Search Header Logo

формулы двойного и половинного углов

Authored by Ольга Харченко

Mathematics

10th Grade

Used 53+ times

формулы двойного и половинного углов
AI

AI Actions

Add similar questions

Adjust reading levels

Convert to real-world scenario

Translate activity

More...

    Content View

    Student View

11 questions

Show all answers

1.

MULTIPLE SELECT QUESTION

45 sec • 1 pt

 cos2α=\cos2\alpha=  

 cos2αsin2α\cos^2\alpha-\sin^2\alpha  

 ctgαtgαctgα+tgα\frac{\operatorname{ctg}\alpha-tg\alpha}{\operatorname{ctg}\alpha+tg\alpha}   

 12sin2α1-2\sin^2\alpha  

 2cos2α12\cos^2\alpha-1  

 2tgα1tg2α\frac{2tg\alpha}{1-tg^2\alpha}  

2.

MULTIPLE SELECT QUESTION

45 sec • 1 pt

 sin2α=\sin2\alpha=  

 cos2αsin2α\cos^2\alpha-\sin^2\alpha  

 2sinαcosα2\sin\alpha\cos\alpha  

 12sin2α1-2\sin^2\alpha  

 2tgα+ctgα\frac{2}{tg\alpha+\operatorname{ctg}\alpha}   

 2tgα1tg2α\frac{2tg\alpha}{1-tg^2\alpha}  

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 sin2α=\sin^2\alpha=  

 1+cos2α2\frac{1+\cos^{ }2\alpha}{2}  

 1cos2α2\frac{1-\cos2\alpha}{2}  

 1cos2α1-\cos2\alpha  

 1+cos2α1+\cos2\alpha  

4.

MULTIPLE SELECT QUESTION

45 sec • 1 pt

 tg2α=tg2\alpha=  

 2tgα1tg2α\frac{2tg\alpha}{1-tg^2\alpha}  

 ctgαtgα2\frac{\operatorname{ctg}\alpha-tg\alpha}{2}  

 2ctgαtgα\frac{2}{\operatorname{ctg}\alpha-tg\alpha}  

 2tgα+ctgα\frac{2}{tg\alpha+\operatorname{ctg}\alpha}   

5.

MULTIPLE SELECT QUESTION

45 sec • 1 pt

 ctg2α=\operatorname{ctg}2\alpha=  

 2tgα1tg2α\frac{2tg\alpha}{1-tg^2\alpha}  

 ctgαtgα2\frac{\operatorname{ctg}\alpha-tg\alpha}{2}  

 2ctgαtgα\frac{2}{\operatorname{ctg}\alpha-tg\alpha}  

 2tgα+ctgα\frac{2}{tg\alpha+\operatorname{ctg}\alpha}   

 ctg2α12ctgα\frac{\operatorname{ctg}^2\alpha-1}{2\operatorname{ctg}\alpha}  

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 cos2α=\cos^2\alpha=  

 1+cos2α2\frac{1+\cos^{ }2\alpha}{2}  

 1cos2α2\frac{1-\cos2\alpha}{2}  

 1cos2α1-\cos2\alpha  

 1+cos2α1+\cos2\alpha  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

  sin α2\ \left|\sin\ \frac{\alpha}{2}\right|  =

 1cosα1+cosα\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}  

 1+cosα2\sqrt{\frac{1+\cos\alpha}{2}}  

 1cosα2\sqrt{\frac{1-\cos\alpha}{2}}  

Access all questions and much more by creating a free account

Create resources

Host any resource

Get auto-graded reports

Google

Continue with Google

Email

Continue with Email

Microsoft

Continue with Microsoft

or continue with

Facebook

Facebook

Apple

Apple

Others

Others

Already have an account?