ข้อ 6 อีกข้อ ไม่แน่ใจว่ามีคนเฉลยหรือยัง เพราะยังไม่ได้ค้น
% MathType!MTEF!2!1!+-
\[\begin{array}{l}
\\
from\quad T\left( x \right) = \sin x - {\cos ^2}x + {\sin ^3}x - {\cos ^4}x + {\sin ^5}x - {\cos ^6}x + ...\\
\quad \quad \quad \quad \quad \; = \left( {\sin x + {{\sin }^3}x + {{\sin }^5}x + ...} \right) - \left( {{{\cos }^2}x + {{\cos }^4}x + {{\cos }^6}x + ...} \right)\\
Geo.\;Series\quad {S_\infty } = \frac{{{a_1}}}{{1 - r}}\\
then\quad T\left( x \right) = \frac{{\sin x}}{{1 - {{\sin }^2}x}} - \frac{{{{\cos }^2}x}}{{1 - {{\cos }^2}x}}\\
3T\left( {\frac{\pi }{3}} \right)\quad = 3\left( {\frac{{6\sqrt 3 - 1}}{3}} \right)
\end{array}\]
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