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Ôn tập công thức lượng giác

Authored by Hải Nguyễn

Mathematics

11th Grade

Used 16+ times

Ôn tập công thức lượng giác
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10 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

11+tan2x=?\frac{1}{1+\tan^2x}=?

sin2x\sin^2x

cos2x\cos^2x

tan2x\tan^2x

cot2x\cot^2x

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 cos23x ba˘ˋng\cos^23x\ bằng 


 1cos6x2\frac{1-\cos6x}{2} 

 1+cos6x2\frac{1+\cos6x}{2} 

 1+sin6x2\frac{1+\sin6x}{2} 

 1cos6x1-\cos6x 

3.

FILL IN THE BLANKS QUESTION

1 min • 1 pt

 A=1+cosx+cos2x+cos3x2cos2x+cosx1=?A=\frac{1+\cos x+\cos2x+\cos3x}{2\cos^2x+\cos x-1}=? 




(a)  

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Biểu thức nào sau đâu đúng?

Sin(a+b)Sin(ab)=Sina+SinbSinaSinb\frac{Sin\left(a+b\right)}{Sin\left(a-b\right)}=\frac{Sina+Sinb}{Sina-Sinb}

Sin(a+b)Sin(ab)=SinaSinbSina+Sinb\frac{Sin\left(a+b\right)}{Sin\left(a-b\right)}=\frac{Sina-Sinb}{Sina+Sinb}

Sin(a+b)Sin(ab)=Tana+TanbTanaTanb\frac{Sin\left(a+b\right)}{Sin\left(a-b\right)}=\frac{Tana+Tanb}{Tana-Tanb}

Sin(a+b)Sin(ab)=Cota+CotbCotaCotb\frac{Sin\left(a+b\right)}{Sin\left(a-b\right)}=\frac{Cota+Cotb}{Cota-Cotb}

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Đẳng thức nào sau đây đúng

Cos(α+π3)=Cosα+12Cos\left(\alpha+\frac{\pi}{3}\right)=Cos\alpha+\frac{1}{2}

Cos(α+π3)=32Sinα12CosαCos\left(\alpha+\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}Sin\alpha-\frac{1}{2}Cos\alpha

Cos(α+π3)=12Sinα32CosαCos\left(\alpha+\frac{\pi}{3}\right)=\frac{1}{2}Sin\alpha-\frac{\sqrt{3}}{2}Cos\alpha

Cos(α+π3)=12Cosα32SinαCos\left(\alpha+\frac{\pi}{3}\right)=\frac{1}{2}Cos\alpha-\frac{\sqrt{3}}{2}Sin\alpha

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 A= Cos54o.Cos4oCos36o.Cos86o=?A=\ Cos54^o.Cos4^o-Cos36^o.Cos86^o=? 

 Cos50oCos50^o  

 Cos58oCos58^o  

 Sin50oSin50^o  

 Sin58oSin58^o  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 Ne^ˊu tan(β2)=4tan(α2) thıˋ tan(βα2) ba˘ˋngNếu\ \tan\left(\frac{\beta}{2}\right)=4\tan\left(\frac{\alpha}{2}\right)\ thì\ \tan\left(\frac{\beta-\alpha}{2}\right)\ bằng  

 3sinα53cosα\frac{3\sin\alpha}{5-3\cos\alpha}  

 3sinα5+3cosα\frac{3\sin\alpha}{5+3\cos\alpha}  

 3cosα53cosα\frac{3\cos\alpha}{5-3\cos\alpha}  

 3cosα5+3cosα\frac{3\cos\alpha}{5+3\cos\alpha}  

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