sin7π
@岳乳4270：sin7兀/12点算 -
仇桑19884947083…… sin (7π/12)=cos(1/2-7/12)π=cos(π/12) cos(π/6)=2cos^2(π/12)-1=根号3/2 所以所求=根号[(根号3/2+1)/2]=根号[(2根号3+4)/8]=(根号6+根号2)/4
@岳乳4270：求三角函数的值 sin7π/4(精确到0.01) sin( - 119°45″) - 作业帮
仇桑19884947083…… [答案] sin(7π/4) = - sin(π/4) = - √2/2 ≈ - 1.414 /2 ≈ - 0.71 sin( -119°45') ≈ - sin( 119.75 * π / 180) ≈ - sin(2.09) ≈ - 0.87
@岳乳4270：sin(7π+a)= - - 1/3,则tan(π+a)的值为 -
仇桑19884947083…… sin(7π+a)=sin(π+a)=-sin(a)= -1/3,可得sin(a)= 1/3;cos(a)=±√8/3 tan(π+a)=tan(a)=sin(a)/cos(a)=±√2/4
@岳乳4270：已知函数fx=sin(7π -
仇桑19884947083…… 已知函数fx=sin(7π/6-2x)-2sin^2x+1求函数周期和单调递增区间 f(x)=sin(7π/6-2x)-2sin^2x+1=sin(π+π/6-2x)+cos2x=sin(2x-π/6)+cos2x=√3/2sin2x+3/2cos2x=√3(1/2sin2x+√3/2cos2x)=√3sin(2x+π/3) 最小正周期T=π ,2kπ-π/2<2x+π/3<2kπ+π/2 , kπ-5π/12<x<kπ+π/12 单调递增区间[kπ-5π/12 , kπ+π/12] ,k∈z
@岳乳4270：为什么sin7π/6= - 1/2 最后那个负号哪来的 - 作业帮
仇桑19884947083…… [答案] sin7π/6=sin(π+π/6)=-sin(π/6)=-1/2
@岳乳4270：sin7兀/12点算 - 作业帮
仇桑19884947083…… [答案] sin (7π/12)=cos(1/2-7/12)π=cos(π/12) cos(π/6)=2cos^2(π/12)-1=根号3/2 所以所求=根号[(根号3/2+1)/2]=根号[(2根号3+4)/8]=(根号6+根号2)/4
@岳乳4270：利用定义求sin7π/3,cos7π/3,tan7π/3. - 作业帮
仇桑19884947083…… [答案] sin7π/3=sin(7π/3-2π)=sinπ/3=2分之根号3 cos7π/3=cos(7π/3-2π)=cosπ/3=2分之1 tan7π/3=tan(7π/3-2π)=tanπ/3=根号3.
@岳乳4270：手算三角函数的过程,如sin7度 -
仇桑19884947083…… 如果要算的话可以用泰勒公式 sin x=x-x^3/3!+x^5/5!+......x为弧度制 计算机都是用这个公式算的
@岳乳4270：sin(7π/12)=?sin(19π/12)=?答案用分数表示 - 作业帮
仇桑19884947083…… [答案] sin(7π/12) = sin(π/2+π/12) = -cos(π/12) = - √{ [1+cos(π/6)]/2} = -√[(4+2√3) /2]sin(19π/12) = sin(3π+π/12) = -sin(π/12) = - √{ [1-cos(π/6)]/2} = -√[(4-2√3) /2]
@岳乳4270：sin7兀/4 -
仇桑19884947083…… sin7π/4=sin(2π-π/4)=sin(-π/4)=-1/√2
仇桑19884947083…… sin (7π/12)=cos(1/2-7/12)π=cos(π/12) cos(π/6)=2cos^2(π/12)-1=根号3/2 所以所求=根号[(根号3/2+1)/2]=根号[(2根号3+4)/8]=(根号6+根号2)/4
@岳乳4270：求三角函数的值 sin7π/4(精确到0.01) sin( - 119°45″) - 作业帮
仇桑19884947083…… [答案] sin(7π/4) = - sin(π/4) = - √2/2 ≈ - 1.414 /2 ≈ - 0.71 sin( -119°45') ≈ - sin( 119.75 * π / 180) ≈ - sin(2.09) ≈ - 0.87
@岳乳4270：sin(7π+a)= - - 1/3,则tan(π+a)的值为 -
仇桑19884947083…… sin(7π+a)=sin(π+a)=-sin(a)= -1/3,可得sin(a)= 1/3;cos(a)=±√8/3 tan(π+a)=tan(a)=sin(a)/cos(a)=±√2/4
@岳乳4270：已知函数fx=sin(7π -
仇桑19884947083…… 已知函数fx=sin(7π/6-2x)-2sin^2x+1求函数周期和单调递增区间 f(x)=sin(7π/6-2x)-2sin^2x+1=sin(π+π/6-2x)+cos2x=sin(2x-π/6)+cos2x=√3/2sin2x+3/2cos2x=√3(1/2sin2x+√3/2cos2x)=√3sin(2x+π/3) 最小正周期T=π ,2kπ-π/2<2x+π/3<2kπ+π/2 , kπ-5π/12<x<kπ+π/12 单调递增区间[kπ-5π/12 , kπ+π/12] ,k∈z
@岳乳4270：为什么sin7π/6= - 1/2 最后那个负号哪来的 - 作业帮
仇桑19884947083…… [答案] sin7π/6=sin(π+π/6)=-sin(π/6)=-1/2
@岳乳4270：sin7兀/12点算 - 作业帮
仇桑19884947083…… [答案] sin (7π/12)=cos(1/2-7/12)π=cos(π/12) cos(π/6)=2cos^2(π/12)-1=根号3/2 所以所求=根号[(根号3/2+1)/2]=根号[(2根号3+4)/8]=(根号6+根号2)/4
@岳乳4270：利用定义求sin7π/3,cos7π/3,tan7π/3. - 作业帮
仇桑19884947083…… [答案] sin7π/3=sin(7π/3-2π)=sinπ/3=2分之根号3 cos7π/3=cos(7π/3-2π)=cosπ/3=2分之1 tan7π/3=tan(7π/3-2π)=tanπ/3=根号3.
@岳乳4270：手算三角函数的过程,如sin7度 -
仇桑19884947083…… 如果要算的话可以用泰勒公式 sin x=x-x^3/3!+x^5/5!+......x为弧度制 计算机都是用这个公式算的
@岳乳4270：sin(7π/12)=?sin(19π/12)=?答案用分数表示 - 作业帮
仇桑19884947083…… [答案] sin(7π/12) = sin(π/2+π/12) = -cos(π/12) = - √{ [1+cos(π/6)]/2} = -√[(4+2√3) /2]sin(19π/12) = sin(3π+π/12) = -sin(π/12) = - √{ [1-cos(π/6)]/2} = -√[(4-2√3) /2]
@岳乳4270：sin7兀/4 -
仇桑19884947083…… sin7π/4=sin(2π-π/4)=sin(-π/4)=-1/√2
