利用和（差）角公式求下列各三角函数的值.1 cos（（-61/12）π）=2 tan((35π)/12)=________

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利用和（差）角公式求下列各三角函数的值.1 cos（（-61/12）π）=____________2 tan((35π)/12)=____________________
利用和（差）角公式求下列各三角函数的值.
1 cos（（-61/12）π）=____________
2 tan((35π)/12)=____________________

利用和（差）角公式求下列各三角函数的值.1 cos（（-61/12）π）=____________2 tan((35π)/12)=____________________
1、
原式=cos(-3*π+11π/12)
=cos(11π/12)
=cos(π/4+2π/3)
=cosπ/4cos2π/3-sinπ/4sin2π/3
=-(√6+√2)/4
2、
原式=tan(3π-π/12)
=tan(-π/12)
=tan(π/4-π/3)
=(tanπ/4-tanπ/3)/(1+tanπ/4tanπ/3)
=-2+√3

第一个等于
四分之根号6加根号2
可以把cos（（-61/12）π）化简成cos15°
然后就可以得到答案了

cos（（-61/12）π）
=cos(-5π-π/12)
=-cos(π/12)
=-cos(π/3-π/4)
=-cos(π/3)cos(π/4)-sin(π/3)sinπ/4
=-√2/4-√6/4
=-(√2+√6)/4
tan((35π)/12)=tan(3π-π/12)=tan(-π/12)
=tan(π/4-π/3)

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cos（（-61/12）π）
=cos(-5π-π/12)
=-cos(π/12)
=-cos(π/3-π/4)
=-cos(π/3)cos(π/4)-sin(π/3)sinπ/4
=-√2/4-√6/4
=-(√2+√6)/4
tan((35π)/12)=tan(3π-π/12)=tan(-π/12)
=tan(π/4-π/3)
=[tan(π/4)-tan(π/3)]/[1+tan(π/4)tan(π/3)]
=(1-√3)/(1+√3)
=-(4-2√3)/2
=√3-2

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cos（（-61/12）π）
原式=cos(-5π-π/12)
=-cos(π/12)
=-cos(π/3-π/4)
=-cos(π/3)cos(π/4)-sin(π/3)sinπ/4
=-√2/4-√6/4
=-(√2+√6)/4
tan((35π)/12）
原式=tan(3π-π/12)=tan(-π/12)
=tan(π/4-π/3)
=[tan(π/4)-tan(π/3)]/[1+tan(π/4)tan(π/3)]
=(1-√3)/(1+√3)
=-(4-2√3)/2
应该是正确的叭.
=√3-2

收起

cos（（-61/12）π）
=cos(-5π-π/12)
=-cos(π/12)
=-cos(π/3-π/4)
=-cos(π/3)cos(π/4)-sin(π/3)sinπ/4
=-√2/4-√6/4
=-(√2+√6)/4
tan((35π)/12)=tan(3π-π/12)=tan(-π/12)
=tan(π/4-π/3)

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收起

cos(a+2kπ)=cos a
sin(s+2kπ)=cos s
tan(a+2kπ)=tan a
sin(a+b)=sin acos b+cos asin b
sin(a-b)=sin acos b-cos asin b
cos(a+b)=cos acos b-sin asin b
cos(a-b)=cos acos b+sin asin b
tan(a-b)=tan a-tan b/1+tan atan b
tan(a+b)=tan a+tan b/1+tan atan b