sina+cosa пж)))...

khadisha1 @khadisha1

July 2022 1 8 Report

sin2a/2cosa; sin2a/sina; 3cos^a/sin2a; cos2a/sina+cosa пж)))

Answers & Comments

UluanaV

Verified answer

$\frac{sin2\alpha}{2cos\alpha}=\frac{2sin\alpha*cos\alpha}{2cos\alpha}=sin\alpha$

$\frac{sin2\alpha}{sin\alpha}=\frac{2sin\alpha*cos\alpha}{sin\alpha}=2cos\alpha$

$\frac{3cos^2\alpha}{sin2\alpha}=\frac{3cos\alpha*cos\alpha}{2sin\alpha*cos\alpha}=\frac{3cos\alpha}{2sin\alpha}=1.5tg\alpha$

$\frac{cos2\alpha}{sin\alpha+cos\alpha}=\frac{cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}=\frac{(cos\alpha-sin\alpha)*(cos\alpha+sin\alpha)}{cos\alpha+sin\alpha}=cos\alpha-sin\alpha$

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