Find the range of ${sin}^{4} x + {cos}^{4} x$

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Solution

As the function is of the form ${sin}^{2 n} x + {cos}^{2 n} x$
Thus,Range $\in [\frac{1}{2^{2 - 1}}, 1] = [\frac{1}{2}, 1]$

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Similar questions

\frac{{cos}^{4} x}{{cos}^{2} y} + \frac{{sin}^{4} x}{{sin}^{2} y} = 1

\frac{{cos}^{4} y}{{cos}^{2} x} + \frac{{sin}^{4} y}{{sin}^{2} x} = ?

\frac{{cos}^{4} x}{{cos}^{2} y} + \frac{{sin}^{4} x}{{sin}^{2} y} = 1

\frac{{cos}^{4} y}{{cos}^{2} x} + \frac{{sin}^{4} y}{{sin}^{2} x} = 1

I = \int \frac{d x}{{sin}^{4} x + {cos}^{4} x}

I

\int \frac{cot x}{\sqrt{{cos}^{4} x + {sin}^{4} x}} d x = . . . . . . + c

\int \frac{s i n x c o s x}{s i n^{4} x + c o s^{4} x} d x =

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