Toán 3 sin^4 – cos^4=1/2 tính 2 sin ^4 -cos^4 02/08/2021 By Everleigh 3 sin^4 – cos^4=1/2 tính 2 sin ^4 -cos^4
Đáp án: 1/4 Giải thích các bước giải: $\begin{array}{l}3{\sin ^4}x – {\cos ^4}x = \frac{1}{2}\\ \Rightarrow 2{\sin ^4}x + {\sin ^4}x – {\cos ^4}x = \frac{1}{2}\\ \Rightarrow \frac{1}{2}.4{\sin ^4}x + \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^2}x – {{\cos }^2}x} \right) = \frac{1}{2}\\ \Rightarrow \frac{1}{2}.{\left( {2{{\sin }^2}x} \right)^2} – 1.\left( {{{\cos }^2}x – {{\sin }^2}x} \right) = \frac{1}{2}\\ \Rightarrow \frac{1}{2}.{\left( {1 – \cos 2x} \right)^2} – \cos 2x = \frac{1}{2}\\ \Rightarrow \frac{1}{2}.{\cos ^2}2x – \cos 2x + \frac{1}{2} – \cos 2x = \frac{1}{2}\\ \Rightarrow \frac{1}{2}{\cos ^2}2x – 2\cos 2x = 0\\ \Rightarrow \cos 2x = 0\left( {do:\cos 2x \le 1\forall x} \right)\\ \Rightarrow 2{\sin ^4}x – {\cos ^4}x\\ = {\sin ^4}x + {\sin ^4}x – {\cos ^4}x\\ = \frac{1}{4}.{\left( {2{{\sin }^2}x} \right)^2} – \cos 2x\\ = \frac{1}{4}.{\left( {1 – \cos 2x} \right)^2} – 0\\ = \frac{1}{4}\end{array}$ Trả lời
Đáp án: 1/4
Giải thích các bước giải:
$\begin{array}{l}
3{\sin ^4}x – {\cos ^4}x = \frac{1}{2}\\
\Rightarrow 2{\sin ^4}x + {\sin ^4}x – {\cos ^4}x = \frac{1}{2}\\
\Rightarrow \frac{1}{2}.4{\sin ^4}x + \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^2}x – {{\cos }^2}x} \right) = \frac{1}{2}\\
\Rightarrow \frac{1}{2}.{\left( {2{{\sin }^2}x} \right)^2} – 1.\left( {{{\cos }^2}x – {{\sin }^2}x} \right) = \frac{1}{2}\\
\Rightarrow \frac{1}{2}.{\left( {1 – \cos 2x} \right)^2} – \cos 2x = \frac{1}{2}\\
\Rightarrow \frac{1}{2}.{\cos ^2}2x – \cos 2x + \frac{1}{2} – \cos 2x = \frac{1}{2}\\
\Rightarrow \frac{1}{2}{\cos ^2}2x – 2\cos 2x = 0\\
\Rightarrow \cos 2x = 0\left( {do:\cos 2x \le 1\forall x} \right)\\
\Rightarrow 2{\sin ^4}x – {\cos ^4}x\\
= {\sin ^4}x + {\sin ^4}x – {\cos ^4}x\\
= \frac{1}{4}.{\left( {2{{\sin }^2}x} \right)^2} – \cos 2x\\
= \frac{1}{4}.{\left( {1 – \cos 2x} \right)^2} – 0\\
= \frac{1}{4}
\end{array}$