by a) \(1 + {\rm{ }}{\tan ^2}x{\rm{ }} = \frac{1}{{{{\cos }^2}x}}\) b) \(1 + {\cot ^2}x = \frac{1}{{{{\sin }^2}x}}\)
c) \({\cos ^4}x-{\rm{ si}}{{\rm{n}}^4}x = 2{\cos ^2}x{\rm{ }} – 1\) d) \({\sin ^6}x + {\cos ^6}x{\rm{ }} = {\rm{ }}1 – {\rm{ }}3{\sin ^2}x.{\cos ^2}x\)
Phương pháp giải:
Sử dụng công thức lượng giác: \(\left\{ \begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}\\\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }}\end{array} \right..\)
Sử dụng hằng đẳng thức.
Lời giải chi tiết:
a) \(1 + {\rm{ }}{\tan ^2}x{\rm{ }} = \frac{1}{{{{\cos }^2}x}}\)
\(VT = 1 + {\rm{ }}{\tan ^2}x\)\( = 1 + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\)\( = \frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}}\)\( = \frac{1}{{{{\cos }^2}x}} = VP\)(đpcm)
b) \(1 + {\cot ^2}x{\rm{ }} = \frac{1}{{{{\sin }^2}x}}\)
\(VT = 1 + {\cot ^2}x{\rm{ }} = 1 + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}\)\( = \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}} = \frac{1}{{{{\sin }^2}x}}\) (đpcm)
c) \({\cos ^4}x–{\sin ^4}x = 2{\cos ^2}x{\rm{ }} – 1\)
\(\begin{array}{l}{\cos ^4}x–{\sin ^4}x = \left( {{{\cos }^2}x–{{\sin }^2}x} \right)\left( {{{\cos }^2}x{\rm{ + }}{{\sin }^2}x} \right)\\ = {\cos ^2}x–{\sin ^2}x = {\cos ^2}x – \left( {1 – {{\cos }^2}x{\rm{ }}} \right)\\ = 2{\cos ^2}x{\rm{ }} – 1\,\,\,\,\left( {dpcm} \right)\end{array}\)
d. \({\sin ^6}x + {\cos ^6}x{\rm{ }} = 1 – 3{\sin ^2}x.{\cos ^2}x\)
\(\begin{array}{l}{\sin ^6}x + {\cos ^6}x = \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x – {{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right)\\ = \left( {{{\sin }^4}x + {{\cos }^4}x} \right) – {\sin ^2}x.{\cos ^2}x\\ = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} – 2{\sin ^2}x.{\cos ^2}x – {\sin ^2}x.{\cos ^2}x\\ = 1 – 3{\sin ^2}x.{\cos ^2}x\,\,\,\,\left( {dpcm} \right)\end{array}\)
