(tan91*tan92*……*tan179)/(sin^21+sin^22+……+sin^289)=

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/25 05:50:13
(tan91*tan92*……*tan179)/(sin^21+sin^22+……+sin^289)=
xSN@&&fwZB[~ h  !I=I ^*)< nKx l{3Ԭ\Jȴм-tqZ[F۾_rtI9:\Jޤ+G,%$^s):җMj:nd*z(JK"\FbT;A0lr\ ZAݾoOF3i)L; (Msb&e{RLUq#x[Yƌz_^zZx똔2,e*~A}[+mleY!bOxU @`)I Zbeq^@йeZY͜b#J(dJ/Ld{hsg"OG?~_}d"ߝcn

(tan91*tan92*……*tan179)/(sin^21+sin^22+……+sin^289)=
(tan91*tan92*……*tan179)/(sin^21+sin^22+……+sin^289)=

(tan91*tan92*……*tan179)/(sin^21+sin^22+……+sin^289)=
tan91*tan92*……*tan179=(-tan89°)*(-tan88°)*……*(-tan1°)=(-cot1°)*(-cot2°)*…*(-cot44°)*(-tan45°)*(cot44°)*…*(-tan1°)=-1
分母=sin^21°+sin^22°+……+sin^244°+sin^245°+cos^244°+……+cos^22°+cos^21° =89/2
所以 最终结果为 -2/89

(tan91*tan92*……*tan179)/(sin^21+sin^22+……+sin^289)
=[tan(180-89)*tan(180-88)*...*tan(180-1)]*(sin²1+sin²2+...+sin²45+cos²44+cos²43+...+cos²1)
=-(tan89*tan88*...*t...

全部展开

(tan91*tan92*……*tan179)/(sin^21+sin^22+……+sin^289)
=[tan(180-89)*tan(180-88)*...*tan(180-1)]*(sin²1+sin²2+...+sin²45+cos²44+cos²43+...+cos²1)
=-(tan89*tan88*...*tan1)*[(sin²1+cos²1)+(sin²2+cos²2)+...+(sin²44+cos²44)+sin²45]
=(cot1*cot2*..*cot44*tan45*tan44*..*tan1)*(44+1/2)
=[(cos1*tan1)*(cot2*tan2)*...*(cot44*tan44)*tan45]*(44.5)
=(1*1*...*1)*(44.5)
=44.5
希望能帮到你O(∩_∩)O

收起

(tan91*tan92*……*tan179)/(sin^21+sin^22+……+sin^289)= 高一三角函数竞赛题lg│tan91°│+│tan92°│+……+lg│tan179°│lg│tan91°│+│tan92°│+……+lg│tan179°│= 计算(tan91°tan92°……tan179°)/sin^2 1°+sin 2°(tan91°tan92°……tan179°)/(tan91°tan92°……tan179°)/sin^2 1°+sin^2 2°……sin^2 89°求详细解题过程 计算,tan91°tan92°……tan179°/sin²1°+sin²2°+……+sin²89°= sin91°<sin92°,cos91°<cos92°,tan91°<tan92°,cot91°<cot92°哪个是正确的? 数学计算:tan1*tan2*……*tan89 tan1'*tan2'*tan3'……tan89'的值. tan89与tan91那个大 Tan1°tan2°+tan2°tan3°+ tan3°tan4°……+tan59°tan60°= 三角函数题,tan1°*tan2°*tan3*…*tan89° --------------------------------求值sin1°+sin2°+sin3°+…+sin89° 求tan1°乘tan2°乘tan3°乘…tan88°乘tan89°=? tan1等于tan1°吗 三角函数题,tan1°*tan2°*tan3°*……*tan89° --------------------------------sin1°+sin2°+sin3°+……+sin89° 高中数学诱导公式习题tan1°+tan2°+tan3°……+tan89°是加法不是乘法!、麻烦给下过程~ 答案89/2 . 三角函数计算 (22 18:57:20)计算:tan1°×tan2°×tan3°×tan4°×tan5°×……×tan87°×tan88°×tan89° 高中三角函数计算题求值:(1-tan1)(1+tan2)(1+tan3)……(1+tan44)(1+tan45)各项之间均为乘. 关于三角函数的(1+tan1°)(1+tan2°)…(1+tan43°)(1+tan44°)=? 一道高中三角函数求值题(1+tan1°)*(1+tan2°)…(1+tan44°)*(1+tan45°)