解答过程如下:①由题意得/()=co2-sin2-0-1=-1;②由题意得f(x)=cos2x-sin2x=cos2x令t=2x,因为4所以-32x≤2所以-长2画出函数y=cost在区间[-r,r上的图象由图象可知,y=cost在区间-5,0]上是增函数,在区间[0,上是减函数,且cos()cos,所以当t=,即x=时,f(x)取得最小值0当t=0,即x=0时,...
cos2x>sin2x⟺cos2x=cos2x−sin2x>0⟺2nπ−2π<2x<2nπ+2πwherenis any integer What are the values and types of the critical points, if any, off(x)=cos2x−sin2x? https://socratic.org/questions/what-are-the-values-and-types-of-the-critical-points-if-any-of-f-x-...
刷刷题APP(shuashuati.com)是专业的大学生刷题搜题拍题答疑工具,刷刷题提供函数y=(cos2x-sin2x)tan2x的最小正周期是( )A.pi;2B.pi;C.2pi;D.4pi;的答案解析,刷刷题为用户提供专业的考试题库练习。一分钟将考试题Word文档/Excel文档/PDF文档转化为在线题库,制作自己的电
cos2x+sin2x−2sinxcosx cos2x−sin2x= (cosx−sinx)2 (cosx−sinx)×(cosx+sinx)═ cosx−sinx cosx+sinx= 1−tanx 1+tanx=右边∴ 1−2sinxcosx cos2x−sin2x= 1−tanx 1+tanx成立 结果一 题目 SectionBDirections:Completethe following passage by using the words in the box. Each...
【解析】10.令 f(x)=|sin2x|+|cos2x| ,首先可以证明 π/(4) 函数f(x)的一个周期=|cos2x|+|sin2x|=f(x) π/(4)是函数的一个周期,若存在 0T_0π/(4) ,使 f(x+T_0)=f(x) 成立,则有 f(T_0)=f(0),即 |sin2To|+|cos2T_0|=1 , sin2T_0⋅cos2T_0=0 cos2T_0=0 或 ...
【解析】对于A: x∈(π/(4),π/(2)) 时显然 y=2cosx 和 y=sin2x 在 (π/(4),π/(2)) 递减,故f()在 (π/(4),π/(2)) 递减,故A正确;对于B: x∈(0,π/(6)) 时显然 y=2cosx 在 (0,π/(6)) 递减,而 y=sin2x 在(0,π/(6)) 递增故B不合题意;对于C :x∈(π/(2)...
☆1 equations [ 1/1 Equation] Work: Find the solution of equation sin2x+cos2x = 0 . Question type: Equation Solution: the solutions is: x=0 There are1solution(s).*Note: Radian. 解方程的详细方法请参阅:《方程的解法》 Your problem has not been solved here? Please go to theHot Proble...
to blow down 相关知识点: 试题来源: 解析 解:(1)∵m∥n,∴-1/2cosx=-√3sinx,∴tanx=(√3)/6,∴cos^2x-sin2x=(cos^2x-2sinxcosx)/(sin^2x+cos^2x)=(1-2tanx)/(tan^2x+1)=(1-(√3)/3)/(1/(12)+1)=(12-4√3)/(13);(2)f(x)=(m+n)⋅m=m^2+m⋅n=cos^...
(1)由f(x)=sin2x+cos2x=√2sin(2x+π4),所以周期 T=2π2=π,令2kπ−π2≤2x+π4≤2kπ+π2,得kπ−3π8≤x≤kπ+π8,(k∈Z),∴f(x)的单调递增区间为[kπ−3π8,kπ+π8],(k∈Z).(2)当x∈[0,π2]时,则2x+π4∈[π4,5π4],所以sin(2x+π4)∈[−√22...
Answer to: Find the slope of the tangent to y = sin2x + cos2x at the point where x = pi. a) 0.112 b) 0 c) 1 d) -3.124 By signing up, you'll get...