chứng minh hệ thức $sin^{6}$($\frac{x}{2}$ ) – $cos^{6}$($\frac{x}{2}$) = $\frac{1}{4}$cos(x) ( $sin^{2}$ – 4 ) 30/10/2021 Bởi Jade chứng minh hệ thức $sin^{6}$($\frac{x}{2}$ ) – $cos^{6}$($\frac{x}{2}$) = $\frac{1}{4}$cos(x) ( $sin^{2}$ – 4 )
Giải thích các bước giải: Ta có: \(\begin{array}{l}{\sin ^6}\frac{x}{2} – {\cos ^6}\frac{x}{2}\\ = \left( {{{\sin }^2}\frac{x}{2} – {{\cos }^2}\frac{x}{2}} \right)\left( {{{\sin }^4}\frac{x}{2} + {{\sin }^2}\frac{x}{2}.{{\cos }^2}\frac{x}{2} + {{\cos }^4}\frac{x}{2}} \right)\\ = – \cos x.\left[ {\left( {{{\sin }^4}\frac{x}{2} + 2{{\sin }^2}\frac{x}{2}.{{\cos }^2}\frac{x}{2} + {{\cos }^4}\frac{x}{2}} \right) – {{\sin }^2}\frac{x}{2}.{{\cos }^2}\frac{x}{2}} \right]\\ = – \cos x.\left[ {{{\left( {{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2}} \right)}^2} – {{\sin }^2}\frac{x}{2}.{{\cos }^2}\frac{x}{2}} \right]\\ = – \cos x.\left[ {1 – \frac{1}{4}{{\left( {2\sin \frac{x}{2}.\cos \frac{x}{2}} \right)}^2}} \right]\\ = – \cos x.\left( {1 – \frac{1}{4}{{\sin }^2}x} \right)\\ = \frac{1}{4}\cos x\left( {{{\sin }^2}x – 4} \right)\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^6}\frac{x}{2} – {\cos ^6}\frac{x}{2}\\
= \left( {{{\sin }^2}\frac{x}{2} – {{\cos }^2}\frac{x}{2}} \right)\left( {{{\sin }^4}\frac{x}{2} + {{\sin }^2}\frac{x}{2}.{{\cos }^2}\frac{x}{2} + {{\cos }^4}\frac{x}{2}} \right)\\
= – \cos x.\left[ {\left( {{{\sin }^4}\frac{x}{2} + 2{{\sin }^2}\frac{x}{2}.{{\cos }^2}\frac{x}{2} + {{\cos }^4}\frac{x}{2}} \right) – {{\sin }^2}\frac{x}{2}.{{\cos }^2}\frac{x}{2}} \right]\\
= – \cos x.\left[ {{{\left( {{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2}} \right)}^2} – {{\sin }^2}\frac{x}{2}.{{\cos }^2}\frac{x}{2}} \right]\\
= – \cos x.\left[ {1 – \frac{1}{4}{{\left( {2\sin \frac{x}{2}.\cos \frac{x}{2}} \right)}^2}} \right]\\
= – \cos x.\left( {1 – \frac{1}{4}{{\sin }^2}x} \right)\\
= \frac{1}{4}\cos x\left( {{{\sin }^2}x – 4} \right)
\end{array}\)
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