To Find: maximum value of sin4x+cos4x Let f(x)=sin4x+cos4x=(sin2x)2+(cos2x)2 =(sin2x)2+(cos2x)2+2sin2xcos2x−2sin2xcos2x =(sin2x+cos2x)2−2sin2xcos2x(∵a2+b2+2ab=(a+b)2) =1−
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×4sin2xcos2x(∵sin2x+cos2x=1) =1−
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×(2‌sin‌x‌cos‌x)2 =1−
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×sin22x(∵2‌sin‌x‌cos‌x=sin‌2‌x) As we know sin2x lies between [0,1] So 0≤sin22x≤1 We need to take the minimum value of sin x as there is a minus sign in front of the sine term. So we need to take it as zero. ∴ Maximum value of f(x)=1−0=1