137页13题答案

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 19:40:53
137页13题答案
xuUn@/utfl'Hu~o(RtEDTvC j)G*$?@;]w=<4`N~tWw4iv'>ӿW9j]6K{"/=OښG"D ¥\qHsh嫡"@Ј`#̮c=:o\ ~ ŏ2M2@@P̰@"=홴HCʵ's% r2@wW/9m05@f(@Y  "Erٗz )̹)-Tr=ݲe2j!fk=՜*tgGc^ 3lR ˄'[ BfKH̢@E<PՐ1"XGQq@!U AW L,kmԡ 4Ʋ-?<|~. 9xl{}?6Fz#\] q=#0RRӏ8mȭ7[:,Sv_wcRc0m{ׁ<-BQE8;x ׉т.י\*.ct:&?eK}l>V;ZaT U3Vç2^Xۃ_"%M\_=LpD/{:<8{߲!aЁ=Km%kMO#0

137页13题答案
137页13题答案

137页13题答案
是不是这个

(1)3根号15sinx+3根号5cosx
=6根号5((根号3)/2sinx+1/2cosx)
=6根号5(cos10°sinx+sin30°cosx)
=(6根号5)sin(x+30°)
(2)3/2cosx-根号3/2sinx
原式=√3(√3/2cosx-1/2sinx)
=√3(sinπ/3cosx-cosπ/3sinx)
=√3sin(π/3-x)
(3)根号3sinx/2+cosx/2
=2(根号3/2sinx/2+1/2cosx/2)
=2(cos30°sinx/2+sin30°cosx/2)
=2sin(x/2+30°)
(4)√2/4sin(π/4-x)+√6/4cos(π/4-x)
=√2/2(1/2*sin(π/4-x)+√3/2cos(π/4-x))
=√2/2(cosπ/3*sin(π/4-x)+sinπ/3*cos(π/4-x))
=√2/2sin(π/4-x+π/3)
=√2/2sin(7π/12-x)
(5)sin347°cos148°+sin77°cos58°
=sin(360°-13°)cos(180°-32°)+cos13°sin32°
=sin13°cos32°+cos13°sin32°
=sin(13+32)°
=sin45°
=根号2/2
(6)sin164°sin224°+sin254°sin314°
=sin16°(-sin44°)+(-sin74°)(-sin46°)
=sin74°sin46°-sin16°sin44°
=sin74°cos44°-cos74°sin44°
=sin(74°-44°)
=sin30
=1/2
(7)(8)特别简单,套正弦余弦公式就Ok了,注意符号
(9)(tan5π/4+tan5π/12)/(1-tan5π/12)
根据tan的和角公式:
tan(x+y)=(tanx+tany)/(1-tanxtany)
再根据诱导公式:tan 5π/4 = tan(π+π/4) =tan π/4
所以
(tan 5π/4 + tan 5π/12)/(1-tan 5π/12) (诱导公式,并利用tan π/4=1)
=(tan π/4 + tan 5π/12)/(1-tan π/4*tan 5π/12) (由tan的和角公式)
=tan(π/4 + 5π/12)
=tan 2π/3
=-根号3
综上,原式= -根号3.
(10)(sin(x+p)-2sinxcosp)/2sinxsinp+cos(x+p)
=(cosxsinp-sinxcosp)/(sinxsinp+cosxcosp)
=sin(p-x)/cos(p-x)
=tan(p-x)