sin+π+α
@於阁6319：三角函数已知,sin(π+α)= - 1/2计算:(1)sin(5π - α);(2)sin(π/2+α);(3)cos(α+3π/2);(4)tan(π/2 - α).不用理由只要解答全过程. - 作业帮
羿伊15039134322…… [答案] 已知sin(π+α)=-1/2,即sinα=1/2 sin(5π-α)=sinα=1/2 sin(π/2+α)=cosα=2 cos(α+3π/2)=sinα=1/2 tan(π/2-α)=cotα=cosα/sinα=4
@於阁6319：已知sin(π+α)= - 1/2,计算: - 作业帮
羿伊15039134322…… [答案] sin(π+α)=-1/2 sin(α)=1/2 α=π/6+2kπ或者5π/6+2kπ(k∈Z)
@於阁6319：化简 sin(π+α)+sin(2π+α)+sin(3π+α)+……+sin(kπ+α) k属于Z - 作业帮
羿伊15039134322…… [答案] sin(π+α)=-sina sina=sin(2π+α)=sin(4π+α). sin(π+α)+sin(2π+α)+sin(3π+α)+……+sin(kπ+α) 所以当k=2n时,k为偶数时 原式=-sina+sina-sina+sina. =0 当k=2n+1时即k为奇数时 原式=-sina+sina-sina+sina. =-sina 你能明白,赞同
@於阁6319：已知α是第二象限角,且sin(π+α)=k - 1/k+1,sin(5π/2+α)=3k - 1/k+1,求α得正切值 -
羿伊15039134322…… 由诱导公式,sin(π+α)=-sinα=(k-1)/(k+1),即sinα=(1-k)/(k+1); sin(5π/2+α)=cosα=(3k-1)/(k+1)sin²α+cos²α=1;即:(1-k)²/(k+1)²+(3k-1)²/(k+1)²=1 (1-k)²+(3k-1)²=(k+1)² 10k²-8k+2=k²+2k+1 9k²-10k+1=0 (9k-1)(k-1)k=1/9 ...
@於阁6319：化简sin(π+α)cos(3π/2+α)+sin(π/2+α)cos(π+α)答案是 - 1 求过程 - 作业帮
羿伊15039134322…… [答案] 根据第一、二象限sin值取正,第一、四象限cos值取正,sin(π+α)cos(3π/2+α)+sin(π/2+α)cos(π+α)=-sinacosa+sina(-cosa)=-2sinacosa=-sin2a,若a=π/4,则-sin2a=-sinπ/2=-1
@於阁6319：已知sin(π+α)=2cos(π - α),计算:(1)2sinα - cosαsinα+2cosα(2)sin2α+sinαcosα - 2cos2α - 作业帮
羿伊15039134322…… [答案] ∵sin(π+α)=2cos(π-α),∴-sinα=-2cosα,∴tanα=2. (1)原式= 2tanα-1 tanα+2= 2*2-1 2+2= 3 4. (2)原式= sin2α+sinαcosα-2cos2α sin2α+cos2α= tan2α+tanα-2 tan2α+1= 22+2-2 22+1= 4 5.
@於阁6319：已知sin(π+α)=3/5,α为第四象限角,求cos(α - 2π). -
羿伊15039134322…… sin(π+α)=3/5,sin(α)=-3/5,α为第四象限角,cos(α-2π)=cos(α)=4/5
@於阁6319：一道诱导公式计算题2sin(π+α)cos(π - α) - cos(π - α)2sin(π+α)cos(π - α) - cos(π - α) 分之1+sin方α+cos(2分之3π+α) - sin方(2分之π+α) - 作业帮
羿伊15039134322…… [答案] 2sinα*cosα+cosα分之1+sin方α+sinα-cosα =cosα(2sinα+1)分之2sin方α+cos方α+sinα-cosα =cosα(2sinα+1)分之sinα(2sinα+1)+cosα(cosα-1) =cotα+2sinα+1分之cosα-1 你这道题只能化简到这里
@於阁6319：sin(4π+a)会等于sin(π+a)吗 -
羿伊15039134322…… 不会.sin(4π+α)=sinα sin(π+α)= -sinα
@於阁6319：化简:[sin(π+α)·cos(π - α)·tan( - a+3π)]/[sin(5π - α)·tan(8π - α)·cos(π+α)] - 作业帮
羿伊15039134322…… [答案] [sin(π+α)·cos(π-α)·tan(-a+3π)]/[sin(5π-α)·tan(8π-α)·cos(π+α)] =[-sinα·(-cosα)·(-tana)]/[sinα·(-tanα)·(-cosα)] =(-sinα·cosα·tana)/[sinα·tanα·cosα) =-1
羿伊15039134322…… [答案] 已知sin(π+α)=-1/2,即sinα=1/2 sin(5π-α)=sinα=1/2 sin(π/2+α)=cosα=2 cos(α+3π/2)=sinα=1/2 tan(π/2-α)=cotα=cosα/sinα=4
@於阁6319：已知sin(π+α)= - 1/2,计算: - 作业帮
羿伊15039134322…… [答案] sin(π+α)=-1/2 sin(α)=1/2 α=π/6+2kπ或者5π/6+2kπ(k∈Z)
@於阁6319：化简 sin(π+α)+sin(2π+α)+sin(3π+α)+……+sin(kπ+α) k属于Z - 作业帮
羿伊15039134322…… [答案] sin(π+α)=-sina sina=sin(2π+α)=sin(4π+α). sin(π+α)+sin(2π+α)+sin(3π+α)+……+sin(kπ+α) 所以当k=2n时,k为偶数时 原式=-sina+sina-sina+sina. =0 当k=2n+1时即k为奇数时 原式=-sina+sina-sina+sina. =-sina 你能明白,赞同
@於阁6319：已知α是第二象限角,且sin(π+α)=k - 1/k+1,sin(5π/2+α)=3k - 1/k+1,求α得正切值 -
羿伊15039134322…… 由诱导公式,sin(π+α)=-sinα=(k-1)/(k+1),即sinα=(1-k)/(k+1); sin(5π/2+α)=cosα=(3k-1)/(k+1)sin²α+cos²α=1;即:(1-k)²/(k+1)²+(3k-1)²/(k+1)²=1 (1-k)²+(3k-1)²=(k+1)² 10k²-8k+2=k²+2k+1 9k²-10k+1=0 (9k-1)(k-1)k=1/9 ...
@於阁6319：化简sin(π+α)cos(3π/2+α)+sin(π/2+α)cos(π+α)答案是 - 1 求过程 - 作业帮
羿伊15039134322…… [答案] 根据第一、二象限sin值取正,第一、四象限cos值取正,sin(π+α)cos(3π/2+α)+sin(π/2+α)cos(π+α)=-sinacosa+sina(-cosa)=-2sinacosa=-sin2a,若a=π/4,则-sin2a=-sinπ/2=-1
@於阁6319：已知sin(π+α)=2cos(π - α),计算:(1)2sinα - cosαsinα+2cosα(2)sin2α+sinαcosα - 2cos2α - 作业帮
羿伊15039134322…… [答案] ∵sin(π+α)=2cos(π-α),∴-sinα=-2cosα,∴tanα=2. (1)原式= 2tanα-1 tanα+2= 2*2-1 2+2= 3 4. (2)原式= sin2α+sinαcosα-2cos2α sin2α+cos2α= tan2α+tanα-2 tan2α+1= 22+2-2 22+1= 4 5.
@於阁6319：已知sin(π+α)=3/5,α为第四象限角,求cos(α - 2π). -
羿伊15039134322…… sin(π+α)=3/5,sin(α)=-3/5,α为第四象限角,cos(α-2π)=cos(α)=4/5
@於阁6319：一道诱导公式计算题2sin(π+α)cos(π - α) - cos(π - α)2sin(π+α)cos(π - α) - cos(π - α) 分之1+sin方α+cos(2分之3π+α) - sin方(2分之π+α) - 作业帮
羿伊15039134322…… [答案] 2sinα*cosα+cosα分之1+sin方α+sinα-cosα =cosα(2sinα+1)分之2sin方α+cos方α+sinα-cosα =cosα(2sinα+1)分之sinα(2sinα+1)+cosα(cosα-1) =cotα+2sinα+1分之cosα-1 你这道题只能化简到这里
@於阁6319：sin(4π+a)会等于sin(π+a)吗 -
羿伊15039134322…… 不会.sin(4π+α)=sinα sin(π+α)= -sinα
@於阁6319：化简:[sin(π+α)·cos(π - α)·tan( - a+3π)]/[sin(5π - α)·tan(8π - α)·cos(π+α)] - 作业帮
羿伊15039134322…… [答案] [sin(π+α)·cos(π-α)·tan(-a+3π)]/[sin(5π-α)·tan(8π-α)·cos(π+α)] =[-sinα·(-cosα)·(-tana)]/[sinα·(-tanα)·(-cosα)] =(-sinα·cosα·tana)/[sinα·tanα·cosα) =-1
