如果tan a = 4/3且cot b = 5 / 12,0 ^度,则找到sin(a + b)的值

如果tan a = 4/3且cot b = 5 / 12,0 ^度,则找到sin(a + b)的值
Anonim

回答:

#sin(A + B)=65分之56#

说明:

鉴于, #tana = 4/3和cotb = 5/12#

#rarrcota = 3/4的#

#rarrsina = 1 / CSCA = 1 / SQRT(1 +婴儿床^ 2A)= 1 / SQRT(1+(3/4)^ 2)= 4/5#

#rarrcosa = SQRT(1-罪^ 2A)= SQRT(1-(4/5)^ 2)= 3/5#

#rarrcotb = 5/12#

#rarrsinb = 1 / CSCB = 1 / SQRT(1 +婴儿床^ 2B)= 1 / SQRT(1+(5/12)^ 2)= 12/13#

#rarrcosb = SQRT(1-罪^ 2B)= SQRT(1-(12/13)^ 2)= 5/13#

现在, #sin(A + B)=新浪* +的CoSb COSA * SINB#

#=(4/5)(5/13)+(3/5)*(12/13)=56/65#

回答:

#sin(A + B)=65分之56#

说明:

这里,

#0 ^ circ <color(violet)(a)<90 ^ circ => I ^(st)Quadrant => color(blue)(All,fns。> 0。#

#0 ^ circ <color(violet)(b)<90 ^ circ => I ^(st)Quadrant => color(blue)(All,fns。> 0#

所以,

#0 ^ circ <颜色(紫色)(a + b)<180 ^ circ => I ^(st)和II ^(nd)象限#

#=>颜色(蓝色)(sin(a + b)> 0#

现在,

#塔纳= 4/3 =>塞卡= + SQRT(1 +黄褐色^ 2A)= SQRT(1 + 16/9)= 5/3#

#:.颜色(红色)(COSA)= 1 /塞卡=颜色(红色)(3/5#

#=>颜色(红色)(新浪)= + SQRT(1-COS ^ 2A)= SQRT(1-9 / 25)=颜色(红色)(4/5#

也,

#cotb = 5/12 => CSCB = + SQRT(1个+婴儿床^ 2B)= SQRT(1 + 25/144)=一十二分之一十三#

#:.颜色(红色)(SINB)= 1 / CSCB =颜色(红色)(12/13#

#=>颜色(红色)(的CoSb)= + SQRT(1-罪^ 2B)= SQRT(1-144 / 169)=颜色(红色)(5/13#

因此,

#sin(A + B)= sinacosb + cosasinb#

#=> SIN(A + B)= 4 / 5xx5 / 13 + 3 / 5xx12 / 13#

#sin(A + B)= 20/65±65分之36=65分之56#