tan+sin+cos+数值表图

曾河园2509sin45° - cos30°+tan60°等于多少?(2)sin60°+tan45° - cos30°(3)cos60° - tan30°+sin30° -
郝沫顷18272639357 ______[答案] 首先我这儿没有公式编辑器,只会手打,谅解. (1)sin45=二分之根号二,cos30=二分之根号三,tan60=根号三.原式=二分之根号二加根号三. (2)sin60=二分之根号三 ,tan45=1,cos30=二分之根号三,也就是sin60=cos30. 原式=tan45=1 (3)cos60...

曾河园2509sin²120°+cos180°+tan45° - cos²330°+sin( - 210°)急用,注意 是cos²330° -
郝沫顷18272639357 ______[答案] sin120°=sin60°=√3/2 cos180° = -1 tan45°=1 cos330° = cos30° = √3/2 sin(-210°) = sin(-360°+150°)=sin150°=sin30°=1/2 原式=3/4 -1 + 1 - 3/4 + 1/2 = 1/2

曾河园25092cos派/2 - tan派/4+3/4tan²派/6 - sin派/6+cos²派/6+sin3派/2 计算 -
郝沫顷18272639357 ______[答案] 2cos(π/2)=0tan(π/4)=1tan(π/6)=1/√3,tan²(π/6)=1/3sin(π/6)=1/2cos(π/6)=√3/2,cos²(π/6)=3/4sin(3π/2)=-1所以原式=2*0-1+(3/4)*(1/3)-(1/2)+(3/4)-1=2-1+(1/4)-(1/2)+(3/4)-1=1/2

曾河园2509化简 - sin(180+a)+sin( - a) - tan(360+a)/tan(a+180)+cos(180 - a)+cos( - a) -
郝沫顷18272639357 ______[答案] -sin(180+a)+sin(-a)-tan(360+a)/tan(a+180)+cos(180-a)+cos(-a) =sina-sina-tana/tana -cosa+cosa =-1

曾河园2509sin2120°+cos180°+tan45° - cos2( - 330°)+sin( - 210°)=______. -
郝沫顷18272639357 ______[答案] 原式= 3 4-1+1- 3 4+ 1 2= 1 2. 故答案为: 1 2

曾河园2509tanθ /2=2/3求1—cosθ +sinθ /1+cosθ +sinθ -
郝沫顷18272639357 ______[答案] tan(θ/2)=2/3 由万能公式有 sinθ=2tan(θ/2)/[1+(tan(θ/2))^2]=(4/3)/[1+(2/3)^2]=12/13 cosθ=[1-(tan(θ/2))^2]/[1+(tan(θ/2))^2]=[1-(2/3)^2]/[1+(2/3)^2]=5/13 故(1-cosθ+sinθ)/(1+cosθ +sinθ) =(1-5/13+12/13)/(1+5/13+12/13) =2/3

曾河园2509化简tan∝+cos∝÷(1+sin∝) -
郝沫顷18272639357 ______[答案] tanx+cosx/(1+sinx) =(sinx)/cosx+cosx(1-sinx)/(cosx)^2 =(sinxcosx+cosx-sinxcosx)/(cosx)^2 =1/cosx

曾河园2509sinα+cosα/2cosα - sinα=1/3 则,tan2α=?cos2α/1+sin2α=? -
郝沫顷18272639357 ______[答案] (sinα+cosα)/(2cosα-sinα)=1/3 则,tan2α=? cos2α/(1+sin2α)=? 由(sinα+cosα)/(2cosα-sinα)=1/3,得(tanα+1)/(2-tanα)=1/3; 即有3(tanα+1)=2-tanα,故得4tanα=-1,tanα=-1/4. 于是得tan2α=2tanα/(1-tan²α)=-(1/2)/[1-(1/16)]=-(1/2)/(15/16)=-8/...

曾河园2509任意角三角函数 化简 tanα(cosα - sinα)+﹙sinα+tanα﹚/(cotα+cosα) -
郝沫顷18272639357 ______[答案] tanα ,oct α拆成Sin a/Cosa形式,通分、利用和差公式慢慢算.你不做永远也不会.

曾河园2509化简:sin2(a+兀)*cos(兀+a)/tan(兀+a)*cos3( - a - 兀)*tan( - a - 2兀)化简:sin2(a+兀)*cos(兀+a)/tan(兀+a)*cos3( - a - 兀)*tan( - a - 2兀) -
郝沫顷18272639357 ______[答案] sin2(a+兀)*cos(兀+a)/tan(兀+a)*cos3(-a-兀)*tan(-a-2兀) =sin^2a*(-cosa)/tana*(-cos^3a)*(-tana) =sin^2a/cos^2a*(-tan^2a) =tan^2a/(-tan^2a) =-1