8-cынып. Геометрия 14.12.2023

Authored by Gulhan Abdigalieva

Mathematics

8th Grade

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Негізгі тригонометриялық тепе-теңдік формуласы?

$a^2+b^2=c^2$

$1+tg^2A=\frac{1}{\cos^2A}$

$\sin^2A+\cos^2A=1$

$tgA=\frac{\sin A}{\cos A}$

DRAW QUESTION

3 mins • 3 pts

Егер $\cos\alpha=\frac{4}{5}$ болса, α сүйір бұрышы үшін sinα өрнегінің мәнін табыңыз.

DRAW QUESTION

3 mins • 2 pts

Өрнекті ықшамда: $\sin^4A+\cos^4A+2\sin^2A\cos^2A\$

DRAW QUESTION

3 mins • 2 pts

Өрнекті ықшамда: $\left(1-\cos^2A\right)\left(1+\cos^2A\right)$

DRAW QUESTION

3 mins • 3 pts

Өрнекті ықшамда: $\frac{5-\sin^2\alpha-\cos^2\alpha}{7\sin^2\alpha+7\cos^2\alpha}$

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8-cынып. Геометрия 14.12.2023

Негізгі тригонометриялық тепе-теңдік формуласы?

Егер cos⁡α=45\cos\alpha=\frac{4}{5}cosα=54​ болса, α сүйір бұрышы үшін sinα өрнегінің мәнін табыңыз.

Өрнекті ықшамда: sin⁡4A+cos⁡4A+2sin⁡2Acos⁡2A \sin^4A+\cos^4A+2\sin^2A\cos^2A\ sin4A+cos4A+2sin2Acos2A

Өрнекті ықшамда: (1−cos⁡2A)(1+cos⁡2A)\left(1-\cos^2A\right)\left(1+\cos^2A\right)(1−cos2A)(1+cos2A)

Өрнекті ықшамда: 5−sin⁡2α−cos⁡2α7sin⁡2α+7cos⁡2α\frac{5-\sin^2\alpha-\cos^2\alpha}{7\sin^2\alpha+7\cos^2\alpha}7sin2α+7cos2α5−sin2α−cos2α​

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Егер $\cos\alpha=\frac{4}{5}$ болса, α сүйір бұрышы үшін sinα өрнегінің мәнін табыңыз.

Өрнекті ықшамда: $\sin^4A+\cos^4A+2\sin^2A\cos^2A\$

Өрнекті ықшамда: $\left(1-\cos^2A\right)\left(1+\cos^2A\right)$

Өрнекті ықшамда: $\frac{5-\sin^2\alpha-\cos^2\alpha}{7\sin^2\alpha+7\cos^2\alpha}$