sin ⁡ α cos ⁡ α = \frac{\sin\alpha}{\cos\alpha}= cos α sin α =

ctg ⁡ α \operatorname{ctg}\alpha c t g α

s i i n α siin\alpha s i i n α

t g α ⋅ ctg ⁡ α = tg\alpha\cdot\operatorname{ctg}\alpha= t g α ⋅ c t g α =

cos ⁡ α sin ⁡ α \frac{\cos\alpha}{\sin\alpha} sin α cos α

sin ⁡ α cos ⁡ α \frac{\sin\alpha}{\cos\alpha} cos α sin α

t g α + 1 tg\alpha+1 t g α + 1

ctg ⁡ α + 1 \operatorname{ctg}\alpha+1 c t g α + 1

Основні тригонометричні тотожності

Authored by Роман Савчин

Mathematics

Professional Development

Used 87+ times

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15 questions

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MULTIPLE CHOICE QUESTION

2 mins • 1 pt

$\left(1-\sin^2x\right)\cdot tg^2x=$

$\cos^2x$

$\sin^2x$

$\operatorname{ctg}^2x$

$tg^2x$

$1$

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

$\frac{1-\cos^2\alpha}{\sin\alpha\cdot\cos\alpha}$

$\cos\alpha$

$\sin\alpha$

$tg\alpha$

$\operatorname{ctg}\alpha$

$1$

FILL IN THE BLANKS QUESTION

2 mins • 1 pt

$\frac{tg\alpha\cdot\cos\alpha}{\sin\alpha}-1=$

(a)

FILL IN THE BLANKS QUESTION

2 mins • 1 pt

$\left(\sin^2\beta\cdot\operatorname{ctg}^2\beta\right)+\sin^2\beta=$

(a)

FILL IN THE BLANKS QUESTION

2 mins • 1 pt

$\frac{1-\sin^2\alpha}{1-\cos^2\alpha}\ \cdot\ tg^2\alpha=$

(a)

FILL IN THE BLANKS QUESTION

2 mins • 1 pt

$tg\alpha\cdot\operatorname{ctg}\beta\cdot\frac{\sin\beta\cos\alpha}{\cos\beta\sin\alpha}+2=$

(a)

FILL IN THE BLANKS QUESTION

2 mins • 1 pt

$\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2=$

(a)

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

$\frac{\sin^2\alpha-1}{\cos^2\alpha-1}+tg\alpha\operatorname{ctg}\alpha=$

$\frac{1}{\sin^2\alpha}$

$\frac{1}{\cos^2\alpha}$

$2$

$\sin^2\alpha$

$\cos^2\alpha$

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

$\frac{1}{1+\cos\alpha}+\frac{1}{1-\cos\alpha}=$

$\frac{1}{\sin^2\alpha}$

$\frac{2}{\cos^2\alpha}$

$\frac{2}{\sin^2\alpha}$

$\frac{1}{\cos^2\alpha}$

10.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

$\cos^2\alpha-tg^2\alpha\cdot\operatorname{ctg}^2\alpha=$

$\sin^2\alpha$

$\cos^2\alpha$

$-\cos^2\alpha$

$tg^2\alpha$

$-\sin^2\alpha$

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Основні тригонометричні тотожності

(1−sin⁡2x)⋅tg2x=\left(1-\sin^2x\right)\cdot tg^2x=(1−sin2x)⋅tg2x=

1−cos⁡2αsin⁡α⋅cos⁡α\frac{1-\cos^2\alpha}{\sin\alpha\cdot\cos\alpha}sinα⋅cosα1−cos2α​

tgα⋅cos⁡αsin⁡α−1=\frac{tg\alpha\cdot\cos\alpha}{\sin\alpha}-1=sinαtgα⋅cosα​−1= (a)

(sin⁡2β⋅ctg⁡2β)+sin⁡2β=\left(\sin^2\beta\cdot\operatorname{ctg}^2\beta\right)+\sin^2\beta=(sin2β⋅ctg2β)+sin2β= (a)

1−sin⁡2α1−cos⁡2α ⋅ tg2α=\frac{1-\sin^2\alpha}{1-\cos^2\alpha}\ \cdot\ tg^2\alpha=1−cos2α1−sin2α​ ⋅ tg2α= (a)

tgα⋅ctg⁡β⋅sin⁡βcos⁡αcos⁡βsin⁡α+2=tg\alpha\cdot\operatorname{ctg}\beta\cdot\frac{\sin\beta\cos\alpha}{\cos\beta\sin\alpha}+2=tgα⋅ctgβ⋅cosβsinαsinβcosα​+2= (a)

(sin⁡α+cos⁡α)2+(sin⁡α−cos⁡α)2=\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2=(sinα+cosα)2+(sinα−cosα)2= (a)

sin⁡2α−1cos⁡2α−1+tgαctg⁡α=\frac{\sin^2\alpha-1}{\cos^2\alpha-1}+tg\alpha\operatorname{ctg}\alpha=cos2α−1sin2α−1​+tgαctgα=

11+cos⁡α+11−cos⁡α=\frac{1}{1+\cos\alpha}+\frac{1}{1-\cos\alpha}=1+cosα1​+1−cosα1​=

cos⁡2α−tg2α⋅ctg⁡2α=\cos^2\alpha-tg^2\alpha\cdot\operatorname{ctg}^2\alpha=cos2α−tg2α⋅ctg2α=

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$\left(1-\sin^2x\right)\cdot tg^2x=$

$\frac{1-\cos^2\alpha}{\sin\alpha\cdot\cos\alpha}$

$\frac{tg\alpha\cdot\cos\alpha}{\sin\alpha}-1=$

(a)

$\left(\sin^2\beta\cdot\operatorname{ctg}^2\beta\right)+\sin^2\beta=$

(a)

$\frac{1-\sin^2\alpha}{1-\cos^2\alpha}\ \cdot\ tg^2\alpha=$

(a)

$tg\alpha\cdot\operatorname{ctg}\beta\cdot\frac{\sin\beta\cos\alpha}{\cos\beta\sin\alpha}+2=$

(a)

$\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2=$

(a)

$\frac{\sin^2\alpha-1}{\cos^2\alpha-1}+tg\alpha\operatorname{ctg}\alpha=$

$\frac{1}{1+\cos\alpha}+\frac{1}{1-\cos\alpha}=$

$\cos^2\alpha-tg^2\alpha\cdot\operatorname{ctg}^2\alpha=$