Tính f'(x) biết \(f\left( x \right) = {\cos ^2}x + {\cos ^2}\left( {{{2\pi } \over 3} + x} \right) + {\cos ^2}\left( {{{2\pi } \over 3} - x} \right).\)
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Lời giải:
Báo sai\(\begin{array}{l}
f\left( x \right) = {\cos ^2}x + {\cos ^2}\left( {\frac{{2\pi }}{3} + x} \right) + {\cos ^2}\left( {\frac{{2\pi }}{3} - x} \right)\\
= {\cos ^2}x + {\left( {\cos \frac{{2\pi }}{3}\cos x - \sin \frac{{2\pi }}{3}\sin x} \right)^2}\\
+ {\left( {\cos \frac{{2\pi }}{3}\cos x + \sin \frac{{2\pi }}{3}\sin x} \right)^2}\\
= {\cos ^2}x + {\left( { - \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x} \right)^2}\\
+ {\left( { - \frac{1}{2}\cos x + \frac{{\sqrt 3 }}{2}\sin x} \right)^2}\\
= {\cos ^2}x + \left( {\frac{1}{4}{{\cos }^2}x + \frac{{\sqrt 3 }}{2}\sin x\cos x + \frac{3}{4}{{\sin }^2}x} \right)\\
+ \left( {\frac{1}{4}{{\cos }^2}x - \frac{{\sqrt 3 }}{2}\sin x\cos x + \frac{3}{4}{{\sin }^2}x} \right)\\
= {\cos ^2}x + \frac{1}{2}{\cos ^2}x + \frac{3}{2}{\sin ^2}x\\
= \frac{3}{2}{\cos ^2}x + \frac{3}{2}{\sin ^2}x\\
= \frac{3}{2}\left( {{{\cos }^2}x + {{\sin }^2}x} \right)\\
= \frac{3}{2}\\
\Rightarrow f'\left( x \right) = 0
\end{array}\)