$\sin\theta + \cos\theta = \frac{1}{4}$のとき、$\sin\theta\cos\theta$の値と、$\sin^3\theta + \cos^3\theta$の値を求める問題です。

1. 問題の内容

$\sin\theta + \cos\theta = \frac{1}{4}$のとき、$\sin\theta\cos\theta$の値と、$\sin^3\theta + \cos^3\theta$の値を求める問題です。

2. 解き方の手順

まず、$\sin\theta\cos\theta$の値を求めます。

$\sin\theta + \cos\theta = \frac{1}{4}$の両辺を2乗します。

$(\sin\theta + \cos\theta)^2 = (\frac{1}{4})^2$

$\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = \frac{1}{16}$

$\sin^2\theta + \cos^2\theta = 1$なので、

$1 + 2\sin\theta\cos\theta = \frac{1}{16}$

$2\sin\theta\cos\theta = \frac{1}{16} - 1$

$2\sin\theta\cos\theta = \frac{1}{16} - \frac{16}{16}$

$2\sin\theta\cos\theta = -\frac{15}{16}$

$\sin\theta\cos\theta = -\frac{15}{32}$

次に、$\sin^3\theta + \cos^3\theta$の値を求めます。

$\sin^3\theta + \cos^3\theta$は因数分解できます。

$\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta)$

$\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$

$\sin\theta + \cos\theta = \frac{1}{4}$、$\sin\theta\cos\theta = -\frac{15}{32}$を代入します。

$\sin^3\theta + \cos^3\theta = (\frac{1}{4})(1 - (-\frac{15}{32}))$

$\sin^3\theta + \cos^3\theta = (\frac{1}{4})(1 + \frac{15}{32})$

$\sin^3\theta + \cos^3\theta = (\frac{1}{4})(\frac{32}{32} + \frac{15}{32})$

$\sin^3\theta + \cos^3\theta = (\frac{1}{4})(\frac{47}{32})$

$\sin^3\theta + \cos^3\theta = \frac{47}{128}$

3. 最終的な答え

$\sin\theta\cos\theta = -\frac{15}{32}$

$\sin^3\theta + \cos^3\theta = \frac{47}{128}$