goniometria

goniometria

9th - 12th Grade

15 Qs

quiz-placeholder

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goniometria

goniometria

Assessment

Quiz

Mathematics

9th - 12th Grade

Medium

Created by

Vincenzina Aurelio

Used 4+ times

FREE Resource

15 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Quale delle seguenti NON è vera:

cos(2α)=12sin2α\cos\left(2\alpha\right)=1-2\sin^2\alpha

sin(2α)=2sinαcosα\sin\left(2\alpha\right)=2\sin\alpha\cos\alpha

cos(2α)=sin2αcos2α\cos\left(2\alpha\right)=\sin^2\alpha-\cos^2\alpha

cos(2α)=2cos2α1\cos\left(2\alpha\right)=2\cos^2\alpha-1

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 sen2αcos2α=sen^2\alpha-\cos^2\alpha=  

 cos2α\cos2\alpha  

 cos2α-\cos2\alpha  

 1+tg2α1+tg2\alpha  

 sen2αcos2αsen2\alpha-\cos2\alpha  

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 sen(α+β)sen(αβ)=?sen\left(\alpha+\beta\right)-sen\left(\alpha-\beta\right)=? 

 2cosαsenβ2\cos\alpha\cdot sen\beta  

 2cosαsenα2\cos\alpha\cdot sen\alpha  

 2cosαsenβ-2\cos\alpha\cdot sen\beta  

 2senαcosβ2sen\alpha\cdot\cos\beta  

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 (senα+cosα)2=\left(sen\alpha+\cos\alpha\right)^2=  

 sen2αcos2αsen^2\alpha-\cos^2\alpha  

 11  

 1+sen2α1+sen2\alpha  

 1+cos2α1+\cos2\alpha  

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Calcolare, in funzione dell'angolo  α\alpha , il valore dell'espressione  cos (π+α)+sin(πα)tan(π2α)cos(π2+α)\cos\ \left(\pi+\alpha\right)+\sin\left(\pi-\alpha\right)-\tan\left(\frac{\pi}{2}-\alpha\right)\cos\left(\frac{\pi}{2}+\alpha\right) 

 sin α-\sin\ \alpha  

 sin α\sin\ \alpha  

 cosα\cos\alpha  

 cosα-\cos\alpha  

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Semplifica la seguente espressione  sin2α1cos2α+tan2α+sin2αtan2α+11+tan2α\sin^2\alpha-\frac{1}{\cos^2\alpha}+\tan^2\alpha+\frac{\sin^2\alpha}{\tan^2\alpha}+\frac{1}{1+\tan^2\alpha}  

esprimendola in funzione di cos  α\alpha  

 cosα\cos\alpha  

  cosα-\ \cos\alpha  

 cos2α\cos^2\alpha  

  cos2α-\ \cos^2\alpha  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 2sen(α2)cos(α2)2\cdot sen\left(\frac{\alpha}{2}\right)\cdot\cos\left(\frac{\alpha}{2}\right)  

 cosα\cos\alpha  

 senαsen\alpha  

 tgαtg\alpha  

 senαcosαsen\alpha\cdot\cos\alpha  

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