\sin^4 x + \cos^4 x = 1 - 2 \left(\frac{1}{4} \sin^2 2x\right) = 1 - \frac{1}{2} \sin^2 2x
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- \(\sin^2 x + \cos^2 x = 1\) なので、この式は次のように簡略化されます:
- \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x
- \sin^2 x \cos^2 x = \left(\frac{1}{2} \sin 2x\right)^2 = \frac{1}{4} \sin^2 2x
- \(\sin^2 2x\) の範囲は \([0, 1]\) です。したがって、\(\frac{1}{2} \sin^2 2x\) の最大値は \(\frac{1}{2}\) であり、\(1 - \frac{1}{2} \sin^2 2x\) の最小値は、\(\sin^2 2x = 1\) のときに達成されます:
- - \frac{1}{2} \cdot 1 = \frac{1}{2}
- したがって、\(\sin^4 x + \cos^4 x\) の最小値は \(\boxed{\frac{1}{2}}\) です。
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