不查表求值:cos5π/12·sinπ/12=
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不查表求值:cos5π/12·sinπ/12=
1.如题
2.根号下(cos4-sin^2(2)+2)=
3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=
4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
1.如题
2.根号下(cos4-sin^2(2)+2)=
3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=
4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
1 ,cos5π/12·sinπ/12
=cos(π/4+π/6)*sin(π/4-π/6)
=(√6-√2)/4*(√6+√2)/4=1/4
2.√[cos4-sin^2(2)+2]
=√[cos^2(2)-2sin^2(2)+2]
=√[cos^2(2)+2(1-sin^2(2)]
=√[3cos^2(2)]=-√3cos2
3;sinα=4/5,cosα=-3/5 ,tanα=-4/3
tan(α+β)=1=(tanα+tanβ)/(1-tanαtanβ)
(1-tanαtanβ)=(tanα+tanβ)
1+4/3tanβ=-4/3tanβ ,tanβ=-3/8
4
=cos(π/4+π/6)*sin(π/4-π/6)
=(√6-√2)/4*(√6+√2)/4=1/4
2.√[cos4-sin^2(2)+2]
=√[cos^2(2)-2sin^2(2)+2]
=√[cos^2(2)+2(1-sin^2(2)]
=√[3cos^2(2)]=-√3cos2
3;sinα=4/5,cosα=-3/5 ,tanα=-4/3
tan(α+β)=1=(tanα+tanβ)/(1-tanαtanβ)
(1-tanαtanβ)=(tanα+tanβ)
1+4/3tanβ=-4/3tanβ ,tanβ=-3/8
4
不查表求值:cos5π/12·sinπ/12=
cos5π/12=sinπ/12
为什么cos5π/12等于sinπ/12
(sin5π/12-sinπ/12)(cos5π/12+cosπ/12)=?
cos5/12πcosπ/12+cosπ/12sinπ/6=?
求值cos5π/8*cosπ8
(sin5π/12+cos5π/12)(sinπ/12-cosπ/12)
反三角函数求值arccos(cos5π/4)=?arctan(tan4π/3)=?arcsin(sin6)=?
求值:sin(π/12)+cos(π/12)
sin5/12π×cos5/12π怎么算?
计算cos5分之π+cos5分之2π+cos5分之3π+cos5分之4π
例题9和10,我知道有解析,但就是看不懂,好像是公式吧,我想知道cos5π/12是怎么变成sin的,周期不是π/2吗,乘