(asinπ/5 +bcosπ/5)/(acosπ/5 -bsinπ/5)=tan8π/15,則b/a等于 A.√3 B.√3/3 C.-√3 D.-√3/3
(asinπ/5 +bcosπ/5)/(acosπ/5 -bsinπ/5)=tan8π/15,則b/a等于 A.√3 B.√3/3 C.-√3 D.-√3/3
(asinπ/5 +bcosπ/5)/(acosπ/5 -bsinπ/5)=tan8π/15,則b/a等于
A.√3B.√3/3C.-√3D.-√3/3
(asinπ/5 +bcosπ/5)/(acosπ/5 -bsinπ/5)=tan8π/15,則b/a等于
A.√3B.√3/3C.-√3D.-√3/3
數(shù)學(xué)人氣：389 ℃時(shí)間：2020-06-03 21:43:01
優(yōu)質(zhì)解答
(asinπ/5 +bcosπ/5)/(acosπ/5 -bsinπ/5)=tan8π/15(tanπ/5+b/a)/(1-b/atanπ/5)=tan(π/5+π/3)(tanπ/5+b/a)/(1-b/atanπ/5)=(tanπ/5+tanπ/3)/(1-tanπ/5tanπ/3)(tanπ/5+b/a)/(1-b/atanπ/5)=(tanπ/5+√3...
我來回答
類似推薦
- 已知非零常數(shù) a,b滿足（acosπ/5-bsinπ/5）/（asinπ/5+bcosπ/5）=1/（tan8π/15）,求b/a.
- 已知非零常數(shù) a,b滿足（acosπ/5-bsinπ/5）/（asinπ/5+bcosπ/5）=1/（tan8π/15）,求b/a. 詳細(xì)過程
- 已知非零函數(shù)a.b滿足：(asin(π/5)+bcos(π/5))/(acos(π/5)-bsin(π/5))=tan8π/5,求b/a的值
- 已知asin(γ+α)=bsin(γ+β),求證tanγ=bsinβ-asinα/acosα-bcosβ
- （ACOSα+BSinα)平方+（Asinα-Bcosα)平方
- My father has lunch at twelve o'clock.(對(duì)劃線部分提問）劃線部分是at twelve o'clock.
- 運(yùn)動(dòng)會(huì)開幕式現(xiàn)場(chǎng)放飛100個(gè)藍(lán)氣球、紅氣球、黃氣球,其中紅氣球比黃氣球多2個(gè),藍(lán)氣球比黃氣球少1個(gè),三種氣球各有多少個(gè)?
- What‘s your favorite TV show怎樣回答
- jony用英文怎么讀,最好把音標(biāo)寫出來,
- 描寫舞蹈的文章
- 學(xué)校要粉刷新教室學(xué)校要粉刷新教室,已知教室的長(zhǎng)為12米,寬為6米,高為35米,除去門窗的面積18平方米,如
- 由FeO、Fe2O3、Fe3O4組成的混合物，測(cè)得其中鐵元素與氧元素的質(zhì)量比為21:8，則這種混合物中FeO、Fe2O3、Fe3O4的物質(zhì)的量之比可能為（ ?。?A.1:2:1 B.2:1:1 C.1:1:1 D.1:1:38
猜你喜歡
- 1《畫蛇添足》的故事中,“為蛇足者”為什么“終亡其酒”?
- 2花費(fèi)時(shí)間take和spend的區(qū)別
- 3醛可以生成醇
- 4一根鐵絲長(zhǎng)24米,要把它圍成一個(gè)長(zhǎng)方形,長(zhǎng)是寬的1.4倍,這個(gè)長(zhǎng)方形的面積是（ )平方米?
- 5請(qǐng)教在五聲調(diào)式體系中,調(diào)號(hào)相同的調(diào)式稱為 （ ） A、同名調(diào) B、平行調(diào) C、同宮調(diào) D、同主音調(diào)
- 6圖書館有甲乙兩個(gè)書架,后來甲書添38本,乙書借出72本,這時(shí)甲架是乙架書的3倍,求甲乙書架各有書多少本?
- 7計(jì)算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/2x+4020
- 8擴(kuò)建前的面積是605萬平方米,比擴(kuò)建后約少40分之27,擴(kuò)建后的面積約是多少萬平方米
- 9Nobody wants to eat them,__ __?怎么填
- 10a familiar visitor的意思!
- 11五年級(jí)下冊(cè)暑假作業(yè)(英語)
- 12千瓦和大卡怎么換算