Chủ đề này có 2 trả lời

[blackhole]

1) Tính giá trị của biểu thức sau:

$$\cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + ..... + \cos \dfrac{{6\pi }}{7}$$

2)Chứng minh cá đẳng thức sau:
a)$$\dfrac{1}{{16}} + \dfrac{1}{{32}}\cos 2x - \dfrac{1}{{16}}\cos 4x - \dfrac{1}{{32}}\cos 6x = {\sin ^2}x.{\cos ^4}x$$

[khanh3570883]

1)
Ta có:
$\begin{array}{l}
c{\rm{os}}0 + \cos \dfrac{\pi }{7} + ... + \cos \dfrac{{6\pi }}{7}\\
= \cos 0 + \dfrac{{2\sin \dfrac{\pi }{{14}}\left( {\cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + ... + \cos \dfrac{{6\pi }}{7}} \right)}}{{2\sin \dfrac{\pi }{{14}}}}\\
= 1 + \dfrac{{\cos \dfrac{\pi }{2}\sin \dfrac{{3\pi }}{7}}}{{\sin \dfrac{\pi }{{14}}}} = 1
\end{array}$

[spiderandmoon]

2)
VT= $\dfrac{1}{16}(1-cos4x) + \dfrac{1}{32}(cos2x-cos6x)$
=$\dfrac{1}{8}sin^2 2x + \dfrac{1}{16}sin2x.sin4x$
=$\dfrac{1}{8}sin^2 2x + \dfrac{1}{8}sin^2 2x.cos 2x$
=$\dfrac{1}{8}sin^2 2x + \dfrac{1}{8}sin^2 2x.cos 2x$
=$\dfrac{1}{8}sin^2 2x*(1 + cos 2x)$
=$\dfrac{1}{4}sin^2 2x*cos^2 x$
=$\sin^2 x.cos^4 x$