=ba+b and cos2⁡x=aa+b∴ sin8⁡xb3+cos8⁡xa3=b4b3(a+b)4+a4a3(a+b)4=1(a+b)3

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