`CMR: \frac{1}{sin^4x}-cot^4x=1+\frac{2}{sin^2x}` 29/08/2021 Bởi Clara `CMR: \frac{1}{sin^4x}-cot^4x=1+\frac{2}{sin^2x}`
Giải thích các bước giải: Ta có: \(\begin{array}{l}\dfrac{1}{{{{\sin }^4}x}} – {\cot ^4}x\\ = \dfrac{1}{{{{\sin }^4}x}} – {\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^4}\\ = \dfrac{1}{{{{\sin }^4}x}} – \dfrac{{{{\cos }^4}x}}{{{{\sin }^4}x}}\\ = \dfrac{{1 – {{\cos }^4}x}}{{{{\sin }^4}x}}\\ = \dfrac{{\left( {1 – {{\cos }^2}x} \right)\left( {1 + {{\cos }^2}x} \right)}}{{{{\sin }^4}x}}\\ = \dfrac{{{{\sin }^2}x.\left( {1 + {{\cos }^2}x} \right)}}{{{{\sin }^4}x}}\,\,\,\,\,\,\left( {{{\sin }^2}x + {{\cos }^2}x = 1} \right)\\ = \dfrac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}\\ = \dfrac{{1 + \left( {1 – {{\sin }^2}x} \right)}}{{{{\sin }^2}x}}\\ = \dfrac{{2 – {{\sin }^2}x}}{{{{\sin }^2}x}}\\ = – 1 + \dfrac{2}{{{{\sin }^2}x}}\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{1}{{{{\sin }^4}x}} – {\cot ^4}x\\
= \dfrac{1}{{{{\sin }^4}x}} – {\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^4}\\
= \dfrac{1}{{{{\sin }^4}x}} – \dfrac{{{{\cos }^4}x}}{{{{\sin }^4}x}}\\
= \dfrac{{1 – {{\cos }^4}x}}{{{{\sin }^4}x}}\\
= \dfrac{{\left( {1 – {{\cos }^2}x} \right)\left( {1 + {{\cos }^2}x} \right)}}{{{{\sin }^4}x}}\\
= \dfrac{{{{\sin }^2}x.\left( {1 + {{\cos }^2}x} \right)}}{{{{\sin }^4}x}}\,\,\,\,\,\,\left( {{{\sin }^2}x + {{\cos }^2}x = 1} \right)\\
= \dfrac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}\\
= \dfrac{{1 + \left( {1 – {{\sin }^2}x} \right)}}{{{{\sin }^2}x}}\\
= \dfrac{{2 – {{\sin }^2}x}}{{{{\sin }^2}x}}\\
= – 1 + \dfrac{2}{{{{\sin }^2}x}}
\end{array}\)
Không CM được.