算一算1+sin30°=(2020.7.5)
算一算1+sin30°=(2020.7.5)
sin 54°cos 24°-sin36°sin 24°=sin 30°.
sin 54°cos 24°-sin36°sin 24°=sin 30°.
已知sin(α+β)=2/1,sin(α-β)=3/1求证:sinαcosβ=5cosαsinβ
已知sin(α+β)=2/1,sin(α-β)=3/1求证:sinαcosβ=5cosαsinβ
sin(-30°)=sin30°
sin(-30°)=sin30°
【计算题】已知sinα+cosα=1,求:(1)sinαcosα; (2)sin α-cos α; (3)sin α-cos α
【计算题】已知sinα+cosα=1,求:(1)sinαcosα; (2)sin α-cos α; (3)sin α-cos α
下列计算正确的是() A: log2cos7π4=-12 B: 若f(cos x)=cos 2x,则f(sin 30°)=12 C: 若sin(π+α)=-12,则sin(4π-α)=-12 D: 设tan(π+α)=2,则sin(α-π)+cos(π-α)sin(π+α)-cos(π-α)=1
下列计算正确的是() A: log2cos7π4=-12 B: 若f(cos x)=cos 2x,则f(sin 30°)=12 C: 若sin(π+α)=-12,则sin(4π-α)=-12 D: 设tan(π+α)=2,则sin(α-π)+cos(π-α)sin(π+α)-cos(π-α)=1
化简三角函数表达式<img src="http://img1.ph.126.net/i0New2KyMz3LiF5MNUZIPA==/6597565646403144931.png" />? TrigFactor[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]|Simplify[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]|TrigReduce[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]|Cancel[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
化简三角函数表达式<img src="http://img1.ph.126.net/i0New2KyMz3LiF5MNUZIPA==/6597565646403144931.png" />? TrigFactor[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]|Simplify[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]|TrigReduce[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]|Cancel[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
以下不能正确计算代数式值的C语言表达式是() A: 1/3*sin(1/2)*sin(1/2) B: sin(0.5)*sin(0.5)/3 C: pow(sin(0.5),2)/3 D: 1/3.0*pow(sin(1.0/2),2)
以下不能正确计算代数式值的C语言表达式是() A: 1/3*sin(1/2)*sin(1/2) B: sin(0.5)*sin(0.5)/3 C: pow(sin(0.5),2)/3 D: 1/3.0*pow(sin(1.0/2),2)
<img src="http://edu-image.nosdn.127.net/2507E32A7888F1F05F34CD6088FE894F.png?imageView&thumbnail=890x0&quality=100" />? AC+AB×cosθ1=BC×cosθ3; AB×sinθ1=BC×sinθ3<br >|AC+AB×cosθ1=BC×cosθ3; AB×cosθ1=BCcos×θ3<br >|AB×sinθ1=BC×cosθ3; AC+AB×cosθ1=BC×sinθ3|;AB×cosθ1=BC×cosθ3; AC+AB×sinθ1=BC×sinθ3<br >
<img src="http://edu-image.nosdn.127.net/2507E32A7888F1F05F34CD6088FE894F.png?imageView&thumbnail=890x0&quality=100" />? AC+AB×cosθ1=BC×cosθ3; AB×sinθ1=BC×sinθ3<br >|AC+AB×cosθ1=BC×cosθ3; AB×cosθ1=BCcos×θ3<br >|AB×sinθ1=BC×cosθ3; AC+AB×cosθ1=BC×sinθ3|;AB×cosθ1=BC×cosθ3; AC+AB×sinθ1=BC×sinθ3<br >
\(\int{\sin 3x\cos 4xdx}\)=( )。 A: \(\frac{1}{2}\sin x-\frac{1}{14}\cos 7x+C\) B: \(\frac{1}{2}\cos x-\frac{1}{14}\cos 7x+C\) C: \(\frac{1}{2}\cos x+\frac{1}{14}\sin 7x+C\) D: \(\frac{1}{2}\sin x+\frac{1}{14}\sin 7x+C\)
\(\int{\sin 3x\cos 4xdx}\)=( )。 A: \(\frac{1}{2}\sin x-\frac{1}{14}\cos 7x+C\) B: \(\frac{1}{2}\cos x-\frac{1}{14}\cos 7x+C\) C: \(\frac{1}{2}\cos x+\frac{1}{14}\sin 7x+C\) D: \(\frac{1}{2}\sin x+\frac{1}{14}\sin 7x+C\)