Cho\(\pi< a< \dfrac{3\pi}{2}\).Trong các khẳng định sau khẳng định nào đúng?
A sin(\(\dfrac{7\pi}{2}+a\))>0
B sin(\(\dfrac{7\pi}{2}+a\))≥0
C sin(\(\dfrac{7\pi}{2}+a\))<0
D sin(\(\dfrac{7\pi}{2}+a\))≤0
Cho cot\(\dfrac{\Pi}{14}\) = a. Tính K theo a:
K = sin\(\dfrac{2\Pi}{7}\) + sin\(\dfrac{4\Pi}{7}\) + sin\(\dfrac{6\Pi}{7}\)
cho \(\dfrac{\pi}{2}\)<α<\(\pi\). tìm khẳng định đúng?
A. sin α<0 B. tan α>0 C. cot α>0 D. cos α<0
giải chi tiết nha
Vì 0 < α < π/2 nên sin α > 0, cos α > 0, tan α > 0, cot α > 0.
`\pi/2 < \alpha < \pi=>\alpha` nằm ở góc phần tư thứ `2`
`=>{(sin \alpha > 0;cos \alpha < 0),(tan \alpha < 0; cot \alpha < 0):}`
`->\bb D`
Tính giá trị biểu thức
\(A=sin\dfrac{\pi}{9}-sin\dfrac{5\pi}{9}+sin\dfrac{7\pi}{9}\)
\(A=sin\left(\dfrac{7}{9}pi\right)+sin\left(\dfrac{pi}{9}\right)-sin\left(\dfrac{5}{9}pi\right)\)
\(=2\cdot sin\left(\dfrac{1}{2}\cdot\dfrac{8}{9}pi\right)\cdot cos\left(\dfrac{1}{2}\cdot\dfrac{6}{9}pi\right)-sin\left(\dfrac{5}{9}pi\right)\)
\(=sin\left(\dfrac{4}{9}pi\right)-sin\left(\dfrac{5}{9}pi\right)\)
\(=2\cdot cos\left(\dfrac{\dfrac{4}{9}pi+\dfrac{5}{9}pi}{2}\right)\cdot sin\left(\dfrac{\dfrac{4}{9}pi-\dfrac{5}{9}pi}{2}\right)\)
=0
1; tính B \(=4sin^4\dfrac{\pi}{16}+2cos\dfrac{\pi}{8}\)
2;tính C= \(\dfrac{\sin\dfrac{\pi}{5}-\sin\dfrac{2\pi}{15}}{\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{15}}\)
3; tính D=\(\sin\dfrac{\pi}{9}-sin\dfrac{5\pi}{9}+sin\dfrac{7\pi}{9}\)
tính F=\(\sin^2\dfrac{\pi}{6}+\sin^2\dfrac{2\pi}{6}+...+\sin^2\dfrac{5\pi}{6}+\sin^2\pi\)
2/ biết \(\sin\beta=\dfrac{4}{5},0< \beta< \dfrac{\pi}{2}\) giá trị của biểu thúc a=\(\dfrac{\sqrt{3}\sin\left(\alpha+\beta\right)-\dfrac{4\cos\left(\alpha+\beta\right)}{\sqrt{3}}}{\sin\alpha}\)
Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)
Trong các khẳng định sau, khẳng định nào là sai?
A. \(\sin \left( {\pi - \alpha } \right) = \sin \alpha \)
B. \(\cos \left( {\pi - a} \right) = \cos \alpha \)
C. \(\sin \left( {\pi + \alpha } \right) = - \sin \alpha \).
D. \(\cos (\pi + \alpha ) = - \cos \alpha \)
Ta có: \(\cos \left( {\pi - \alpha } \right) = - \cos \alpha \)
Vậy ta chọn đáp án B
tính \(A=\sin^2\dfrac{\pi}{9}+\sin^2\dfrac{2\pi}{9}+\sin\dfrac{\pi}{9}\cdot\sin\dfrac{2\pi}{9}\)
Tính
\(sin\dfrac{\pi}{30}.sin\dfrac{7\pi}{30}.sin\dfrac{13\pi}{30}.sin\dfrac{19\pi}{30}.sin\dfrac{25\pi}{30}\)
Chứng minh đẳng thức: \(\dfrac{tan\left(\alpha-\dfrac{\pi}{2}\right).cos\left(\dfrac{3\pi}{2}+\alpha\right)-sin^3\left(\dfrac{7\pi}{2}-\alpha\right)}{cos\left(\alpha-\dfrac{\pi}{2}\right).tan\left(\dfrac{3\pi}{2}+\alpha\right)}=sin^2\alpha\)
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)