若f(x)在[0,\pi ]上有且只有3个零点,则\begin{cases} \dfrac{7\pi }{3\omega }\leqslant \pi \\\dfrac{10\pi }{3\omega }>\pi \\\end{cases},得\frac{7}{3}\leqslant \omega < \frac{10}{3},故答案为:\left[ \frac{7}{3},\frac{10}{3} \right)....
确实,\(\cos\left(\frac{\pi}{6} - \alpha\right)\) 并不等同于 \(\sin(\alpha)\),但它们之间有一个特定的关系。根据余弦的减法公式,我们有:\[\cos\left(\frac{\pi}{6} - \alpha\right) = \cos\left(\frac{\pi}{6}\right)\cos(\alpha) + \sin\left(\frac{\pi}{6}\...
Solving this type of equations may require the use of trigonometric identities and the inverse of the previously mentioned trigonometric functions. Answer and Explanation: Let the missing angle be θ and write: $$\displaystyle \begin{aligned} \cos\left(\frac{\pi}{6} \right) ...
Answer to: In the exercise, find all solutions of the equation in the interval ~[0, 2π). \cos\left ( x+\frac{\pi}{6} \right )-\cos\left (...
\frac{19}{17} 相关知识点: 试题来源: 解析 A 令t=\alpha +\frac{ \pi }{6},\alpha \in \left( \frac{ \pi }{6},\frac{2 \pi }{3} \right),则\cos t=\frac{4}{5},\sin t>0,\sin t=\frac{3}{5},\tan t=\frac{3}{4},\tan \left( 2\alpha +\frac{ \pi ...
∵\sin \left( \alpha -\frac{ \pi }{6} \right)=\frac{1}{3},∴\cos \left( \frac{ \pi }{3}+\alpha \right)=\cos \left[ \left( \alpha -\frac{ \pi }{6} \right)+\frac{ \pi }{2} \right]=-\sin \left( \alpha -\frac{ \pi }{6} \right)=-\frac{1}{3}...
Given: The trigonometric function is {eq}\cos \left(-\frac{\pi }{6}\right) {/eq}. Let us find the exact value of the trigonometric function: $$\be...Become a member and unlock all Study Answers Start today. Try it now Create an account Ask a questi...
(6)函数y=Asin(\omega x+\varphi)图象的对称轴为 x=\frac{k\pi}{\omega}-\frac{\varphi}{\omega}+\frac{\pi}{2\omega}(k\in Z) ,对称中心为 \left(\frac{k\pi}{\omega},0 \right)(k\in Z) ;函数y=Acos(\omega x+\varphi) 图象的对称轴为 x=\frac{k\pi}{\omega}-\frac{\varphi...
{eq}\cos \frac{\pi}{4} {/eq}Trigonometric Components:To evaluate values inside trigonometric components such as {eq}\sin(x) {/eq} and {eq}\cos(x) {/eq} the unit circle table of values can be utilized for common angles of radian measures. If this doesn't help, then the ...
-\frac{7}{8} 因为\cos \left( \frac{\pi }{4}-x \right)=\frac{1}{4}, 所以;\cos \left( \frac{\pi }{2}-2x \right)=\cos 2\left( \frac{\pi }{4}-x \right)=2{{\cos }^{2}}\left( \frac{\pi }{4}-x \right)-1=-\frac{7}{8}, 故\sin 2x=\cos \le...