\frac{4}{9} 相关知识点: 试题来源: 解析 B 由\cos \left( x-\frac{ \pi }{4} \right)=\frac{\sqrt{2}}{3},\cos \left( 2x-\frac{ \pi }{2} \right)=1-2\cos \left( x-\frac{ \pi }{4} \right)=1-2\times \frac{2}{9}=\frac{5}{9},由诱导公式,\cos \left...
1. 因为\alpha \in \left( 0,\frac{ \pi }{2} \right),所以\alpha +\frac{ \pi }{4}\in \left( \frac{ \pi }{4},\frac{3 \pi }{4} \right), 所以\sin \left( \alpha +\frac{ \pi }{4} \right)=\sqrt{1-{{\cos }^{2}}\left( \alpha +\frac{ \pi }{4} \right...
{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 ...
\int_{}^{}\frac{1}{a^{2}-x^{2}}dx=\frac{1}{2a}ln\left| \frac{x+a}{x-a}\right|+C ( 分母中如果是x^{2}-a^{2} 就把等式右边的ln中的分子分母颠倒) \int_{}^{}\frac{1}{\sqrt{a^{2}-x^{2}}}dx=arcsin\frac{x}{a}+C \int_{}^{}\frac{1}{\sqrt{a^{2}+x...
\displaystyle \rightarrow r^2=a^2cos(2\theta-\frac{\pi}{2}) \rightarrow pic3:r^2=a^2sin2\theta 接下来关于双纽线相关数据计算,只考虑采用极坐标形式。 在计算整条曲线弧长、面积、旋转体积、旋转表面积之前优先考虑其对称性。 2.弧长(不用掌握,了解即可!) 记图2的弧长为L,其中 \displaystyle...
右移 \frac{\pi }{12} 个单位 相关知识点: 试题来源: 解析 B y=\cos 3x\xrightarrow{向左平移\frac{\pi }{12}}y=\cos \left[ 3\left( x+\frac{\pi }{12} \right) \right]=\cos \left( 3x+\frac{\pi }{4} \right)故选\text{B}.反馈 收藏 ...
-\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...
{eq}y = x \cos x; \quad y' = \cos x - y \tan x; \quad y\left(\frac{\pi}{4}\right)=\frac{\pi}{4\sqrt 2} {/eq} Differential Equations: In this problem we want to verify that a given function i...
Suppose {eq}y {/eq} is a Quatient of two functions {eq}f(x) {/eq} and {eq}g(x) {/eq} such that {eq}y=\dfrac{f(x)}{g(x)} {/eq}, where {eq}g\left( x \right)\ne 0 {/eq}. Then, the derivative of {eq}y {/eq} is calculated as {eq}\dfrac{...
Taking log, \begin{align*} \log \left( \prod_{i=0}^{n-1} \left( 2 + \cos \left( \frac{i \pi}{ n } \right ) \right )^{\frac \pi n}\right ) &= \sum_{i=0}^{n-1} \frac ...