Toán chứng minh: sin2a+sin5a-sin3a/1+cosa-2sin^binh2a=2sina 05/10/2021 By Eva chứng minh: sin2a+sin5a-sin3a/1+cosa-2sin^binh2a=2sina
Giải thích các bước giải: Ta có: \(\begin{array}{l}\dfrac{{\sin 2a + \sin 5a – \sin 3a}}{{1 + \cos a – 2{{\sin }^2}2a}}\\ = \dfrac{{\sin 2a + \left( {\sin 5a – \sin 3a} \right)}}{{\cos a + \left( {1 – 2{{\sin }^2}2a} \right)}}\\ = \dfrac{{\sin 2a + 2.cos\dfrac{{5a + 3a}}{2}.\sin \dfrac{{5a – 3a}}{2}}}{{\cos a + \cos 4a}}\\ = \dfrac{{2\sin a.\cos a + 2\cos 4a.\sin a}}{{\cos a + \cos 4a}}\\ = \dfrac{{2\sin a.\left( {\cos a + \cos 4a} \right)}}{{\cos a + \cos 4a}}\\ = 2\sin a\end{array}\) Trả lời
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{\sin 2a + \sin 5a – \sin 3a}}{{1 + \cos a – 2{{\sin }^2}2a}}\\
= \dfrac{{\sin 2a + \left( {\sin 5a – \sin 3a} \right)}}{{\cos a + \left( {1 – 2{{\sin }^2}2a} \right)}}\\
= \dfrac{{\sin 2a + 2.cos\dfrac{{5a + 3a}}{2}.\sin \dfrac{{5a – 3a}}{2}}}{{\cos a + \cos 4a}}\\
= \dfrac{{2\sin a.\cos a + 2\cos 4a.\sin a}}{{\cos a + \cos 4a}}\\
= \dfrac{{2\sin a.\left( {\cos a + \cos 4a} \right)}}{{\cos a + \cos 4a}}\\
= 2\sin a
\end{array}\)