SIN(-7/6π),cos(15/4π),tan(-8/3π),tan(33/4π)是 负六分之七,四分之十五,负三分之八,四分之三十三啊

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SIN(-7/6π),cos(15/4π),tan(-8/3π),tan(33/4π)是 负六分之七,四分之十五,负三分之八,四分之三十三啊
SIN(-7/6π),cos(15/4π),tan(-8/3π),tan(33/4π)
是 负六分之七,四分之十五,负三分之八,四分之三十三啊

SIN(-7/6π),cos(15/4π),tan(-8/3π),tan(33/4π)是 负六分之七,四分之十五,负三分之八,四分之三十三啊
sin(-7/6π)
=sin(2π-7π/6)
=sin5π/6
=1/2
cos(15/4π)
=cos(4π-π/4)
=sin(-π/4)
=cos(v/4)
=√2/2
tan(-8/3π)
=tan(-3π+π/3)
=tan(π/3)
=√3
tan(33/4π)
=tan(8π+π/4)
=tan(π/4)
=1

cos^6(π/8)-sin^6(π/8)=求值,[cos^2(π/8)-sin^2(π/8)][cos^4(π/8)+sin^4(π/8)+cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-cos^2(π/8)sin^2(π/8)]=cos(π/4)[1-[sin^2(π/4)]/4]==7√2/16 cos(-π/4)*sin(5/6π)=? sin(-19π/6)=;sin(-16π/3)=;cos(-79π/6)=;tan(-26π/3)=;sin(15π/4)=;cos(-60°)-sin(-210°)=? 利用公式求下列三角函数値:(l).cos(-420°);(2).sin(-7/6π);(3).sin(-1300°);(4).cos(-79/6π)化简(1)sin(α+180°)cos(-α)sin(-α-180°)(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊! 利用和差角公式化简 (2)sin(π/3+α)+sin(π/3-α)(2)sin(π/3+α)+sin(π/3-α)(3)cos(π/4+α)-cos(π/4-α)(4)cos(60°+α)+cos(60°-α)(5)sin(α-β)cosβ+cos(α-β)sinβ(6)cos(α+β)cosβ+sin(α+β)sinβ 由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证:1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a) [sin(a-6π)cos(7π/2-a)]/[sin(3π+a)+sin(a-π)】= [Sin(6π+a)Cos(7π-a)]/[Sin(3π+a)+Sin(-π+a) 化简sin(x+7π/4)+cos(x-3π/4)步骤我已经找到撒sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2- cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6) 已知13sinα+5cosβ=9,13cosα+5sinβ=15,则sin(α+β)=_.9、已知cos(α-π/6)+sinα=(4√3已知13sinα+5cosβ=9,13cosα+5sinβ=15,则sin(α+β)=_.9、已知cos(α-π/6)+sinα=(4√3)/5,则sin(α+7π/6)= 已知tan(α+π/4)=3,计算2sinαcosα+6cos方α-3/5cos方α-6sinαcosα-5sin方α cos(9π/4)+tan(-7π/6)+sin(21π)= 设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π+α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2(-sinα)(-cosα)+cosα/[1+sin² :(1)sin(2π-α)cos(π+α)cos(π/2+α)cos(13π/2-α)/cos(π-α)sin(5π-α)sin(-π-α)sin(9π/2+α)(2)sin(25π/6)+cos(25π/3)+tan(-27π/4) 已知6sin²α+sinαcosα-2cos²α=0,α∈(π/2,π),求值1).(sinα-3cosα)/(sinα-cosα);2).sinαcosα-sin²α;3).sin²α-3cosαsinα-2 sin(-19π/6)=?已知[3Sin(π+a)+Cos(π-a)]/4Sin(-a)-Sin(5π/2+a)=2,求tan a 已知α=7/12π,那么cosα√(1-sinα)/(1+sinα)+sinα√(1-cosα)/(1-cosα)=